Probability
Short Answer Type Questions
1. For a loaded die, the probabilities of outcomes are given as under:
$P(1)=P(2)=0.2, P(3)=P(5)=P(6)=0.1$ and $P(4)=0.3$
The die is thrown, two times. Let $A$ and $B$ be the events, ‘same number each time’; and ‘a total score is 10 or more’; respectively. Determine whether or not A and B are independent.
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Solution
A loaded die is thrown such that
$P(1)=P(2)=0.2, P(3)=P(5)=P(6)=0.1$ and $P(4)=0.3$ and die is thrown two times. Also given that:
$A=$ Same number each time and
$B=$ Total score is 10 or more.
$ \begin{aligned} & \text{ So, } P(A)=[P(1,1)+P(2,2)+P(3,3)+P(4,4)+P(5,5)+P(6,6)] \\ & =P(1) \cdot P(1)+P(2) \cdot P(2)+P(3) \cdot P(3)+P(4) \cdot P(4) \\ & +P(5) \cdot P(5)+P(6) \cdot P(6) \\ & =0.2 \times 0.2+0.2 \times 0.2+0.1 \times 0.1+0.3 \times 0.3+0.1 \times 0.1 \\ & +0.1 \times 0.1 \\ & =0.04+0.04+0.01+0.09+0.01+0.01=0.20 \end{aligned} $
Now $B=[(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)]$
$P(B)=[P(4) \cdot P(6)+P(6) \cdot P(4)+P(5) \cdot P(5)+P(5) \cdot P(6)$ $+P(6) \cdot P(5)+P(6) \cdot P(6)$
$=0.3 \times 0.1+0.1 \times 0.3+0.1 \times 0.1+0.1 \times 0.1+0.1 \times 0.1$ $+0.1 \times 0.1$
$=0.03+0.03+0.01+0.01+0.01+0.01=0.10$
$A$ and $B$ both events will be independent if
$$ \begin{equation*} P(A \cap B)=P(A) \cdot P(B) \tag{i} \end{equation*} $$
Here,
$ (A \cap B)={(5,5),(6,6)} $
$ \therefore \quad P(A \cap B)=P(5,5)+P(6,6)=P(5) \cdot P(5)+P(6) \cdot P(6) $
$ =0.1 \times 0.1+0.1 \times 0.1=0.02 $
From eq. (i) we get
$ \begin{aligned} & 0.02=0.20 \times 0.10 \\ & 0.02=0.02 \end{aligned} $
Hence, $A$ and $B$ are independent events.
2. Refer to Exercise 1 above. If the die were fair, determine whether or not, the events A and B are independent.
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Solution
According to the solution of Q. 1 , we have
$A={(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}$
$\therefore n(A)=6$ and $n(S)=6 \times 6=36$
So, $P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}$
and $\quad B={(4,6),(6,4),(5,5),(5,6),(6,5),(6,6)}$
$n(B)=6$ and $n(S)=36$
$\therefore \quad P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}$
$A \cap B={(5,5),(6,6)}$
$\therefore P(A \cap B)=\frac{2}{36}=\frac{1}{18}$
Therefore, if $A$ and $B$ are independent, then
$P(A \cap B)=P(A) \cdot P(B)$
$\Rightarrow \quad \frac{1}{18} \neq \frac{1}{6} \times \frac{1}{6} \Rightarrow \frac{1}{18} \neq \frac{1}{36}$
Hence, $A$ and $B$ are not independent events.
3. The probability that atleast one of the two events A and B occurs is 0.6 . If $A$ and $B$ occurs simultaneously with probability 0.3 , evaluate $P(\overline{A})+P(\overline{B})$.
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Solution
We know that:
$A \cup B$ denotes that atleast one of the events occurs and $A \cap B$ denotes that the two events occur simultaneously.
$ \begin{matrix} \text{ So, } P(A \cup B) =P(A)+P(B)P(A \cap B) \\ \Rightarrow 0.6 =P(A)+P(B)0.3 \\ \Rightarrow 0.9 =P(A)+P(B) \\ \Rightarrow 0.9 =1P(\overline{A})+1P(\overline{B}) \\ \Rightarrow P(\overline{A})+P(\overline{B}) =20.9=1.1 \end{matrix} $
Hence, the required answer is 1.1.
4. A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that atleast one of the three marbles drawn be black, if the first marble is red?
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Solution
Let red marbles be represented with $R$ and black marble with B. The following three conditions are possible, if atleast one of the three marbles drawn be black and the first marble is red.
(i) $E_1$ : II ball is black and III is red
(ii) $E_2$ : II ball is black and III is also black
(iii) $E_3:$ II ball is red and III is black
$ \begin{aligned} & \therefore \quad P(E_1)=P(R_1) \cdot P(B_1 / R_1) \cdot P(R_2 / R_1 B_1)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}=\frac{60}{336}=\frac{5}{28} \\ & \quad P(E_2)=P(R_1) \cdot P(B_1 / R_1) \cdot P(B_2 / R_1 B_1)=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}=\frac{30}{336}=\frac{5}{56} \\ & \text{ and } P(E_3)=P(R_1) \cdot P(R_2 / R_1) \cdot B(B_1 / R_1 R_2)=\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}=\frac{60}{336}=\frac{5}{28} \\ & \therefore \quad P(E)=P(E_1)+P(E_2)+P(E_3)=\frac{5}{28}+\frac{5}{56}+\frac{5}{28}=\frac{25}{56} \end{aligned} $
Hence the required probability is $\frac{25}{56}$.
5. Two dice are thrown together and the total score is noted. The events $E, F$ and $G$ are ‘a total of 4’ ‘a total of 9 or more’, and ‘a total divisible by $5^{\prime}$, respectively. Calculate $P(E), P(F)$ and $P(G)$ and decide which pairs of events, if any, are independent.
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Solution
Two dice are thrown together
$\therefore n(S)=36$
$E=\text{ A total of } 4={(2,2),(1,3),(3,1)} \quad \therefore n(E)=3$
$F=\text{ A total of } 9 \text{ or more } $
$=\{(3,6),(6,3),(5,4),(4,5),(5,5),(4,6),(6,4),(5,6),(6,5),(6,6)\}$
$\therefore n(F)=10 ; G=\text{ A total divisible by } 5 $
$={(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)} \therefore n(G)=7$
Here, we see that $(E \cap F)=\phi$ and $(E \cap G)=\phi$
and $ (F \cap G)={(4,6),(6,4),(5,5)} $
$ \begin{aligned} \therefore \quad n(F & \cap G)=3 \text{ and }(E \cap F \cap G)=\phi \\ \therefore \quad P(E) & =\frac{n(E)}{n(S)}=\frac{3}{36}=\frac{1}{12} \\ P(F) & =\frac{n(F)}{n(S)}=\frac{10}{36}=\frac{5}{18} ; P(G)=\frac{n(G)}{n(S)}=\frac{7}{36} \\ P(F \cap G) & =\frac{3}{36}=\frac{1}{12} \text{ and } P(F) \cdot P(G)=\frac{5}{18} \cdot \frac{7}{36}=\frac{35}{648} \end{aligned} $
Since, $P(F \cap G) \neq P(F) . P(G)$
Hence, there is no pair of independent events.
6. Explain why the experiment of tossing a coin three times is said to have binomial distribution.
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Solution
We know that random variable $X$ takes values $0,1,2,3, \ldots, n$ is said to be binomial distribution having parameters $n$ and $p$, if the probability is given by $P(X=r)={ }^{n} C_r p^{r} q^{nr}$, where $q=1p$ and $r=0,1,2,3, \ldots$
Similarly in case of tossing a coin 3 times,
$n=3$ and $X$ has the values $0,1,2,3$ with $p=\frac{1}{2}, q=\frac{1}{2}$.
Hence, it is said to have a binomial distribution.
7. $A$ and $B$ are two events such that $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A \cap B)=\frac{1}{4}$. Find:
(i) $P(A / B)$
(ii) $P(B / A)$
(iii) $P(A^{\prime} / B)$
(iv) $P(A^{\prime} / B^{\prime})$
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Solution
We have $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A \cap B)=\frac{1}{4}$
$ \begin{aligned} P(A^{\prime}) & =1\frac{1}{2}=\frac{1}{2}, P(B^{\prime})=1\frac{1}{3}=\frac{2}{3} \\ P(A^{\prime} \cap B^{\prime}) & =1P(A \cup B)=1[P(A)+P(B)P(A \cap B)] \\ & =1[\frac{1}{2}+\frac{1}{3}\frac{1}{4}]=1[\frac{6+43}{12}]=1\frac{7}{12}=\frac{5}{12} \end{aligned} $
(i) $P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{1 / 3}=\frac{3}{4}$
(ii) $P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{1 / 4}{1 / 2}=\frac{1}{2}$
(iii) $P(A^{\prime} / B)=\frac{P(A^{\prime} \cap B)}{P(B)}=\frac{P(B)P(A \cap B)}{P(B)}$
$ =1\frac{P(A \cap B)}{P(B)}=1\frac{1 / 4}{1 / 3}=1\frac{3}{4}=\frac{1}{4} $
(iv) $P(A^{\prime} / B^{\prime})=\frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}=\frac{5 / 12}{2 / 3}=\frac{5}{12} \times \frac{3}{2}=\frac{5}{8}$
8. Three events A, B and C have probabilities $\frac{2}{5}, \frac{1}{3}$ and $\frac{1}{2}$ respectively. Given that $P(A \cap C)=\frac{1}{5}$ and $P(B \cap C)=\frac{1}{4}$, find the values of $P(C / B)$ and $P(A^{\prime} \cap C^{\prime})$.
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Solution
We have $P(A)=\frac{2}{5}, P(B)=\frac{1}{3}$ and $P(C)=\frac{1}{2}$
$ \begin{aligned} & P(A \cap C)=\frac{1}{5} \text{ and } P(B \cap C)=\frac{1}{4} \\ & \therefore \quad P(C / B)=\frac{P(B \cap C)}{P(B)}=\frac{1 / 4}{1 / 3}=\frac{3}{4} \\ & P(A^{\prime} \cap C^{\prime})=1P(A \cup C) \\ & =1[P(A)+P(C)P(A \cap C)] \end{aligned} $
$ =1[\frac{2}{5}+\frac{1}{2}\frac{1}{5}]=1\frac{7}{10}=\frac{3}{10} $
Hence, the required probabilities are $\frac{3}{4}$ and $\frac{3}{10}$.
9. Let $E_1$ and $E_2$ be two independent events such that $P(E_1)=p_1$ and $P(E_2)=p_2$. Describe in words of the events whose probabilities are:
(i) $p_1 p_2$
(ii) $(1p_1) p_2$
(iii) $1(1p_1)(1p_2)$
(iv) $p_1+p_22 p_1 p_2$
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Solution
Here,
$ P(E_1)=p_1 \text{ and } P(E_2)=p_2 $
(i)
$ p_1 p_2=P(E_1) \cdot P(E_2)=P(E_1 \cap E_2) $
So, $E_1$ and $E_2$ occur.
(ii)
$ (1p_1) \cdot p_2=P(E_1)^{\prime} \cdot P(E_2)=P(E_1^{\prime} \cap E_2) $
So, $E_1$ does not occur but $E_2$ occurs.
(iii) $1(1p_1)(1p_2)=1P(E_1)^{\prime} P(E_2)^{\prime}=1P(E_1^{\prime} \cap E_2^{\prime})$
$ =1[1P(E_1 \cup E_2)]=P(E_1 \cup E_2) $
So, either $E_1$ or $E_2$ or both $E_1$ and $E_2$ occur.
(iv)
$ \begin{aligned} p_1+p_22 p_1 p_2 & =P(E_1)+P(E_2)2 P(E_1) \cdot P(E_2) \\ & =P(E_1)+P(E_2)2 P(E_1 \cap E_2) \\ & =P(E_1 \cup E_2)2 P(E_1 \cap E_2) \end{aligned} $
So, either $E_1$ or $E_2$ occurs but not both.
10. A discrete random variable $X$ has the probability distribution given as below:
$\begin{array}{lllll} \hline X & 0.5 & 1 & 1.5 & 2 \\ \hline P(X) & k & k^2 & 2k^2 & k \\ \hline \end{array}$
(i) Find the value of $k$.
(ii) Determine the mean of the distribution.
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Solution
For a probability distribution, we know that if $P_i \geq 0$
(i) $\sum _{i=1}^{n} P_i=1 \Rightarrow k+k^{2}+2 k^{2}+k=1$
$ \begin{aligned} & \Rightarrow \quad 3 k^{2}+2 k1=0 \Rightarrow 3 k^{2}+3 kk1=0 \\ & \Rightarrow \quad 3 k(k+1)1(k+1)=0 \Rightarrow(3 k1)(k+1)=0 \\ & \therefore \quad k=\frac{1}{3} \text{ and } k=1 \end{aligned} $
But $k \geq 0 \quad \therefore k=\frac{1}{3}$
(ii) Mean of the distribution
$ E(X)=\sum _{i=1}^{n} X_i P_i=0.5 k+1 . k^{2}+1.5(2 k^{2})+2 k $
11. Prove that
$ \begin{aligned} & =\frac{k}{2}+k^{2}+3 k^{2}+2 k=4 k^{2}+\frac{5}{2} k \\ & =4(\frac{1}{3})^{2}+\frac{5}{2}(\frac{1}{3})=\frac{4}{9}+\frac{5}{6}=\frac{23}{18} \end{aligned} $
(i) $P(A)=P(A \cap B)+P(A \cap \overline{B})$
(ii) $P(A \cup B)=P(A \cap B)+P(A \cap \overline{B})+P(\overline{A} \cap B)$
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Solution
(i) To prove: $P(A)=P(A \cap B)+P(A \cap \overline{B})$ R.H.S. $=P(A \cap B)+P(A \cap \bar{{}B})$
$ \begin{aligned} & =P(A) \cdot P(B)+P(A) \cdot P(\overline{B})=P(A)[P(B)+P(\overline{B})] \\ & =P(A) \cdot 1 \\ & =P(A)=\text{ L.H.S. Hence proved. } \end{aligned} $
(ii) To prove: $P(A \cup B)=P(A \cap B)+P(A \cap \overline{B})+P(\overline{A} \cap B)$
R.H.S. $=P(A) \cdot P(B)+P(A) \cdot P(\bar{{}B})+P(\bar{{}A}) \cdot P(B)$
$ \begin{aligned} & =P(A) \cdot P(B)+P(A)[1P(B)]+[1P(A)] \cdot P(B) \\ & =P(A) \cdot P(B)+P(A)P(A) \cdot P(B)+P(B)P(A) \cdot P(B) \\ & =P(A)+P(B)P(A \cap B) \\ & =P(A \cup B)=\text{ L.H.S. Hence proved. } \end{aligned} $
12. If $X$ is the number of tails in three tosses of a coin, determine the standard deviation of $X$.
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Solution
Given that: $X=0,1,2,3$
$\therefore P(X=r)={ }^{n} C_r p^{r} q^{nr}$
where $n=3, p=\frac{1}{2}, q=\frac{1}{2}$ and $r=0,1,2,3$
$P(X=0)=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8} ; P(X=1)=3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{3}{8}$
$P(X=2)=3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{3}{8} ; P(X=3)=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$
Probability distribution table is:
$\begin{array}{lllll} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array}$
$ \begin{aligned} & E(X)=0+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2} \\ & E(X^{2})=0+1 \times \frac{3}{8}+4 \times \frac{3}{8}+9 \times \frac{1}{8}=\frac{3}{8}+\frac{12}{8}+\frac{9}{8}=\frac{24}{8}=3 \end{aligned} $
We know that $Var(X)=E(X^{2})[E(X)]^{2}=3(\frac{3}{2})^{2}=3\frac{9}{4}=\frac{3}{4}$
$\therefore$ Standard deviation $=\sqrt{Var(X)}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$.
13. In a dice game, a player pays a stake of $Re 1$ for each throw of a die. She receives ₹ 5 if the die shows a 3, ₹ 2 , if the die shows a 1 or 6 , and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?
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Solution
Let $X$ be the random variable of profit per throw.
$\begin{array}{lllll} \hline X & 1 & 1 & 4 \\ \hline P(X) & \frac{1}{2} & \frac{1}{2} & \frac{1}{6} \\ \hline \end{array}$
Since, she loses ₹ 1 for getting any of 2, 4, 5 .
So, $P(X=1)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
$ \begin{matrix} P(X=1)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} & (\because \text{ die showing } 1 \text{ or } 6) \\ P(X=4)=\frac{1}{6} & (\because \text{ die shows only a } 3) \end{matrix} $
Player’s expected profit $=\sum p_1 x_i$
$ =1 \times \frac{1}{2}+1 \times \frac{1}{3}+4 \times \frac{1}{6}=\frac{1}{2}+\frac{1}{3}+\frac{2}{3}=\frac{1}{2}=₹ 0.50 $
14. Three dice are thrown at the same time. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
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Solution
The dice is thrown three times
$\therefore$ Sample space $n(S)=(6)^{3}=216$
Let $E_1$ be the event when the sum of numbers on the dice was six and $E_2$ be the event when three two’s occur.
$ \begin{aligned} \Rightarrow E_1=\{(1,1,4),(1,2,3),(1,3,2),(1,4,1),(2,1,3),(2,2,2),(2,3,1),(3,1,2),\\(3,2,1),(4,1,1)\}[\therefore E_2=\{2,2,2\}] \end{aligned} $
$\begin{aligned}\Rightarrow n(E_1)=10 \text{ and } n(E_2)=1\end{aligned}$
$\begin{aligned}\therefore P(E_2 / E_1)=\frac{P(E_1 \cap E_2)}{P(E_1)}=\frac{1 / 216}{10 / 216}=\frac{1}{10} . \end{aligned} $
15. Suppose 10,000 tickets are sold in a lottery each for Re. 1. First prize is of ₹ 3,000 and the second prize is of ₹ 2,000. There are three third prizes of ₹ 500 each. If you buy one ticket, what is your expectation?
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Solution
Let $X$ be the random variable where $X=0,500,2000$ and 3000
$\begin{array}{llllll} \hline X & 0 & 500 & 2000 & 3000 \\ \hline P(X) & \frac{9995}{10000} & \frac{3}{10000} & \frac{1}{10000} & \frac{1}{10000} \\ \hline \end{array}$
Hence, expectation is ₹ 0.65 .
16. A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.
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Solution
Let $W_1$ and $W_2$ be two bags containing ( $4 W, 5 B)$ and (9 W, 7 B) balls respectively.
Let $E_1$ be the event that the
transferred ball from the bag
$W_1$ to $W_2$ is white and $E_2$ the event that the transferred ball is black.
And $E$ be the event that the ball drawn from the second bag is white.
$ \begin{aligned} \therefore P(E / E_1) & =\frac{10}{17}, \quad P(E / E_2)=\frac{9}{17} \\ P(E_1) & =\frac{4}{9} \text{ and } P(E_2)=\frac{5}{9} \\ \therefore \quad P(E) & =P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2) \\ & =\frac{4}{9} \times \frac{10}{17}+\frac{5}{9} \times \frac{9}{17}=\frac{40}{153}+\frac{45}{153}=\frac{85}{153}=\frac{5}{9} \end{aligned} $
Hence, the required probability is $\frac{5}{9}$.
17. Bat I contains 3 black and 2 white balls, bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
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Solution
Given that $bag I={3 B, 2 W}$ and $\quad$ bag II $={2 B, 4 W}$
Let $E_1=$ The event that bag I is selected
$ E_2=\text{ The event that bag II is selected } $
and $E=$ The event that a black ball is selected
$ \begin{aligned} \therefore \quad P(E_1) & =\frac{1}{2}, P(E_2)=\frac{1}{2}, P(E _{/} E_1)=\frac{3}{5} \text{ and } P(E_2 E_2)=\frac{1}{3} \\ P(E) & =P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2) \\ & =\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{1}{3}=\frac{3}{10}+\frac{1}{6}=\frac{9+5}{30}=\frac{14}{30}=\frac{7}{15} \end{aligned} $
Hence, the required probability is $\frac{7}{15}$.
18. A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?
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Solution
Given that the box has 5 blue and 4 red balls.
Let $E_1$ be the event that first ball drawn is blue $E_2$ be the event that first ball drawn is red
and $E$ is the event that second ball drawn is blue.
$ \begin{aligned} \therefore \quad P(E) & =P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2) \\ & =\frac{5}{9} \times \frac{4}{8}+\frac{4}{9} \times \frac{5}{8}=\frac{20}{72}+\frac{20}{72}=\frac{40}{72}=\frac{5}{9} \end{aligned} $
Hence, the required probability is $\frac{5}{9}$.
19. Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are Kings?
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Solution
Let $E_1, E_2, E_3$ and $E_4$ be the events that first, second, third and fourth card is King respectively.
$\therefore P(E_1 \cap E_2 \cap E_3 \cap E_4)$
$=P(E_1) \cdot P(E_2 / E_1) \cdot P[\frac{E_3}{(E_1 \cap E_2)}] \cdot P[\frac{E_4}{(E_1 \cap E_2 \cap E_3 \cap E_4)}]$
$=\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49}=\frac{24}{52 \cdot 51 \cdot 50 \cdot 49}=\frac{1}{13 \cdot 17 \cdot 25 \cdot 49}=\frac{1}{27075}$
Hence, the required probability is $\frac{1}{27075}$.
20. A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.
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Solution
Here, $p=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} \Rightarrow q=1\frac{1}{2}=\frac{1}{2}$ and $n=5$
$ \begin{aligned} \therefore \quad P(x & =r)={ }^{n} C_r p^{r} q^{nr} \\ & ={ }^{5} C_3(\frac{1}{2})^{3}(\frac{1}{2})^{53}=\frac{5 !}{3 ! 2 !} \cdot(\frac{1}{2})^{3} \cdot(\frac{1}{2})^{2}=10 \cdot \frac{1}{8} \cdot \frac{1}{4}=\frac{5}{16} \end{aligned} $
Hence, the required probability is $\frac{5}{16}$.
21. Ten coins are tossed. What is the probability of getting atleast 8 heads?
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Solution
Here, $n=10, p=\frac{1}{2}, q=\frac{1}{2}=\frac{1}{2}$
$ \begin{aligned} & P(X \geq 8)=P(x=8)+P(x=9)+P(x=10) \\ & ={ }^{10} C_8(\frac{1}{2})^{8}(\frac{1}{2})^{108}+{ }^{10} C_9(\frac{1}{2})^{9}(\frac{1}{2})^{109}+{ }^{10} C _{10}(\frac{1}{2})^{10}(\frac{1}{2})^{0} \\ & =\frac{10 !}{8 ! 2 !} \cdot(\frac{1}{2})^{8} \cdot(\frac{1}{2})^{2}+\frac{10 !}{9 ! 1 !}(\frac{1}{2})^{9}(\frac{1}{2})+(\frac{1}{2})^{10} \\ & =45 \cdot(\frac{1}{2})^{10}+10 \cdot(\frac{1}{2})^{10}+(\frac{1}{2})^{10}=(\frac{1}{2})^{10} \cdot(45+10+1) \\ & =56(\frac{1}{2})^{10}=56 \times \frac{1}{1024}=\frac{7}{128} \end{aligned} $
Hence, the required probability is $\frac{7}{128}$.
22. The probability of a man hitting a target is 0.25 . He shoots 7 times. What is the probability of his hitting at least twice?
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Solution
Here $n=7, \quad p=0.25=\frac{25}{100}=\frac{1}{4} \quad$ and $q=1\frac{1}{4}=\frac{3}{4}$
$ \begin{aligned} P(X \geq 2) & =1[P(X=0)+P(X=1)] \\ & =1[{ }^{7} C_0(\frac{1}{4})^{0}(\frac{3}{4})^{7}+{ }^{7} C_1(\frac{1}{4})^{1}(\frac{3}{4})^{6}] \\ & =1[(\frac{3}{4})^{7}+\frac{7}{4}(\frac{3}{4})^{6}]=1(\frac{3}{4})^{6}(\frac{3}{4}+\frac{7}{4}) \\ & =1(\frac{3}{4})^{6}(\frac{10}{4})=1\frac{729}{4096} \times \frac{10}{4}=1\frac{7290}{16384} \\ & =\frac{163847290}{16384}=\frac{9094}{16384}=\frac{4547}{8192} \end{aligned} $
Hence, the required probability is $\frac{4547}{8192}$.
23. A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be atleast one defective watch?
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Solution
Probability of defective watch out of 100 watches $=\frac{10}{100}=\frac{1}{10}$.
Here, $n=8, p=\frac{1}{10}$ and $q=1\frac{1}{10}=\frac{9}{10}$ and $r \geq 1$
$P(X \geq 1)=1P(x=0)=1{ }^{8} C_0(\frac{1}{10})^{0}(\frac{9}{10})^{80}=1(\frac{9}{10})^{8}$
Hence, the required probability is $1(\frac{9}{10})^{8}$.
24. Consider the probability distribution of a random variable X:
$\begin{array}{llllll} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \end{array}$
Calculate:
(i) $V(\frac{X}{2})$
(ii) Variance of $X$.
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Solution
Here, we have
$\begin{array}{llllll} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 0.1 & 0.25 & 0.3 & 0.2 & 0.15 \\ \hline \end{array}$
$=E(X^{2})[E(X)]^{2}$ 
where $E(X)=\sum _{i=1}^{n} x_i p_i$ and $E(X^{2})=\sum _{i=1}^{n} p_i x_i^{2}$
$\therefore \quad E(X)=0 \times 0.1+1 \times 0.25+2 \times 0.3+3 \times 0.2+4 \times 0.15$
$=0+0.25+0.6+0.6+0.6=2.05$
$E(X^{2})=0 \times 0.1+1 \times 0.25+4 \times 0.3+9 \times 0.2+16 \times 0.15$
$=0+0.25+1.2+1.8+2.40=5.65$
(i) $V(\frac{X}{2})=\frac{1}{4} V(X)=\frac{1}{4}[5.65(2.05)^{2}]=\frac{1}{4}[5.654.2025]$
$=\frac{1}{4} \times 1.4475=0.361875$
$[\because V(\frac{X}{2})=\frac{1}{4} V(X)]$
(ii) $\quad Var(X)=1.4475$
25. The probability distribution of a random variable $X$ is given below:
$\begin{array}{llllll} \hline x & 0 & 1 & 2 & 3 \\ \hline y & k & \frac{k}{2} & \frac{k}{4} & \frac{k}{8} \\ \hline \end{array}$
(i) Determine the value of $k$.
(ii) Determine $P(X \leq 2)$ and $P(X>2)$
(iii) Find $P(X \leq 2)+P(X>2)$.
Show Answer
Solution
(i) We know that $P(0)+P(1)+P(2)+P(3)=1$
$ \begin{matrix} \Rightarrow k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8} =1 \\ \Rightarrow \quad \frac{8 k+4 k+2 k+k}{8} =1 \Rightarrow 15 k=8 \\ \therefore \quad k =\frac{1}{15} \end{matrix} $
(ii)
$ \begin{aligned} P(X \leq 2) & =P(X=0)+P(X=1)+P(X=2) \\ & =k+\frac{k}{2}+\frac{k}{4}=\frac{7 k}{4}=\frac{7}{4} \times \frac{8}{15}=\frac{14}{15} \end{aligned} $
and $P(X>2)=P(X=3)=\frac{k}{8}=\frac{1}{8} \times \frac{8}{15}=\frac{1}{15}$
(iii) $P(X \leq 2)+P(X>2)=\frac{14}{15}+\frac{1}{15}=\frac{14+1}{15}=\frac{15}{15}=1$.
26. For the following probability distribution, determine standard deviation of the random variable $X$.
$\begin{array}{llll} \hline x & 2 & 3 & 4 \\ \hline y & 0.2 & 0.5 & 0.3 \\ \hline \end{array}$
Show Answer
Solution
We know that: Standard deviation (S.D.) $=\sqrt{\text{ Variance }}$
$ \begin{aligned} & \therefore \quad Var(X)=E(X^{2})[E(X)]^{2} \\ & E(X)=2 \times 0.2+3 \times 0.5+4 \times 0.3=0.4+1.5+1.2=3.1 \\ & E(X^{2})=4 \times 0.2+9 \times 0.5+16 \times 0.3=0.8+4.5+4.8=10.1 \\ & V(X)=10.1(3.1)^{2}=10.19.61=0.49 \\ & \therefore \quad \text{ S.D. }=\sqrt{Var(X)}=\sqrt{0.49}=0.7 \end{aligned} $
27. A biased die is such that $P(4)=\frac{1}{10}$ and other scores being equally likely. The die is tossed twice. If $X$ is the ’number of fours seen’, find the variance of the random variable $X$.
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Solution
Here, random variable $X=0,1,2$
$ \begin{aligned} P(4) & =\frac{1}{10}, P(\overline{4})=1\frac{1}{10}=\frac{9}{10} \\ P(X=0) & =P(\overline{4}) \cdot P(\overline{4})=\frac{9}{10} \times \frac{9}{10}=\frac{81}{100} \\ P(X=1) & =P(\overline{4}) \cdot P(4)+P(4) \cdot P(\overline{4})=\frac{9}{10} \times \frac{1}{10}+\frac{1}{10} \times \frac{9}{10}=\frac{18}{100} \end{aligned} $
$P(X=2)=P(4) \cdot P(4)=\frac{1}{10} \times \frac{1}{10}=\frac{1}{100}$
$\begin{array}{llll} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{81}{100} & \frac{18}{100} & \frac{1}{100} \\ \hline \end{array}$
We know that $V(X)=E(X^{2})[E(X)]^{2}$
$ \begin{aligned} E(X) & =0 \times \frac{81}{100}+1 \times \frac{18}{100}+\frac{2}{100}=\frac{20}{100}=\frac{1}{5} \\ E(X^{2}) & =0 \times \frac{81}{100}+1 \times \frac{18}{100}+4 \times \frac{1}{100}=\frac{22}{100}=\frac{11}{50} \\ \therefore \quad Var(X) & =\frac{11}{50}(\frac{1}{5})^{2}=\frac{11}{50}\frac{1}{25}=\frac{9}{50}=0.18 \end{aligned} $
Hence, the required variance $=0.18$.
28. A die is thrown three times. Let $X$ be the ’number of twos seen’, find the expectation of $X$.
Show Answer
Solution
Here, we have $X=0,1,2,3 \quad[\because$ die is thrown 3 times] and $p=\frac{1}{6}, q=\frac{5}{6}$
$\therefore P(X=0)=P(not 2) \cdot P(not 2) \cdot P(not 2)=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}=\frac{125}{216}$
$P(X=1)=P(2) \cdot P(not 2) \cdot P(not 2)+P(not 2) \cdot P(2) \cdot P(not 2)$
$+P($ not 2$) . P(not 2) \cdot P(2)$
$=\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6}+\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}=\frac{25}{216}+\frac{25}{216}+\frac{25}{216}=\frac{75}{216}$
$P(X=2)=P(2) \cdot P(2) \cdot P(not 2)+P(2) \cdot P(not 2) \cdot P(2)$
$+P($ not 2$) \cdot P(2) \cdot P(2)$
$=\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6}+\frac{1}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=\frac{5}{216}+\frac{5}{216}+\frac{5}{216}=\frac{15}{216}$
$P(X=3)=P(2) \cdot P(2) \cdot P(2)=\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6}=\frac{1}{216}$
Now $E(X)=\sum _{i=1}^{n} p_i x_i$
$ \begin{aligned} & i=1 \\ & 0 \times \frac{125}{216}+1 \times \frac{75}{216}+2 \times \frac{15}{216}+3 \times \frac{1}{216} \\ = & 0+\frac{75}{216}+\frac{30}{216}+\frac{3}{216}=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2} \end{aligned} $
Hence, the required expectation is $\frac{1}{2}$.
29. Two biased dice are thrown together. For the first die $P(6)=\frac{1}{2}$, the other scores being equally likely while for the second die, $P(1)=\frac{2}{5}$ and the other scores are equally likely. Find the probability distribution of ’the number of ones seen’.
Show Answer
Solution
Given that: for the first die, $P(6)=\frac{1}{2}$ and $P(\overline{6})=1\frac{1}{2}=\frac{1}{2}$
$\Rightarrow P(1)+P(2)+P(3)+P(4)+P(5)=\frac{1}{2}$
But $P(1)=P(2)=P(3)=P(4)=P(5)$
$\therefore 5 . P(1)=\frac{1}{2} \Rightarrow P(1)=\frac{1}{10}$ and $P(\overline{1})=1\frac{1}{10}=\frac{9}{10}$
For the second die, $P(1)=\frac{2}{5}$ and $P(\overline{1})=1\frac{2}{5}=\frac{3}{5}$
Let $X$ be the number of one’s seen
$ \begin{aligned} \therefore X=0,1,2 \\ \Rightarrow \quad \begin{aligned} P(X=0) =P(\overline{1}) \cdot P _{II}(\overline{1})=\frac{9}{10} \cdot \frac{3}{5}=\frac{27}{50}=0.54 \\ P(X=1) =P(\overline{1}) \cdot P_I(1)+P(1) \cdot P _{(\overline{1})} \\ =\frac{9}{10} \cdot \frac{2}{5}+\frac{1}{10} \cdot \frac{3}{5}=\frac{18+3}{50}=\frac{21}{50}=0.42 \\ P(X=2) =\underset{I \hspace{2mm} II}{P(1) \cdot P(1)}=\frac{1}{10} \cdot \frac{2}{5}=\frac{2}{50}=0.04 \end{aligned} \end{aligned} $
Hence, the required probability distribution is
$\begin{array}{llll} \hline X & 0 & 1 & 2 \\ \hline P(X) & 0.54 & 0.42 & 0.04 \\ \hline \end{array}$
30. Two probability distributions of the discrete random variable $X$ and $Y$ are given below.
$\begin{array}{lllll} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \\ \hline \end{array}$
$\begin{array}{lllll} \hline (Y) & 0 & 1 & 2 & 3 \\ \hline P(Y) & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \frac{1}{10} \\ \hline \end{array}$
Prove that: $E(Y^{2})=2 E(X)$.
Show Answer
Solution
First probability distribution is given by
$\begin{array}{lllll} \hline X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{5} & \frac{2}{5} & \frac{1}{5} & \frac{1}{5} \\ \hline \end{array}$
We know that, $E(X)=\sum _{i=1}^{n} P_i X_i$
$\Rightarrow E(X)=0 \cdot \frac{1}{5}+1 \cdot \frac{2}{5}+2 \cdot \frac{1}{5}+3 \cdot \frac{1}{5}=0+\frac{2}{5}+\frac{2}{5}+\frac{3}{5}=\frac{7}{5}$
For the second probability distribution,
$\begin{array}{lllll} \hline (Y) & 0 & 1 & 2 & 3 \\ \hline P(Y) & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \frac{1}{10} \\ \hline \end{array}$
$ \begin{aligned} E(Y^{2}) & =0 \cdot \frac{1}{5}+1 \cdot \frac{3}{10}+4 \cdot \frac{2}{5}+9 \cdot \frac{1}{10} \\ & =0+\frac{3}{10}+\frac{8}{5}+\frac{9}{10}=\frac{28}{10}=\frac{14}{5} \end{aligned} $
Now $E(Y^{2})=\frac{14}{5}$ and $2 E(X)=2 \cdot \frac{7}{5}=\frac{14}{5}$
Hence, $E(Y^{2})=2 E(X)$.
31. A factory produces bulbs. The probability that any one bulb is defective is $\frac{1}{50}$ and they are packed in boxes of 10 . From the single box, find the probability that (i) none of the bulbs is defective (ii) exactly two bulbs are defective (iii) more than 8 bulbs work properly.
Show Answer
Solution
Let $X$ be the random variable denoting a bulb to be defective.
Here, $n=10, p=\frac{1}{50}, q=1\frac{1}{50}=\frac{49}{50}$
We know that $P(X=r)={ }^{n} C_r p^{r} q^{nr}$
(i) None of the bulbs is defective, i.e., $r=0$
$ P(x=0)={ }^{10} C_0(\frac{1}{50})^{0}(\frac{49}{50})^{100}=(\frac{49}{50})^{10} $
(ii) Exactly two bulbs are defective
$ \begin{aligned} \therefore \quad P(x=2) & ={ }^{10} C_2(\frac{1}{50})^{2}(\frac{49}{50})^{102} \\ & =45 \cdot \frac{(49)^{8}}{(50)^{10}}=45 \times(\frac{1}{50})^{10} \times(49)^{8} \end{aligned} $
(iii) More than 8 bulbs work properly
We can say that less than 2 bulbs are defective
$ P(x<2)=P(x=0)+P(x=1) $
$={ }^{10} C_0(\frac{1}{50})^{0}(\frac{49}{50})^{10}+{ }^{10} C_1(\frac{1}{50})^{1}(\frac{49}{50})^{9}=(\frac{49}{50})^{10}+\frac{1}{5}(\frac{49}{50})^{9}$
$ =(\frac{49}{50})^{9}(\frac{49}{50}+\frac{1}{5})=(\frac{49}{50})^{9}(\frac{59}{50})=\frac{59(49)^{9}}{(50)^{10}} . $
32. Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2headed coin. If you take one out, toss it and get a head, what is the probability that it was a fair coin?
Show Answer
Solution
Let $E_1=$ Event that the coin is fair
$ E_2=\text{ Event that the coin is 2headed } $
and $H=$ Event that the tossed coin gets head.
$ P(E_1)=\frac{1}{2}, \quad P(E_2)=\frac{1}{2}, \quad P(H / E_1)=\frac{1}{2}, \quad P(H / E_2)=1 $
$\therefore$ Using Bayes’ Theorem, we get
$ \begin{aligned} P(E_1 / H) & =\frac{P(E_1) \cdot P(H / E_1)}{P(E_1) \cdot P(H / E_1)+P(E_2) \cdot P(H / E_2)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot 1}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}=\frac{\frac{1}{3}}{\frac{3}{4}}=\frac{1}{3} \end{aligned} $
Hence the required probability is $\frac{1}{3}$.
33. Suppose that 6% of the people with blood group O are left handed and $10 %$ of those with other blood groups are left handed, $30 %$ of the people have blood group O. If a left handed person is selected at random, what is the probability that he/she will have blood group $O$ ?
Show Answer
Solution
Let $E_1=$ The event that a person selected is of blood group $O$ $E_2=$ The event that the person selected is of other group and $H=$ The event that selected person is left handed
$ \begin{matrix} \therefore \quad P(E_1) & =0.30 \text{ and } & P(E_2)=0.70 \\ P(H / E_1) & =0.06 & P(H / E_2)=0.10 \end{matrix} $
So, from Bayes’ Theorem
$ \begin{aligned} & \quad P(\frac{E_1}{H})=\frac{P(E_1) \cdot P(H / E_1)}{P(E_1) \cdot P(H / E_1)+P(E_2) \cdot P(H / E_2)} \\ & =\frac{0.30 \times 0.06}{0.30 \times 0.06+0.70 \times 0.10}=\frac{0.018}{0.018+0.070}=\frac{0.018}{0.088}=\frac{9}{44} \\ & \text{ Hence, the required probability is } \frac{9}{44} . \end{aligned} $
34. Two natural numbers $r$ and $s$ are drawn one at a time, without replacement from the set $S={1,2,3, \ldots, n}$. Find $P(r \leq p / s \leq p)$, where $p \in S$.
Show Answer
Solution
Given that: $S={1,2,3, \ldots, n}$
$ \therefore \quad P(r \leq p / s \leq p)=\frac{P(P \cap S)}{P(S)}=\frac{p1}{n} \times \frac{n}{n1}=\frac{p1}{n1} $
Hence, the required probability is $\frac{p1}{n1}$.
35. Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Show Answer
Solution
Let $X$ be the random variable scores when a die is thrown twice.
and
$ \begin{aligned} X & =1,2,3,4,5,6 \\ S & =\{(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),..,(6,6)\} \end{aligned} $
So, $\quad P(X=1)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$
$ \begin{aligned} & P(X=2)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{3}{36} \\ & P(X=3)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{5}{36} \end{aligned} $
Similarly
$ P(X=4)=\frac{7}{36} ; P(X=5)=\frac{9}{36} \text{ and } P(X=6)=\frac{11}{36} $
So, the required distribution is
$\begin{array}{lllllll} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(X) & \frac{1}{36} & \frac{3}{36} & \frac{5}{36} & \frac{7}{36} & \frac{9}{36} &\frac{11}{36} \\ \hline \end{array}$
Now, the mean $E(X)=\sum _{i=1}^{n} x_i p_i$
$ \begin{aligned} & =1 \times \frac{1}{36}+2 \times \frac{3}{36}+3 \times \frac{5}{36}+4 \times \frac{7}{36}+5 \times \frac{9}{36}+6 \times \frac{11}{36} \\ & =\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}=\frac{161}{36} \end{aligned} $
Hence, the required mean $=\frac{161}{36}$.
36. The random variable $X$ can take only the values $0,1,2$. If $P(X=0)=P(X=1)=p$ and $E(X^{2})=E(X)$, then find the value of $p$.
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Solution
Given that: $\quad X=0,1,2$ and $P(X)$ at $X=0$ and 1 is $p$. Let $P(X)$ at $X=2$ is $x$
$\Rightarrow \quad p+p+x=1 \quad \Rightarrow \quad x=12 p$
Now we have the following distributions.
$\begin{array}{llll} \hline X & 0 & 1 & 2 \\ \hline P(X) & p & p & 12p \\ \hline \end{array}$
$\therefore \quad E(X)=0 . p+1 . p+2(12 p)=p+24 p=23 p$
and $\quad E(X^{2})=0 . p+1 . p+4(12 p)=p+48 p=47 p$
Given that: $\quad E(X^{2})=E(X)$
$\therefore \quad 47 p=23 p \Rightarrow 4 p=2 \Rightarrow p=\frac{1}{2}$
Hence, the required value of $p$ is $\frac{1}{2}$.
37. Find the variance of the distribution:
$\begin{array}{lllllll} \hline X & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(X) & \frac{1}{6} & \frac{5}{18} & \frac{2}{9} & \frac{1}{6} & \frac{1}{9} &\frac{1}{18} \\ \hline \end{array}$
Show Answer
Solution
We know that:
$ \begin{aligned} & \text{ Variance }(X)=E(X^{2})[E(X)]^{2} \\ & E(X)=\sum _{i=1}^{n} p_i x_i \\ &=0 \times \frac{1}{6}+1 \times \frac{5}{18}+2 \times \frac{2}{9}+3 \times \frac{1}{6}+4 \times \frac{1}{9}+5 \times \frac{1}{18} \\ &=0+\frac{5}{18}+\frac{4}{9}+\frac{3}{6}+\frac{4}{9}+\frac{5}{18}=\frac{5+8+9+8+5}{18}=\frac{35}{18} \\ & E(X^{2})=0 \times \frac{1}{6}+1 \times \frac{5}{18}+4 \times \frac{2}{9}+9 \times \frac{1}{6}+16 \times \frac{1}{9}+25 \times \frac{1}{18} \\ &=\frac{5}{18}+\frac{8}{9}+\frac{9}{6}+\frac{16}{9}+\frac{25}{18}=\frac{5+16+27+32+25}{18}=\frac{105}{18} \\ & \therefore \text{ Var }(X)=\frac{105}{18}\frac{35}{18} \times \frac{35}{18}=\frac{18901225}{324}=\frac{665}{324} \\ & \text{ Hence, the required variance is } \frac{665}{324} . \end{aligned} $
38. A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and $B$ wins if she gets a total of 7. If A starts the game, find the probability of winning the game by $A$ in third throw of the pair of dice.
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Solution
Let $A_1$ be the event of getting a total of 6
$ ={(2,4),(4,2),(1,5),(5,1),(3,3)} $
and $B_1$ be the event of getting a total of 7
$ ={(2,5),(5,2),(1,6),(6,1),(3,4),(4,3)} $
Let $P(A_1)$ is the probability, if $A$ wins in a throw $=\frac{5}{36}$
and $P(B_1)$ is the probability, if $B$ wins in a throw $=\frac{1}{6}$
$\therefore$ The required probability of winning $A$ in his third throw
$ =P(\overline{A}_1) \cdot P(\overline{B}_1) \cdot P(A_1)=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{7776} . $
39. Two dice are tossed. Find whether the following two events $A$ and $B$ are independent. $A={(x, y): x+y=11}$ and $B={(x, y): x \neq 5}$ where $(x, y)$ denotes a typical sample point.
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Solution
Given that:
$A=\{(x, y): x+y=11\} \text{ and } B=\{(x, y): x \neq 5\}$
$\therefore \quad A=\{(5,6),(6,5)\},$
$B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),$ $(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),$ $ (6,3),(6,4),(6,5),(6,6)\}$
$ \begin{aligned} & \Rightarrow n(A)=2, n(B)=30 \text{ and } n(A \cap B)=1 \\ & \therefore \quad P(A)=\frac{2}{36}=\frac{1}{18} \text{ and } P(B)=\frac{30}{36}=\frac{5}{6} \\ & \Rightarrow \quad P(A) \cdot P(B)=\frac{1}{18} \cdot \frac{5}{6}=\frac{5}{108} \text{ and } P(A \cap B)=\frac{1}{36} \end{aligned} $
Since $P(A).P(B) \neq P(P \cap B)$
Hence, A and B are not independent.
40. An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$.
Show Answer
Solution
Let $A$ be the event having $m$ white and $n$ black balls
$ \begin{aligned} & E_1={\text{ first ball drawn of white colour }} \\ & E_2={\text{ first ball drawn of black colour }} \end{aligned} $
$ \begin{aligned} & E_3={\text{ second ball drawn of white colour }} \\ & \therefore \quad P(E_1)=\frac{m}{m+n} \text{ and } P(E_2)=\frac{n}{m+n} \\ & P(E_3 / E_1)=\frac{m+k}{m+n+k} \text{ and } P(E_3 / E_2)=\frac{m}{m+n+k} \\ & \text{ Now } P(E_3)=P(E_1) \cdot P(E_3 / E_1)+P(E_2) \cdot P(E_3 / E_2) \\ &=\frac{m}{m+n} \times \frac{m+k}{m+n+k}+\frac{n}{m+n} \times \frac{m}{m+n+k} \\ &=\frac{m}{m+n+k}[\frac{m+k}{m+n}+\frac{n}{m+n}] \\ &=\frac{m}{m+n+k}[\frac{m+n+k}{m+n}]=\frac{m}{m+n} \end{aligned} $
Hence, the probability of drawing a white ball does not depend upon $k$.
Long Answer Type Questions
41. Three bags contain a number of red and white balls as follows, Bag I: 3 red balls, Bag II: 2 red balls and 1 white ball and Bag III: 3 white balls. The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}$, where $i=1,2,3$.
What is the probability that (i) a red ball will be selected (ii) a white ball is selected?
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Solution
Given that:
Bag I: 3 red balls and no white ball
Bag II: 2 red balls and 1 white ball
Bag III: no red ball and 3 white balls
Let $E_1, E_2$ and $E_3$ be the events of choosing Bag I, Bag II and Bag III respectively and a ball is drawn from it.
$\therefore P(E_1)=\frac{1}{6}, P(E_2)=\frac{2}{6}$ and $P(E_3)=\frac{3}{6}$
(i) Let $E$ be the event that red ball is selected
$ \begin{aligned} \therefore P(E) & =P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2)+P(E_3) \cdot P(E / E_3) \\ & =\frac{1}{6} \cdot \frac{3}{3}+\frac{2}{6} \cdot \frac{2}{3}+\frac{3}{6} \cdot 0=\frac{3}{18}+\frac{4}{18}=\frac{7}{18} \end{aligned} $
(ii) Let $F$ be the event that a white ball is selected
$ \begin{matrix} \therefore P(F) & =1P(E) & {[P(E)+P(F)=1]} \\ & =1\frac{7}{18}=\frac{11}{18} & \end{matrix} $
Hence, the required probabilities are $\frac{7}{18}$ and $\frac{11}{18}$.
42. Refer to Exercise Q. 41 above. If a white ball is selected, what is the probability that it came from (i) Bag II (ii) Bag III?
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Solution
Referring to Exercise Q.41, we will use here Bayes’ Theorem
$ \begin{aligned} & \text{ (i) } \begin{aligned} P(E_2 / F) & =\frac{P(E_2) \cdot P(F / E_2)}{P(E_1) \cdot P(F / E_1)+P(E_2) \cdot P(F / E_2)+P(E_3) \cdot P(F / E_3)} \\ & =\frac{\frac{2}{6} \cdot \frac{1}{3}}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1}=\frac{\frac{2}{18}}{\frac{2}{18}+\frac{3}{6}}=\frac{2}{11} \\ \text{ (ii) } P(E_3 / F) & =\frac{P(E_3) \cdot P(F / E_3)}{P(E_1) \cdot P(F / E_1)+P(E_2) \cdot P(F / E_2)+P(E_3) \cdot P(F / E_3)} \\ & =\frac{\frac{3}{6} \cdot 1}{\frac{1}{6} \cdot 0+\frac{2}{6} \cdot \frac{1}{3}+\frac{3}{6} \cdot 1}=\frac{\frac{3}{6}}{\frac{2}{18}+\frac{3}{6}}=\frac{3}{6} \times \frac{18}{11}=\frac{9}{11} \end{aligned} \\ & \text{ Hence, the required probabilities are } \frac{2}{11} \text{ and } \frac{9}{11} . \end{aligned} $
43. A shopkeeper sells three types of flower seeds $A_1, A_2$ and $A_3$. They are sold as a mixture, where the proportions are $4: 4: 2$, respectively. The germination rates of the three types of seeds are $45 %, 60 %$ and $35 %$. Calculate the probability
(i) of a randomly chosen seed to germinate
(ii) that it will not germinate given that the seed is of type $A_3$
(iii) that it is of the type $A_2$ given that a randomly chosen seed does not germinate.
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Solution
Given that $A_1: A_2: A_3=4: 4: 2$
$\therefore P(A_1)=\frac{4}{10}, P(A_2)=\frac{4}{10}$ and $P(A_3)=\frac{2}{10}$
where $A_1, A_2$ and $A_3$ are the three types of seeds.
Let $E$ be the event that a seed germinates and $\overline{E}$ be the event that a seed does not germinate
$\therefore P(\frac{E}{A_1})=\frac{45}{100}, P(\frac{E}{A_2})=\frac{60}{100}$ and $P(\frac{E}{A_3})=\frac{35}{100}$
and $P(\frac{\overline{E}}{A_1})=\frac{55}{100}, P(\frac{\overline{E}}{A_2})=\frac{40}{100}$ and $P(\frac{\overline{E}}{A_3})=\frac{65}{100}$ (i)
$ \begin{aligned} P(E) & =P(A_1) \cdot P(\frac{E}{A_1})+P(A_2) \cdot P(\frac{E}{A_2})+P(A_3) \cdot P(\frac{E}{A_3}) \\ & =\frac{4}{10} \cdot \frac{45}{100}+\frac{4}{10} \cdot \frac{60}{100}+\frac{2}{10} \cdot \frac{35}{100} \\ & =\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49 \end{aligned} $
(ii) $P(\overline{E} / A_3)=1P(E / A_3)=1\frac{35}{100}=\frac{65}{100}=0.65$
(iii) Using Bayes’ Theorem, we get
$ \begin{aligned} & P(A_2 / \overline{E})=\frac{P(A_2) \cdot P(\overline{E} / A_2)}{P(A_1) \cdot P(\overline{E} / A_1)+P(A_2) \cdot P(\overline{E} / A_2)+P(A_3) \cdot P(\overline{E} / A_3)} \\ &= \frac{\frac{4}{10} \cdot \frac{40}{100}}{\frac{4}{10} \cdot \frac{55}{100}+\frac{4}{10} \cdot \frac{40}{100}+\frac{2}{10} \cdot \frac{65}{100}} \\ &=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}=\frac{160}{220+160+130}=\frac{160}{510}=\frac{16}{51}=0.314 \end{aligned} $
Hence, the required probability is $\frac{16}{51}$ or 0.314 .
44. A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from TATA NAGAR?
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Solution
Let $E_1$ : The event that the letter comes from TATANAGAR and $E_2$ : The event that the letter comes from CALCUTTA Also $E_3$ : The event that on the letter, two consecutive letters TA are visible
$\therefore P(E_1)=\frac{1}{2}$ and $P(E_2)=\frac{1}{2}$ and $P(\frac{E_3}{E_1})=\frac{2}{8}$ and $P(\frac{E_3}{E_2})=\frac{1}{7}$
$[\because$ For TATA NAGAR, the two consecutive letters visible are TA, AT, TA, AN, NA, AG, GA, AR] $\therefore P(E_3 / E_1)=\frac{2}{8}$ and [For CALCUTTA, the two consecutive letters visible are $CA, AL, LC, CU, UT, TT$ and TA] So, $P(E_3 / E_2)=\frac{1}{7}$ Now using Bayes’ Theorem, we have
$ P(E_1 / E_3)=\frac{P(E_1) \cdot P(E_3 / E_1)}{P(E_1) \cdot P(E_3 / E_1)+P(E_2) \cdot P(E_3 / E_2)} $
$ =\frac{\frac{1}{2} \cdot \frac{2}{8}}{\frac{1}{2} \cdot \frac{2}{8}+\frac{1}{2} \cdot \frac{1}{7}}=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}=\frac{\frac{1}{8}}{\frac{7+4}{56}}=\frac{7}{11} $
Hence, the required probability is $\frac{7}{11}$.
45. There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3 , a ball is taken from the first bag but if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.
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Solution
Let $E_1$ be the event of selecting Bag I and $E_2$ be the event of selecting Bag II
Let $E_3$ be the event that black ball is selected
$ \begin{aligned} \therefore \quad P(E_1) & =\frac{2}{6}=\frac{1}{3} \text{ and } P(E_2)=1\frac{1}{3}=\frac{2}{3} \\ P(E_3 / E_1) & =\frac{3}{7} \text{ and } P(E_3 / E_2)=\frac{4}{7} \\ \therefore \quad P(E_3) & =P(E_1) \cdot P(E_3 / E_1)+P(E_2) \cdot P(E_3 / E_2) \\ & =\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}=\frac{3+8}{21}=\frac{11}{21} \end{aligned} $
Hence, the required probability is $\frac{11}{21}$.
46. There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
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Solution
We have 3 urns:
$\therefore$ Probabilities of choosing either of the urns are
$ P(U_1)=P(U_2)=P(U_3)=\frac{1}{3} $
Let $H$ be the event of drawing white ball from the chosen urn.
$ \begin{aligned} & \therefore P(H / U_1)=\frac{2}{5}, P(H / U_2)=\frac{3}{5} \text{ and } P(H / U_3)=\frac{4}{5} \\ & \therefore P(U_2 / H)=\frac{P(U_2) \cdot P(H / U_2)}{P(U_1) \cdot P(H / U_1)+P(U_2) \cdot P(H / U_2)+P(U_3) \cdot P(H / U_3)} \\ & \quad=\frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{3}{5}+\frac{1}{3} \cdot \frac{4}{5}}=\frac{\frac{3}{5}}{\frac{2}{5}+\frac{3}{5}+\frac{4}{5}}=\frac{3}{9}=\frac{1}{3} \end{aligned} $
Hence, the required probability is $\frac{1}{3}$.
47. By examining the chest $X$ray, the probability that $TB$ is detected when a person is actually suffering is 0.99 . The probability of an healthy person diagnosed to have TB is 0.001 . In a certain city, 1 in 1000 people suffer from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
Show Answer
Solution
Let $E_1$ : Event that a person has TB
$ E_2 \text{ : Event that a person does not have TB } $
and $H$ : Event that the person is diagnosed to have TB.
$ \begin{aligned} & \therefore \quad P(E_1)=\frac{1}{1000}=0.001, P(E_2)=1\frac{1}{1000}=\frac{999}{1000}=0.999 \\ & P(H / E_1)=0.99, P(H / E_2)=0.001 \\ & \therefore P(E_1 / H)=\frac{P(E_1) \cdot P(H / E_1)}{P(E_1) \cdot P(H / E_1)+P(E_2) \cdot P(H / E_2)} \\ & =\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.001}=\frac{0.99}{0.99+0.999} \\ & =\frac{0.990}{0.990+0.999}=\frac{990}{1989}=\frac{110}{221} \end{aligned} $
Hence, the required probability is $\frac{110}{221}$.
48. An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, $50 %$ are manufactured on machine A, $30 %$ on B and $20 %$ on C. $2 %$ of items produced on A and $2 %$ of items produced on B are defective and 3% of these produced on machine $C$ are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?
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Solution
Let $E_1$ : The event that the item is manufactured on machine A
$E_2$ : The event that the item is manufactured on machine $B$
$E_3$ : The event that the item is manufactured on machine C
Let $H$ be the event that the selected item is defective.
$\therefore$ Using Bayes’ Theorem,
$ \begin{aligned} & P(E_1)=\frac{50}{100}, P(E_2)=\frac{30}{100}, P(E_3)=\frac{20}{100} \\ & P(H / E_1)=\frac{2}{100}, P(H / E_2)=\frac{2}{100} \text{ and } P(H / E_3)=\frac{3}{100} \\ & \therefore P(E_1 / H)=\frac{P(E_1) \cdot P(H / E_1)}{P(E_1) \cdot P(H / E_1)+P(E_2) \cdot P(H / E_2)+P(E_3) \cdot P(H / E_3)} \\ & =\frac{\frac{50}{100} \times \frac{2}{100}}{\frac{50}{100} \times \frac{2}{100}+\frac{30}{100} \times \frac{2}{100}+\frac{20}{100} \times \frac{3}{100}} \\ & =\frac{100}{100+60+60}=\frac{100}{220}=\frac{10}{22}=\frac{5}{11} \end{aligned} $
49. Let $X$ be a discrete random variable whose probability distribution is defined as follows:
$ P(X=x)=\begin{matrix} k(x+1) & \text{ for } x=1,2,3,4 \\ 2 k x & \text{ for } x=5,6,7 \\ 0, & \text{ otherwise } \end{matrix} . $
where $k$ is a constantCalculate:
(i) the value of $k$ (ii) $E(X)$ (iii) Standard deviation of $X$.
Show Answer
Solution
(i) Here, $P(X=x)=k(x+1)$ for $x=1,2,3,4$
So, $P(X=1)=k(1+1)=2 k ; P(X=2)=k(2+1)=3 k$
$ P(X=3)=k(3+1)=4 k ; P(X=4)=k(4+1)=5 k $
Also, $P(X=x)=2 k x$ for $x=5,6,7$
$ \begin{aligned} & P(X=5)=2(5) k=10 k ; P(X=6)=2(6) k=12 k \\ & P(X=7)=2(7) k=14 k \end{aligned} $
and for otherwise it is 0 .
$\therefore$ The probability distribution is given by
$\begin{array}{lllllllll} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text{otherwise} \\ \hline P(X) & 2k & 3k & 4k & 5k & 10k & 12k & 14k & 0 \\ \hline \end{array}$
$ \begin{aligned} & \text{ We know that } \sum _{i=1}^{n} P(X_i)=1 \\ & \text{ So, } 2 k+3 k+4 k+5 k+10 k+12 k+14 k=1 \\ & \Rightarrow \quad 50 k=1 \Rightarrow k=\frac{1}{50} \\ & \text{ Hence, the value of } k \text{ is } \frac{1}{50} \end{aligned} $
(ii) Now the probability distribution is
$\begin{array}{llllllll} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(X) & \frac{2}{50} & \frac{3}{50} & \frac{4}{50} & \frac{5}{50} & \frac{10}{50} & \frac{12}{50} &\frac{14}{50} \\ \hline \end{array}$
$ \begin{aligned} E(X)=1 \times \frac{2}{50}+2 \times \frac{3}{50}+3 \times \frac{4}{50}+4 \times \frac{5}{50}+5 \times \frac{10}{50}+ & 6 \times \frac{12}{50} \\ & +7 \times \frac{14}{50} \end{aligned} $
$ =\frac{2}{50}+\frac{6}{50}+\frac{12}{50}+\frac{20}{50}+\frac{50}{50}+\frac{72}{50}+\frac{98}{50}=\frac{260}{50}=\frac{26}{5}=5.2 $
(iii) We know that Standard deviation (SD) $=\sqrt{\text{ Variance }}$ Variance $=E(X^{2})[E(X)]^{2}$
$ \begin{aligned} & \begin{aligned} & E(X^{2})= 1 \times \frac{2}{50}+4 \times \frac{3}{50}+9 \times \frac{4}{50}+16 \times \frac{5}{50}+25 \times \frac{10}{50} \\ &=\frac{2}{50}+\frac{12}{50}+\frac{36}{50}+\frac{80}{50}+\frac{250}{50}+\frac{432}{50}+\frac{686}{50}=\frac{1498}{50} \end{aligned} \\ & \therefore \text{ Variance }(X)=\frac{1498}{50}(\frac{26}{5})^{2} \\ & =\frac{1498}{50}\frac{676}{25}=\frac{14981352}{50}=\frac{146}{50}=2.92 \\ & \therefore \quad \text{ S.D }=\sqrt{2.92}=1.7 \text{ (approx.) } \end{aligned} $
50. The probability distribution of a discrete random variable $X$ is given as under:
$\begin{array}{lllllll} \hline X & 1 & 2 & 4 & 2A & 3A & 5A \\ \hline P(X) & \frac{1}{2} & \frac{1}{5} & \frac{3}{25} & \frac{1}{10} & \frac{1}{25} & \frac{1}{25} \\ \hline \end{array}$
Calculate: (i) The value of $A$ if $E(X)=2.94$; (ii) Variance of $X$
Show Answer
Solution
(i) We know that: $E(X)=\sum _{i=1}^{n} P_i X_i$
$ \begin{matrix} \therefore & E(X) & =1 \times \frac{1}{2}+2 \times \frac{1}{5}+4 \times \frac{3}{25}+2 A \times \frac{1}{10}+3 A \times \frac{1}{25}+5 A \times \frac{1}{25} \\ & 2.94 & =\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{A}{5}+\frac{3 A}{25}+\frac{A}{5} \\ \Rightarrow & 2.94 & =0.5+0.4+0.48+\frac{13 A}{25}=1.38+\frac{13 A}{25} \\ \Rightarrow & 2.94 & 1.38=\frac{13 A}{25} \Rightarrow 1.56=\frac{13 A}{25} \\ \Rightarrow \quad & A & =\frac{1.56 \times 25}{13}=0.12 \times 25 \\ & \therefore \quad & A & =3 \end{matrix} $
(ii) Now the distribution becomes
$\begin{array}{lllllll} \hline X & 1 & 2 & 4 & 6 & 9 & 15 \\ \hline P(X) & \frac{1}{2} & \frac{1}{5} & \frac{3}{25} & \frac{1}{10} & \frac{1}{25} & \frac{1}{25} \\ \hline \end{array}$
$E(X^{2})=1 \times \frac{1}{2}+4 \times \frac{1}{5}+16 \times \frac{3}{25}+36 \times \frac{1}{10}+81 \times \frac{1}{25}+225 \times \frac{1}{25}$
$ \begin{aligned} & =\frac{1}{2}+\frac{4}{5}+\frac{48}{25}+\frac{36}{10}+\frac{81}{25}+\frac{225}{25} \\ & =0.5+0.8+1.92+3.6+3.24+9.00=19.06 \end{aligned} $
Variance $(X)=E(X^{2})[E(X)]^{2}$
$ =19.06(2.94)^{2}=19.068.64=10.42 $
51. The probability distribution of a random variable $X$ is given as under:
where $k$ is a constant. Calculate:
$ P(X=x)=\begin{cases} k x^{2} & \text{ for } x=1,2,3 \\ 2 k x & \text{ for } x=4,5,6 \\ 0 & \text{ otherwise } \end{cases} . $
(i) $E(X)$ (ii) $E(3 X^{2})$ (iii) $P(X \geq 4)$
Show Answer
Solution
Given that: $\quad P(X=x)=\begin{cases} k x^{2} & \text{ for } x=1,2,3 \\ 2 k x & \text{ for } x=4,5,6 \\ 0 & \text{ otherwise } \end{cases} .$
$\therefore$ Probability distribution of random variable $X$ is
$\begin{array}{lllllllll} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & \text{otherwise} \\ \hline P(X) & k & 4k & 9k & 8k & 10k & 0 \\ \hline \end{array}$
We know that $\sum _{i=1}^{n} P(X_i)=1$
$\therefore k+4 k+9 k+8 k+10 k+12 k=1 \Rightarrow 44 k=1 \Rightarrow k=\frac{1}{44}$
(i) $E(X)=\sum _{i=1}^{n} P_i X_i=1 \times k+2 \times 4 k+3 \times 9 k+4 \times 8 k+5 \times 10 k+6 \times 12 k$ $=k+8 k+27 k+32 k+50 k+72 k=190 k$ $=190 \times \frac{1}{44}=\frac{95}{22}=4.32$ (approx.)
(ii) $E(3 X^{2})=3[k+4 \times 4 k+9 \times 9 k+16 \times 8 k+25 \times 10 k+36 \times 12 k]$
$=3[k+16 k+81 k+128 k+250 k+432 k]=3[908 k]$
$=3 \times 908 \times \frac{1}{44}=\frac{2724}{44}=61.9$ (approx.)
(iii) $P(X \geq 4)=P(X=4)+P(X=5)+P(X=6)$
$=8 k+10 k+12 k=30 k$
$=30 \times \frac{1}{44}=\frac{15}{22}$.
52. A bag contains $(2 n+1)$ coins. It is known that $n$ of these coins have a head on both sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is $\frac{31}{42}$, determine the value of $n$.
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Solution
Given that $n$ coins are two headed coins and the remaining $(n+1)$ coins are fair.
Let $E_1$ : the event that unfair coin is selected
$E_2$ : the event that the fair coin is selected
$E$ : the event that the toss results in a head
$\therefore \quad P(E_1)=\frac{n}{2 n+1}$ and $P(E_2)=\frac{n+1}{2 n+1}$
$P(E / E_1)=1$ (sure event) and $P(E / E_2)=\frac{1}{2}$
$\therefore \quad P(E)=P(E_1) \cdot P(E / E_1)+P(E_2) \cdot P(E / E_2)$
$=\frac{n}{2 n+1} \cdot 1+\frac{n+1}{2 n+1} \cdot \frac{1}{2}=\frac{1}{2 n+1}(n+\frac{n+1}{2})$
$=\frac{1}{2 n+1}(\frac{2 n+n+1}{2})=\frac{3 n+1}{2(2 n+1)}$
But $\quad P(E)=\frac{31}{42}$ (given)
$ \begin{matrix} \therefore \frac{3 n+1}{2(2 n+1)}=\frac{31}{42} & \Rightarrow & \frac{3 n+1}{2 n+1} & =\frac{31}{21} \\ & \Rightarrow & 63 n+21 & =62 n+31 \\ & \Rightarrow & & n & =10 \end{matrix} $
Hence, the required value of $n$ is 10 .
53. Two cards are drawn successively without replacement from a well shuffled deck of cards. Find the mean and standard deviation of the random variable $X$ where $X$ is the number of aces.
Show Answer
Solution
Let $X$ be the random variable such that $X=0,1,2$
and $E=$ the event of drawing an ace
and $F=$ the event of drawing nonace.
$\therefore P(E)=\frac{4}{52}$ and $P(\overline{E})=\frac{48}{52}$
Now $P(X=0)=P(\overline{E}) \cdot P(\overline{E})=\frac{48}{52} \cdot \frac{47}{51}=\frac{188}{221}$
$ \begin{gathered} P(X=1)=P(E) \cdot P(\bar{{}E})+P(\bar{{}E}) \cdot P(E)=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}=\frac{32}{221} \\ P(X=2)=P(E) \cdot P(E)=\frac{4}{52} \cdot \frac{3}{51}=\frac{1}{221} \end{gathered} $
We have Distribution Table:
$\begin{array}{llll} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \\ \hline \end{array}$
Now, Mean $E(X)=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221}=\frac{32}{221}+\frac{2}{221}=\frac{34}{221}=\frac{2}{13}$
$ E(X^{2})=0 \times \frac{188}{221}+1 \times \frac{32}{221}+4 \times \frac{1}{221}=\frac{32}{221}+\frac{4}{221}=\frac{36}{221} $
$\therefore \quad$ Variance $=E(X^{2})[E(X)]^{2}$
$ =\frac{36}{221}(\frac{2}{13})^{2}=\frac{36}{221} \quad \frac{4}{169}=\frac{46868}{13 \times 221}=\frac{400}{2873} $
Standard deviation $=\sqrt{\frac{400}{2873}}=0.377$ (approx.)
54. A die is tossed twice. A ‘success’ is getting an even number on a toss. Find the variance of the number of successes.
Show Answer
Solution
Let $E$ be the event of getting even number on tossing a die.
$\therefore P(E)=\frac{3}{6}=\frac{1}{2}$ and $P(\overline{E})=1\frac{1}{2}=\frac{1}{2}$
Here $X=0,1,2$
$ \begin{gathered} P(X=0)=P(\bar{{}E}) \cdot P(\bar{{}E})=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4} \\ P(X=1)=P(E) \cdot P(\bar{{}E})+P(\bar{{}E}) \cdot P(E)=\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4} \\ P(X=2)=P(E) \cdot P(E)=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4} \end{gathered} $
$\therefore$ Probability distribution table is
$\begin{array}{llll} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{1}{4} & \frac{2}{4} & \frac{1}{4} \\ \hline \end{array}$
$ \begin{aligned} E(X) & =0 \times \frac{1}{4}+1 \times \frac{2}{4}+2 \times \frac{1}{4}=\frac{2}{4}+\frac{2}{4}=1 \\ E(X^{2}) & =0 \times \frac{1}{4}+1 \times \frac{2}{4}+4 \times \frac{1}{4}=\frac{3}{2} \end{aligned} $
$\therefore$ Variance $(X)=E(X^{2})[E(X)]^{2}=\frac{3}{2}1=\frac{1}{2}=0.5$
55. There are 5 cards numbered 1 to 5 , one number on one card. Two cards are drawn at random without replacement. Let $X$ denotes the sum of the numbers on two cards drawn. Find the mean and variance of $X$.
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Solution
Here, sample space $S=\{(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)$, $(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3)$, $(3,5),(5,3),(5,4),(4,5)\}$
$\therefore n(S)=20$
Let $X$ be the random variable denoting the sum of the numbers on two cards drawn.
$ \therefore \quad X=3,4,5,6,7,8,9 $
So, $\quad P(X=3)=\frac{2}{20}$
$ \begin{aligned} & P(X=4)=\frac{2}{20} \\ & P(X=6)=\frac{4}{20} \\ & P(X=8)=\frac{2}{20} \end{aligned} $
$ \begin{aligned} & P(X=5)=\frac{4}{20} \\ & P(X=7)=\frac{4}{20} \end{aligned} $
$ \begin{matrix} P(X=9)=\frac{2}{20} \\ \therefore \text{ Mean, } E(X)=3 \times \frac{2}{20}+4 \times \frac{2}{20}+5 \times \frac{4}{20}+6 \times \frac{4}{20}+7 \times \frac{4}{20} \\ +8 \times \frac{2}{20}+9 \times \frac{2}{20} \\ =\frac{6}{20}+\frac{8}{20}+\frac{20}{20}+\frac{24}{20}+\frac{28}{20}+\frac{16}{20}+\frac{18}{20}=\frac{120}{20}=6 \\ E(X^{2})=9 \times \frac{2}{20}+16 \times \frac{2}{20}+25 \times \frac{4}{20}+36 \times \frac{4}{20}+49 \times \frac{4}{20}+64 \times \frac{2}{20}+81 \times \frac{2}{20} \\ =\frac{18}{20}+\frac{32}{20}+\frac{100}{20}+\frac{144}{20}+\frac{196}{20}+\frac{128}{20}+\frac{162}{20}=\frac{780}{20}=39 \\ \therefore \text{ Variance }(X)=E(X^{2})[E(X)]^{2}=39(6)^{2}=3936=3 \end{matrix} $
Objective Type Questions
56. If $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$, then $P(B / A)$ is equal to
(a) $\frac{1}{10}$
(b) $\frac{1}{8}$
(c) $\frac{7}{8}$
(d) $\frac{17}{20}$
Show Answer
Solution
Given that: $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$
$ \therefore \quad P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{7 / 10}{4 / 5}=\frac{7}{8} $
Hence, the correct option is (c).

Option (a) $\frac{1}{10}$: This option is incorrect because it does not satisfy the formula for conditional probability. The correct calculation for $P(B / A)$ is $\frac{P(A \cap B)}{P(A)}$, which results in $\frac{7/10}{4/5} = \frac{7}{8}$, not $\frac{1}{10}$.

Option (b) $\frac{1}{8}$: This option is incorrect because it also does not satisfy the formula for conditional probability. The correct calculation for $P(B / A)$ is $\frac{P(A \cap B)}{P(A)}$, which results in $\frac{7/10}{4/5} = \frac{7}{8}$, not $\frac{1}{8}$.

Option (d) $\frac{17}{20}$: This option is incorrect because it does not follow the correct calculation for conditional probability. The correct calculation for $P(B / A)$ is $\frac{P(A \cap B)}{P(A)}$, which results in $\frac{7/10}{4/5} = \frac{7}{8}$, not $\frac{17}{20}$.
57. If $P(A \cap B)=\frac{7}{10}$ and $P(B)=\frac{17}{20}$, then $P(A / B)$ equals
(a) $\frac{14}{17}$
(b) $\frac{17}{20}$
(c) $\frac{7}{8}$
(d) $\frac{1}{8}$
Show Answer
Solution
Given that: $P(A \cap B)=\frac{7}{10}$ and $P(B)=\frac{17}{20}$
$ \therefore \quad P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{7 / 10}{17 / 20}=\frac{14}{17} $
Hence, the correct option is $(a)$.

Option (b) $\frac{17}{20}$: This option is incorrect because it represents the probability of event B, not the conditional probability ( P(A / B) ). The conditional probability ( P(A / B) ) is calculated using the formula ( \frac{P(A \cap B)}{P(B)} ), which does not simplify to ( \frac{17}{20} ).

Option (c) $\frac{7}{8}$: This option is incorrect because it does not match the result of the calculation for ( P(A / B) ). The correct calculation is ( \frac{P(A \cap B)}{P(B)} = \frac{7/10}{17/20} = \frac{14}{17} ), not ( \frac{7}{8} ).

Option (d) $\frac{1}{8}$: This option is incorrect because it is not the result of the calculation for ( P(A / B) ). The correct calculation is ( \frac{P(A \cap B)}{P(B)} = \frac{7/10}{17/20} = \frac{14}{17} ), not ( \frac{1}{8} ).
58. If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then
$P(B / A)+P(A / B)$ equals to
(a) $\frac{1}{4}$
(b) $\frac{1}{3}$
(c) $\frac{5}{12}$
(d) $\frac{7}{12}$
Show Answer
Solution
Here, $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$
$ \begin{aligned} & P(A \cup B)=P(A)+P(B)P(A \cap B) \\ & \Rightarrow \quad \frac{3}{5}=\frac{3}{10}+\frac{2}{5}P(A \cap B) \\ & \Rightarrow \quad P(A \cap B)=\frac{3}{10}+\frac{2}{5}\frac{3}{5}=\frac{3+46}{10}=\frac{1}{10} \\ & \text{ Now } P(A / B)+P(B / A)=\frac{P(A \cap B)}{P(B)}+\frac{P(A \cap B)}{P(A)} \\ & =\frac{1 / 10}{2 / 5}+\frac{1 / 10}{3 / 10}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12} \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) (\frac{1}{4}): This option is incorrect because it does not account for the correct calculation of (P(A / B) + P(B / A)). The correct calculation involves finding the probabilities of (P(A \cap B)) divided by (P(B)) and (P(A \cap B)) divided by (P(A)), which sum to (\frac{7}{12}), not (\frac{1}{4}).

Option (b) (\frac{1}{3}): This option is incorrect because it also does not match the correct sum of (P(A / B) + P(B / A)). The individual probabilities (\frac{P(A \cap B)}{P(B)}) and (\frac{P(A \cap B)}{P(A)}) do not add up to (\frac{1}{3}).

Option (c) (\frac{5}{12}): This option is incorrect because it is not the result of the correct calculation. The sum of (\frac{P(A \cap B)}{P(B)}) and (\frac{P(A \cap B)}{P(A)}) is (\frac{7}{12}), not (\frac{5}{12}).
59. If $P(A)=\frac{2}{5}, P(B)=\frac{3}{10}$ and $P(A \cap B)=\frac{1}{5}$, then $P(A^{\prime} / B^{\prime}) \cdot P(B^{\prime} / A^{\prime})$ is equal to
(a) $\frac{5}{6}$
(b) $\frac{5}{7}$
(c) $\frac{25}{42}$
(d) 1
Show Answer
Solution
Given that: $P(A)=\frac{2}{5}, P(B)=\frac{3}{10}$ and $P(A \cap B)=\frac{1}{5}$
$ \begin{aligned} & P(A^{\prime})=1\frac{2}{5}=\frac{3}{5}, P(B^{\prime})=1\frac{3}{10}=\frac{7}{10} \\ & \text{ and } P(A^{\prime} \cap B^{\prime})=1P(A \cup B)=1[P(A)+P(B)P(A \cap B)] \\ &=1[\frac{2}{5}+\frac{3}{10}\frac{1}{5}]=1[\frac{1}{5}+\frac{3}{10}]=1\frac{5}{10}=\frac{1}{2} \\ & \therefore \quad P(A^{\prime} / B^{\prime})=\frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}=\frac{1 / 2}{7 / 10}=\frac{5}{7} \\ & \text{ and } P(B^{\prime} / A^{\prime})=\frac{P(A^{\prime} \cap B^{\prime})}{P(A^{\prime})}=\frac{1 / 2}{3 / 5}=\frac{5}{6} \\ & \therefore P(A^{\prime} / B^{\prime}) \cdot P(B^{\prime} / A^{\prime})=\frac{5}{7} \times \frac{5}{6}=\frac{25}{42} \end{aligned} $
Hence, the correct option is (c).

Option (a) (\frac{5}{6}): This option is incorrect because it represents the value of (P(B^{\prime} / A^{\prime})) alone, not the product of (P(A^{\prime} / B^{\prime})) and (P(B^{\prime} / A^{\prime})).

Option (b) (\frac{5}{7}): This option is incorrect because it represents the value of (P(A^{\prime} / B^{\prime})) alone, not the product of (P(A^{\prime} / B^{\prime})) and (P(B^{\prime} / A^{\prime})).

Option (d) 1: This option is incorrect because the product of (P(A^{\prime} / B^{\prime})) and (P(B^{\prime} / A^{\prime})) is not equal to 1. The correct product is (\frac{25}{42}).
60. If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A / B)=\frac{1}{4}$, then $P(A^{\prime} \cap B^{\prime})$ equals
(a) $\frac{1}{12}$
(b) $\frac{3}{4}$
(c) $\frac{1}{4}$
(d) $\frac{3}{16}$
Show Answer
Solution
Given that: $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(A / B)=\frac{1}{4}$
$ \begin{aligned} P(A / B) & =\frac{P(A \cap B)}{P(B)} \\ \frac{1}{4} & =\frac{P(A \cap B)}{1 / 3} \Rightarrow P(A \cap B)=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12} \end{aligned} $
Now $\quad P(A^{\prime} \cap B^{\prime})=1P(A \cup B)$
$ =1[P(A)+P(B)P(A \cap B)] $
$=1[\frac{1}{2}+\frac{1}{3}\frac{1}{12}]=1[\frac{5}{6}\frac{1}{12}]=1\frac{9}{12}=\frac{3}{12}=\frac{1}{4}$
Hence, the correct option is (c).

Option (a) $\frac{1}{12}$: This option is incorrect because it represents the probability of the intersection of events $A$ and $B$ ($P(A \cap B)$), not the probability of the intersection of their complements ($P(A^{\prime} \cap B^{\prime})$).

Option (b) $\frac{3}{4}$: This option is incorrect because it does not correctly account for the calculation of $P(A^{\prime} \cap B^{\prime})$. The correct calculation involves subtracting the union of $A$ and $B$ from 1, which results in $\frac{1}{4}$, not $\frac{3}{4}$.

Option (d) $\frac{3}{16}$: This option is incorrect because it does not follow from the given probabilities and the correct formula for $P(A^{\prime} \cap B^{\prime})$. The correct calculation yields $\frac{1}{4}$, not $\frac{3}{16}$.
61. If $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$ then $P(A \cup B)$ is equal to
(a) 0.24
(b) 0.3
(c) 0.48
(d) 0.96
Show Answer
Solution
Given that: $P(A)=0.4, P(B)=0.8$ and $P(B / A)=0.6$
$ \begin{aligned} P(B / A)=\frac{P(A \cap B)}{P(A)} & \Rightarrow 0.6=\frac{P(A \cap B)}{0.4} \\ \therefore \quad P(A \cap B) & =0.6 \times 0.4=0.24 \\ P(A \cup B) & =P(A)+P(B)P(A \cap B) \\ & =0.4+0.80.24=1.200.24=0.96 \end{aligned} $
Hence, the correct option is (d).

Option (a) 0.24 is incorrect because it represents the probability of the intersection of events A and B, ( P(A \cap B) ), not the union ( P(A \cup B) ).

Option (b) 0.3 is incorrect because it does not correspond to any relevant probability calculation given the provided data. It is neither the probability of A, B, nor their intersection or union.

Option (c) 0.48 is incorrect because it does not match the correct calculation for the union of events A and B. It might be a miscalculation or an unrelated probability.
62. If $A$ and $B$ are two events and $A \neq \phi, B \neq \phi$, then
(a) $P(A / B)=P(A) \cdot P(B)$
(b) $P(A / B)=\frac{P(A \cap B)}{P(B)}$
(c) $P(A / B) \cdot P(B / A)=1$
(d) $P(A / B)=\frac{P(A)}{P(B)}$
Show Answer
Solution
Given that: $A=\phi$ and $B \neq \phi$,
then
$ P(A / B)=\frac{P(A \cap B)}{P(B)} $
Hence, the correct option is (b).

Option (a) $P(A / B)=P(A) \cdot P(B)$: This is incorrect because the conditional probability ( P(A / B) ) is defined as the probability of event ( A ) occurring given that event ( B ) has occurred. It is not simply the product of the probabilities of ( A ) and ( B ). The correct formula for conditional probability is ( P(A / B) = \frac{P(A \cap B)}{P(B)} ).

Option (c) $P(A / B) \cdot P(B / A)=1$: This is incorrect because the product of the conditional probabilities ( P(A / B) ) and ( P(B / A) ) does not necessarily equal 1. The correct relationship between these probabilities is given by Bayes’ theorem: ( P(A / B) = \frac{P(B / A) \cdot P(A)}{P(B)} ).

Option (d) $P(A / B)=\frac{P(A)}{P(B)}$: This is incorrect because the conditional probability ( P(A / B) ) is not simply the ratio of the probabilities of ( A ) and ( B ). The correct formula for conditional probability is ( P(A / B) = \frac{P(A \cap B)}{P(B)} ). The given formula would only be correct if ( A ) and ( B ) were independent events, which is not stated in the problem.
63. A and $B$ are events such that $P(A)=0.4, P(B)=0.3$ and $P(A \cup B)$ $=0.5$. Then $P(B^{\prime} \cap A)$ equals
(a) $\frac{2}{3}$
(b) $\frac{1}{2}$
(c) $\frac{3}{10}$
(d) $\frac{1}{5}$
Show Answer
Solution
Given that: $P(A)=0.4, P(B)=0.3$ and $P(A \cup B)=0.5$
$ \begin{aligned} P(A \cup B) & =P(A)+P(B)P(A \cap B) \\ 0.5 & =0.4+0.3P(A \cap B) \\ P(A \cap B) & =0.4+0.30.5=0.2 \\ \therefore \quad P(B^{\prime} \cap A) & =P(A)P(A \cap B) \\ & =0.40.2=0.2=\frac{1}{5} \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) $\frac{2}{3}$: This option is incorrect because the calculation of $P(B^{\prime} \cap A)$ is based on the given probabilities and the formula $P(B^{\prime} \cap A) = P(A)  P(A \cap B)$. The correct value is $0.2$, which is not equal to $\frac{2}{3}$.

Option (b) $\frac{1}{2}$: This option is incorrect because, as per the given probabilities and the formula used, $P(B^{\prime} \cap A) = 0.2$. The value $\frac{1}{2}$ (or 0.5) does not match the calculated value of $0.2$.

Option (c) $\frac{3}{10}$: This option is incorrect because the correct calculation of $P(B^{\prime} \cap A)$ results in $0.2$. The value $\frac{3}{10}$ (or 0.3) is not equal to the calculated value of $0.2$.
64. If $A$ and $B$ are two events such that $P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$, then $P(A)$ equals
(a) $\frac{3}{10}$
(b) $\frac{1}{5}$
(c) $\frac{1}{2}$
(d) $\frac{3}{5}$
Show Answer
Solution
Given that: $P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$
We know that $P(A / B)=\frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{2}=\frac{P(A \cap B)}{3 / 5}$
$\therefore \quad P(A \cap B)=\frac{3}{10}$
Now $\quad P(A \cup B)=P(A)+P(B)P(A \cap B)$
$ \begin{aligned} \frac{4}{5} & =P(A)+\frac{3}{5}\frac{3}{10} \\ \Rightarrow \quad P(A) & =\frac{4}{5}\frac{3}{5}+\frac{3}{10}=\frac{1}{5}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2} \end{aligned} $
Hence, the correct option is (c).

Option (a) $\frac{3}{10}$ is incorrect because when we solve for $P(A)$ using the given probabilities and the formula for the union of two events, we find that $P(A) = \frac{1}{2}$, not $\frac{3}{10}$.

Option (b) $\frac{1}{5}$ is incorrect because the calculation for $P(A)$ involves subtracting $P(B)$ and adding $P(A \cap B)$ to $P(A \cup B)$. This results in $P(A) = \frac{1}{2}$, not $\frac{1}{5}$.

Option (d) $\frac{3}{5}$ is incorrect because $P(A)$ is derived from the equation $P(A \cup B) = P(A) + P(B)  P(A \cap B)$. Solving this equation with the given values results in $P(A) = \frac{1}{2}$, not $\frac{3}{5}$.
65. In Exercise 64 above, $P(B / A^{\prime})$ is equal to
(a) $\frac{1}{5}$
(b) $\frac{3}{10}$
(c) $\frac{1}{2}$
(d) $\frac{3}{5}$
Show Answer
Solution
According to Exercise 64, we have
$ \begin{aligned} & P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2}, P(A \cup B)=\frac{4}{5} \\ & P(B / A^{\prime})=\frac{P(B \cap A^{\prime})}{P(A^{\prime})}=\frac{P(B)P(A \cap B)}{1P(A)}=\frac{\frac{3}{5}\frac{3}{10}}{1\frac{1}{2}}=\frac{\frac{3}{10}}{\frac{1}{2}}=\frac{3}{5} \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) $\frac{1}{5}$ is incorrect because the calculation of $P(B / A^{\prime})$ involves the difference $P(B)  P(A \cap B)$ and the division by $1  P(A)$. The correct calculation yields $\frac{3}{5}$, not $\frac{1}{5}$.

Option (b) $\frac{3}{10}$ is incorrect because it represents the value of $P(A \cap B)$, not $P(B / A^{\prime})$. The correct calculation for $P(B / A^{\prime})$ yields $\frac{3}{5}$.

Option (c) $\frac{1}{2}$ is incorrect because it does not match the result of the calculation for $P(B / A^{\prime})$. The correct calculation yields $\frac{3}{5}$, not $\frac{1}{2}$.
66. If $P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$, then
$P(A \cup B)^{\prime}+P(A^{\prime} \cup B)=$
(a) $\frac{1}{5}$
(b) $\frac{4}{5}$
(c) $\frac{1}{2}$
(d) 1
Show Answer
Solution
Given that: $P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2}$ and $P(A \cup B)=\frac{4}{5}$
$ \begin{aligned} P(A / B) & =\frac{P(A \cap B)}{P(B)} \\ \Rightarrow \quad \frac{1}{2} & =\frac{P(A \cap B)}{3 / 5} \Rightarrow P(A \cap B)=\frac{3}{10} \\ P(A \cup B) & =P(A)+P(B)P(A \cap B) \\ \frac{4}{5} & =P(A)+\frac{3}{5}\frac{3}{10} \\ \therefore \quad P(A) & =\frac{4}{5}\frac{3}{5}+\frac{3}{10}=\frac{1}{5}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2} \end{aligned} $
Now $P(A \cup B)^{\prime}+P(A^{\prime} \cup B)$
$ \begin{aligned} & =1P(A \cup B)+1P(A \cap B^{\prime}) \\ & =2\frac{4}{5}P(A) \cdot P(B^{\prime}) \\ & =\frac{6}{5}\frac{1}{2} \cdot(1\frac{3}{5})=\frac{6}{5}\frac{1}{2} \times \frac{2}{5}=\frac{6}{5}\frac{1}{5}=\frac{5}{5}=1 \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) $\frac{1}{5}$: This option is incorrect because the calculation of $P(A \cup B)^{\prime} + P(A^{\prime} \cup B)$ involves subtracting the probabilities and adding the complements, which results in a value of 1, not $\frac{1}{5}$.

Option (b) $\frac{4}{5}$: This option is incorrect because it does not account for the correct combination of probabilities and their complements. The correct calculation shows that the sum is 1, not $\frac{4}{5}$.

Option (c) $\frac{1}{2}$: This option is incorrect because it misrepresents the result of the combined probabilities and their complements. The correct calculation yields a total of 1, not $\frac{1}{2}$.
67. Let $P(A)=\frac{7}{13}, P(B)=\frac{9}{13}$ and $P(A \cap B)=\frac{4}{13}$. Then $P(A^{\prime} / B)$ is equal to
(a) $\frac{6}{13}$
(b) $\frac{4}{13}$
(c) $\frac{4}{9}$
(d) $\frac{5}{9}$
Show Answer
Solution
Given that: $P(A)=\frac{7}{13}, P(B)=\frac{9}{13}$ and $P(A \cap B)=\frac{4}{13}$
$ P(A^{\prime} / B)=\frac{P(A^{\prime} \cap B)}{P(B)}=\frac{P(B)P(A \cap B)}{P(B)}=\frac{\frac{9}{13}\frac{4}{13}}{\frac{9}{13}}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9} $
Hence, the correct option is $(d)$.

Option (a) $\frac{6}{13}$: This option is incorrect because it does not account for the conditional probability formula. The correct calculation involves dividing the difference between $P(B)$ and $P(A \cap B)$ by $P(B)$, which results in $\frac{5}{9}$, not $\frac{6}{13}$.

Option (b) $\frac{4}{13}$: This option is incorrect because it represents the probability of $A \cap B$ rather than the conditional probability $P(A^{\prime} / B)$. The correct calculation for $P(A^{\prime} / B)$ involves using the formula $\frac{P(B)  P(A \cap B)}{P(B)}$.

Option (c) $\frac{4}{9}$: This option is incorrect because it results from an incorrect calculation. The correct calculation for $P(A^{\prime} / B)$ is $\frac{P(B)  P(A \cap B)}{P(B)} = \frac{5}{9}$, not $\frac{4}{9}$.
68. If $A$ and $B$ are such events that $P(A)>0$ and $P(B) \neq 1$ then $P(A^{\prime} / B^{\prime})$ equals
(a) $1P(A / B)$
(b) $1P(A^{\prime} / B)$
(c) $\frac{1P(A \cup B)}{P(B^{\prime})}$
(d) $\frac{P(A^{\prime})}{P(B^{\prime})}$
Show Answer
Solution
Given that: $P(A)>0$ and $P(B) \neq 1$
$ \therefore \quad P(A^{\prime} / B^{\prime})=\frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}=\frac{1P(A \cup B)}{P(B^{\prime})} $
Hence, the correct option is (c).

Option (a): (1  P(A / B)) is incorrect because (P(A^{\prime} / B^{\prime})) is not simply the complement of (P(A / B)). The relationship between (P(A^{\prime} / B^{\prime})) and (P(A / B)) involves the joint probabilities of (A) and (B) and their complements, not just a straightforward complement operation.

Option (b): (1  P(A^{\prime} / B)) is incorrect because (P(A^{\prime} / B^{\prime})) is not the complement of (P(A^{\prime} / B)). The conditional probabilities involving complements of events do not follow this simple complement rule.

Option (d): (\frac{P(A^{\prime})}{P(B^{\prime})}) is incorrect because (P(A^{\prime} / B^{\prime})) is the conditional probability of (A^{\prime}) given (B^{\prime}), which depends on the joint probability of (A^{\prime}) and (B^{\prime}). The correct expression involves the joint probability (P(A^{\prime} \cap B^{\prime})), not just the marginal probability (P(A^{\prime})).
69. If $A$ and $B$ are two independent events with $P(A)=\frac{3}{5}$ and $P(B)=\frac{4}{9}$, then $P(A^{\prime} \cap B^{\prime})$ equals
(a) $\frac{4}{15}$
(b) $\frac{8}{45}$
(c) $\frac{1}{3}$
(d) $\frac{2}{9}$
Show Answer
Solution
Given that: A and B are independent events
such that
$ \begin{aligned} & P(A)=\frac{3}{5} \quad \therefore P(A^{\prime})=1\frac{3}{5}=\frac{2}{5} \\ & P(B)=\frac{4}{9} \quad \therefore P(B^{\prime})=1\frac{4}{9}=\frac{5}{9} \end{aligned} $
$ \therefore \quad P(A^{\prime} \cap B^{\prime})=P(A^{\prime}) \cdot P(B^{\prime})=\frac{2}{5} \cdot \frac{5}{9}=\frac{2}{9} $
Hence, the correct option is $(d)$.

Option (a) (\frac{4}{15}): This option is incorrect because it does not correctly represent the product of the probabilities of the complements of (A) and (B). The correct calculation is (\frac{2}{5} \cdot \frac{5}{9} = \frac{2}{9}), not (\frac{4}{15}).

Option (b) (\frac{8}{45}): This option is incorrect because it also does not correctly represent the product of the probabilities of the complements of (A) and (B). The correct calculation is (\frac{2}{5} \cdot \frac{5}{9} = \frac{2}{9}), not (\frac{8}{45}).

Option (c) (\frac{1}{3}): This option is incorrect because it does not correctly represent the product of the probabilities of the complements of (A) and (B). The correct calculation is (\frac{2}{5} \cdot \frac{5}{9} = \frac{2}{9}), not (\frac{1}{3}).
70. If two events are independent, then
(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) (a) and (b) both are correct
(d) none of the above is correct
Show Answer
Solution
For independent events $A$ and $B, P(A) \cdot P(B)=P(A \cap B)$
So, they will not be mutually exclusive.
If $P(A)+P(B)=1$, they are exhaustive events and for independent events $A$ and $P(A \cap B) \neq 0$.
Hence, the correct option is $(d)$.

Option (a): If two events are mutually exclusive, it means that they cannot occur simultaneously, i.e., ( P(A \cap B) = 0 ). However, for independent events, ( P(A \cap B) = P(A) \cdot P(B) ). Therefore, if ( P(A) \cdot P(B) \neq 0 ), the events cannot be mutually exclusive. Hence, independence does not imply mutual exclusivity.

Option (b): The sum of the probabilities of two independent events does not necessarily equal 1. The sum of their probabilities can be any value between 0 and 2, depending on the individual probabilities of the events. Therefore, independence does not imply that ( P(A) + P(B) = 1 ).

Option (c): Since both (a) and (b) are incorrect, this option is also incorrect.
71. Let $A$ and $B$ be two events such that $P(A)=\frac{3}{8}, P(B)=\frac{5}{8}$ and $P(A \cup B)=\frac{3}{4}$. Then $P(A / B) \cdot P(A^{\prime} / B)$ is equal to
(a) $\frac{2}{5}$
(b) $\frac{3}{8}$
(c) $\frac{3}{20}$
(d) $\frac{6}{25}$
Show Answer
Solution
Given that: $P(A)=\frac{3}{8}, P(B)=\frac{5}{8}$ and $P(A \cup B)=\frac{3}{4}$
$ \begin{matrix} \therefore & P(A \cup B) & =P(A)+P(B)P(A \cap B) \\ \frac{3}{4} & =\frac{3}{8}+\frac{5}{8}P(A \cap B) \\ \Rightarrow \quad P(A \cap B) & =\frac{3}{8}+\frac{5}{8}\frac{3}{4}=\frac{1}{4} \end{matrix} $
$ \begin{aligned} \text{ Now } P(A / B) \cdot P(A^{\prime} / B) & =\frac{P(A \cap B)}{P(B)} \cdot \frac{P(A^{\prime} \cap B)}{P(B)} \\ & =\frac{P(A \cap B)}{P(B)} \cdot \frac{P(B)P(A \cap B)}{P(B)} \\ & =\frac{\frac{1}{5}}{\frac{5}{8}} \cdot \frac{(\frac{5}{8}\frac{1}{4})}{\frac{5}{8}}=\frac{2}{5} \cdot \frac{3}{5}=\frac{6}{25} \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) $\frac{2}{5}$: This option is incorrect because it does not account for the correct calculation of ( P(A / B) \cdot P(A^{\prime} / B) ). The correct calculation involves multiplying ( \frac{1/4}{5/8} ) by ( \frac{(5/8  1/4)}{5/8} ), which results in ( \frac{6}{25} ), not ( \frac{2}{5} ).

Option (b) $\frac{3}{8}$: This option is incorrect because it does not follow from the correct application of the formula for conditional probabilities. The correct product of ( P(A / B) ) and ( P(A^{\prime} / B) ) is ( \frac{6}{25} ), not ( \frac{3}{8} ).

Option (c) $\frac{3}{20}$: This option is incorrect because it results from an incorrect calculation of the probabilities involved. The correct product of ( P(A / B) ) and ( P(A^{\prime} / B) ) is ( \frac{6}{25} ), not ( \frac{3}{20} ).
72. If the events $A$ and $B$ are independent, then $P(A \cap B)$ is equal to
(a) $P(A)+P(B)$
(b) $P(A)P(B)$
(c) $P(A) \cdot P(B)$
(d) $\frac{P(A)}{P(B)}$
Show Answer
Solution
Since $A$ and $B$ are two independent events
$ \therefore \quad P(A \cap B)=P(A) \cdot P(B) $
Hence, the correct option is (c).

Option (a) $P(A) + P(B)$: This option is incorrect because the sum of the probabilities of two events does not account for their intersection. For independent events, the probability of both events occurring together is the product of their individual probabilities, not the sum.

Option (b) $P(A)  P(B)$: This option is incorrect because subtracting the probability of one event from another does not represent the probability of their intersection. The intersection of two independent events is calculated by multiplying their probabilities, not by subtraction.

Option (d) $\frac{P(A)}{P(B)}$: This option is incorrect because dividing the probability of one event by the probability of another does not yield the probability of their intersection. For independent events, the correct operation is multiplication, not division.
73. Two events $E$ and $F$ are independent. If $P(E)=0.3$ and $P(E \cup F)=0.5$, then $P(E / F)P(F / E)$ equals
(a) $\frac{2}{7}$
(b) $\frac{3}{35}$
(c) $\frac{1}{70}$
(d) $\frac{1}{7}$
Show Answer
Solution
Given that: $E$ and $F$ are independent events such that $P(E)=0.3$ and $P(E \cup F)=0.5$
$ \begin{matrix} P(E \cup F) =P(E)+P(F)P(E \cap F) \\ 0.5 =0.3+P(F)P(E) \cdot P(F) \\ \Rightarrow \quad 0.50.3 =P(F)[1P(E)] \Rightarrow 0.2=P(F)(10.3) \\ \Rightarrow \quad 0.2 =P(F) \cdot(0.7) \\ \therefore \quad P(F) =\frac{0.2}{0.7}=\frac{2}{7} \\ \\ \text{ Now } P(E / F)P(F / E) =\frac{P(E \cap F)}{P(F)}\frac{P(E \cap F)}{P(E)} \\ =\frac{P(E) \cdot P(F)}{P(F)}\frac{P(E) \cdot P(F)}{P(E)} =P(E)P(F)=\frac{3}{10}\frac{2}{7}=\frac{1}{70} \end{matrix} $
Hence, the correct option is (c).

Option (a) (\frac{2}{7}): This option is incorrect because it represents the probability (P(F)), not the difference (P(E / F)  P(F / E)). The calculation shows that (P(F) = \frac{2}{7}), but this is not the value we are looking for.

Option (b) (\frac{3}{35}): This option is incorrect because it does not match the calculated difference (P(E / F)  P(F / E)). The correct difference is (\frac{1}{70}), and (\frac{3}{35}) is a different value that does not correspond to any part of the given problem.

Option (d) (\frac{1}{7}): This option is incorrect because it is not the result of the difference (P(E / F)  P(F / E)). The correct calculation shows that the difference is (\frac{1}{70}), and (\frac{1}{7}) is a larger value that does not fit the derived result.
74. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability of getting exactly one red ball is
(a) $\frac{45}{196}$
(b) $\frac{135}{392}$
(c) $\frac{15}{56}$
(d) $\frac{15}{29}$
Show Answer
Solution
Given that: Bag contains 5 red and 3 blue balls.
Probability of getting exactly one red ball if 3 balls are randomly drawn without replacement
$=P(R) \cdot P(B) \cdot P(B)+P(B) \cdot P(R) \cdot P(B)+P(B) \cdot P(B) \cdot P(R)$
$=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}=\frac{30}{336}+\frac{30}{336}+\frac{30}{336}=\frac{90}{336}=\frac{15}{56}$
Hence, the correct option is (c).

Option (a) $\frac{45}{196}$ is incorrect because the correct probability calculation yields $\frac{15}{56}$, not $\frac{45}{196}$. The denominator 196 does not align with the total number of ways to draw 3 balls from 8, which is 56.

Option (b) $\frac{135}{392}$ is incorrect because the correct probability calculation yields $\frac{15}{56}$. The numerator 135 and the denominator 392 do not match the correct values derived from the combination calculations.

Option (d) $\frac{15}{29}$ is incorrect because the correct probability calculation yields $\frac{15}{56}$. The denominator 29 is not the correct total number of ways to draw 3 balls from 8, which is 56.
75. Refer to Question 74 above. The probability that exactly two of the three balls were red, the first ball being red is
(a) $\frac{1}{3}$
(b) $\frac{4}{7}$
(c) $\frac{15}{28}$
(d) $\frac{5}{28}$
Show Answer
Solution
According to Question 74,
Let $E_1$ be the event that first ball is red.
$E_2$ be the event that exactly two of the three balls are red.
$\therefore \quad P(E_1)=P(R) \cdot P(R) \cdot P(B)+P(R) \cdot P(R) \cdot P(R)+P(R) \cdot P(B) \cdot P(R)$ $+P(R) \cdot P(B) \cdot P(B)$
$=\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}+\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}+\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}+\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}$
$=\frac{60}{336}+\frac{60}{336}+\frac{60}{336}+\frac{30}{336}=\frac{210}{336}$
$P(E_1 \cap E_2)=P(R) \cdot P(B) \cdot P(R)+P(R) \cdot P(R) \cdot P(B)$
$=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{4}{6}+\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6}=\frac{60}{336}+\frac{60}{336}=\frac{120}{336}$
$\therefore P(E_2 / E_1)=\frac{P(E_1 \cap E_2)}{P(E_1)}=\frac{120 / 336}{210 / 336}=\frac{4}{7}$
Hence, the correct option is $(b)$.

Option (a) $\frac{1}{3}$: This option is incorrect because the calculated probability of exactly two of the three balls being red, given that the first ball is red, is $\frac{4}{7}$. The value $\frac{1}{3}$ does not match this probability and is not derived from the given conditions and calculations.

Option (c) $\frac{15}{28}$: This option is incorrect because the correct probability, as calculated, is $\frac{4}{7}$. The fraction $\frac{15}{28}$ simplifies to $\frac{15}{28}$, which is not equal to $\frac{4}{7}$. Therefore, it does not represent the correct probability.

Option (d) $\frac{5}{28}$: This option is incorrect because the correct probability is $\frac{4}{7}$. The fraction $\frac{5}{28}$ does not match the calculated probability and is not derived from the given conditions and calculations.
76. Three persons A, B and C fire at a target in turn, starting with A. Their probabilities of hitting the target are $0.4,0.3$ and 0.2 respectively. The probabilities of two hits is
(a) 0.024
(b) 0.188
(c) 0.336
(d) 0.452
Show Answer
Solution
Given that: $P(A)=0.4, P(B)=0.3$ and $P(C)=0.2$
Also $P(\overline{A})=10.4=0.6, P(\overline{B})=10.3=0.7$
and $P(\bar{{}C})=10.2=0.8$
$\therefore$ Probabilities of two hits
$=P(A) \cdot P(B) \cdot P(\overline{C})+P(A) \cdot P(\overline{B}) \cdot P(C)+P(\overline{A}) \cdot P(B) \cdot P(C)$
$=0.4 \times 0.3 \times 0.8+0.4 \times 0.7 \times 0.2+0.6 \times 0.3 \times 0.2$
$=0.096+0.056+0.036=0.188$
Hence, the correct option is $(b)$.

Option (a) 0.024: This value is too low to represent the probability of two hits given the probabilities of hitting the target for A, B, and C. The calculation of two hits involves multiple combinations of hits and misses, and the sum of these probabilities is significantly higher than 0.024.

Option (c) 0.336: This value is too high. The correct calculation involves summing the probabilities of three different scenarios where exactly two out of the three persons hit the target. The sum of these probabilities is 0.188, which is much lower than 0.336.

Option (d) 0.452: This value is also too high. The combined probability of two hits from the three shooters, considering their individual probabilities of hitting and missing the target, results in a total probability of 0.188, which is significantly less than 0.452.
77. Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has atleast one girl is
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{4}{7}$
Show Answer
Solution
Let $G$ denotes the girl and $B$ denotes the boy of the given family.
$\text{So}n(S)=\{(BGG),(GBG),(GGB),(GBB),(BGB),(BBG),(BBB), (GGG)\}$
Let $E_1$ be the event that the family has alteast one girl.
$\therefore E_1=\{(BGG),(GBG),(GGB),(GBB),(BGB),(BBG),(GGG)\}$
Let $E_2$ be the event that the eldest child is a girl.
$ \begin{aligned} & \therefore \quad E_2={(GBG),(GGB),(GBB),(GGG)} \\ & (E_1 \cap E_2)={(GBB),(GGB),(GBG),(GGG)} \\ & \therefore P(E_2 / E_1)=\frac{P(E_1 \cap E_2)}{P(E_1)}=\frac{4 / 8}{7 / 8}=\frac{4}{7} \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) $\frac{1}{2}$: This option is incorrect because it assumes that the probability of the eldest child being a girl is independent of the condition that the family has at least one girl. However, the condition changes the sample space, and the correct probability is not simply $\frac{1}{2}$.

Option (b) $\frac{1}{3}$: This option is incorrect because it does not correctly account for the conditional probability given that there is at least one girl in the family. The correct calculation involves considering the reduced sample space where at least one girl is present, which leads to a different probability.

Option (c) $\frac{2}{3}$: This option is incorrect because it likely misinterprets the problem or miscalculates the conditional probability. The correct approach involves finding the intersection of the events and then dividing by the probability of having at least one girl, which results in $\frac{4}{7}$, not $\frac{2}{3}$.
78. If a die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is
(a) $\frac{1}{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{3}{4}$
Show Answer
Solution
Let $E_1$ be the event of getting even number on the die. $E_2$ be the event of selecting a spade card.
$\therefore P(E_1)=\frac{3}{6}=\frac{1}{2}$ and $P(E_2)=\frac{13}{52}=\frac{1}{4}$
So $\quad P(E_1 \cap E_2)=P(E_1) \cdot P(E_2)=\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}$
Hence, the correction option is (c).

Option (a) $\frac{1}{2}$ is incorrect because it assumes that the probability of both events happening together is the same as the probability of one of the events happening alone, which is not the case. The correct probability is the product of the individual probabilities of the two independent events.

Option (b) $\frac{1}{4}$ is incorrect because it only considers the probability of one of the events (selecting a spade card) and does not account for the combined probability of both events (getting an even number on the die and selecting a spade card).

Option (d) $\frac{3}{4}$ is incorrect because it overestimates the combined probability of the two independent events. The correct combined probability is much lower, as calculated by multiplying the individual probabilities of the two events.
79. A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2 green balls and one blue ball is
(a) $\frac{3}{28}$
(b) $\frac{2}{21}$
(c) $\frac{1}{28}$
(d) $\frac{167}{168}$
Show Answer
Solution
Probability of drawing 2 green and 1 blue balls $=P(G) \cdot P(G) \cdot P(B)+P(G) \cdot P(B) \cdot P(G)+P(B) \cdot P(G) \cdot P(G)$
$=\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{2}{6}+\frac{2}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}=\frac{12}{336}+\frac{12}{336}+\frac{12}{336}=\frac{36}{336}=\frac{3}{28}$
Hence, the correct option is $(a)$.

Option (b) $\frac{2}{21}$ is incorrect because the calculated probability of drawing 2 green balls and 1 blue ball is $\frac{3}{28}$, not $\frac{2}{21}$. The value $\frac{2}{21}$ does not match the correct probability derived from the combinations and calculations.

Option (c) $\frac{1}{28}$ is incorrect because the correct probability of drawing 2 green balls and 1 blue ball is $\frac{3}{28}$. The value $\frac{1}{28}$ is too low and does not match the correct probability derived from the combinations and calculations.

Option (d) $\frac{167}{168}$ is incorrect because it is significantly higher than the correct probability of $\frac{3}{28}$. The value $\frac{167}{168}$ is almost 1, which would imply an almost certain event, whereas the actual probability is much lower.
80. A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested then the probability that both are dead is
(a) $\frac{33}{56}$
(b) $\frac{9}{64}$
(c) $\frac{1}{14}$
(d) $\frac{3}{28}$
Show Answer
Solution
Required probability $=P($ dead $) \cdot P($ dead $)$
$ =\frac{3}{8} \cdot \frac{2}{7}=\frac{3}{28} $
Hence, the correct option is $(d)$.

Option (a) $\frac{33}{56}$: This option is incorrect because it does not correctly represent the probability of selecting two dead batteries out of 8, where 3 are dead. The correct calculation involves multiplying the probabilities of selecting a dead battery first and then another dead battery without replacement, which results in $\frac{3}{28}$, not $\frac{33}{56}$.

Option (b) $\frac{9}{64}$: This option is incorrect because it incorrectly assumes that the probability of selecting two dead batteries is the product of $\frac{3}{8}$ and $\frac{3}{8}$, which would be the case if the batteries were replaced after each selection. However, since the selection is without replacement, the correct probability is $\frac{3}{8} \cdot \frac{2}{7} = \frac{3}{28}$.

Option (c) $\frac{1}{14}$: This option is incorrect because it does not accurately reflect the probability of selecting two dead batteries without replacement. The correct probability is $\frac{3}{8} \cdot \frac{2}{7} = \frac{3}{28}$, which simplifies to $\frac{3}{28}$, not $\frac{1}{14}$.
81. If eight coins are tossed together, then the probability of getting exactly 3 heads is
(a) $\frac{1}{256}$
(b) $\frac{7}{32}$
(c) $\frac{5}{32}$
(d) $\frac{3}{32}$
Show Answer
Solution
Here, $n=8, p=\frac{1}{2}, q=1\frac{1}{2}=\frac{1}{2}$ and $r=3$
We know that $P(x=r)={ }^{n} C_r p^{r} . q^{nr}$
$ \begin{aligned} \therefore \quad P(x=3) & ={ }^{8} C_3(\frac{1}{2})^{3}(\frac{1}{2})^{83} \\ & =\frac{8 !}{3 ! \cdot 5 !} \cdot(\frac{1}{2})^{3}(\frac{1}{2})^{5}=56 \cdot(\frac{1}{2})^{8}=56 \cdot \frac{1}{256}=\frac{7}{32} \end{aligned} $
Hence, the correct option is $(b)$.

Option (a) $\frac{1}{256}$ is incorrect because it represents the probability of getting exactly 0 heads or 8 heads when tossing 8 coins, not 3 heads.

Option (c) $\frac{5}{32}$ is incorrect because it does not match the calculated probability of getting exactly 3 heads, which is $\frac{7}{32}$.

Option (d) $\frac{3}{32}$ is incorrect because it also does not match the calculated probability of getting exactly 3 heads, which is $\frac{7}{32}$.
82. Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 , the probability of getting a sum 3 , is
(a) $\frac{1}{18}$
(b) $\frac{5}{18}$
(c) $\frac{1}{5}$
(d) $\frac{2}{5}$
Show Answer
Solution
Let $E_1$ be the event showing the sum of the numbers on the two dice was less than 6 and $E_2$ be the event that the sum of the numbers is 3 .
$ \begin{aligned} & \therefore \quad E_1=\{(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1), \\ &(2,2),(2,3),(3,2)\} \\ & \text{ and } \quad n(E_1)=10 \\ & E_2={(1,2),(2,1)} \Rightarrow n(E_2)=2 \text{ and } n(E_1 \cap E_2)=2 \end{aligned} $
$\therefore$ Required probability
$ P(E_2 / E_1)=\frac{n(E_1 \cap E_2)}{n(E_1)}=\frac{2}{10}=\frac{1}{5} $
Hence, the correct option is (c).

Option (a) $\frac{1}{18}$ is incorrect because it does not correctly represent the ratio of favorable outcomes to the total outcomes where the sum of the numbers on the dice is less than 6. The correct ratio is $\frac{2}{10}$, which simplifies to $\frac{1}{5}$.

Option (b) $\frac{5}{18}$ is incorrect because it overestimates the probability. The correct probability is $\frac{1}{5}$, which is derived from the ratio of favorable outcomes (sum of 3) to the total outcomes where the sum is less than 6.

Option (d) $\frac{2}{5}$ is incorrect because it overestimates the probability. The correct probability is $\frac{1}{5}$, which is derived from the ratio of favorable outcomes (sum of 3) to the total outcomes where the sum is less than 6.
83. Which one is not a requirement of a Binomial distribution?
(a) There are two outcomes for each trial. (b) There is a fixed number of trials.
(c) The outcomes must be dependent on each other.
(d) The probability of success must be the same for all trials.
Show Answer
Solution
We know that for a Binomial distribution, the outcomes must not be dependent on each other.
Hence, the correct option is (c).

(a) There are two outcomes for each trial.
 This is a requirement of a Binomial distribution because each trial must result in one of two possible outcomes, typically labeled as “success” and “failure.”

(b) There is a fixed number of trials.
 This is a requirement of a Binomial distribution because the number of trials must be predetermined and constant.

(d) The probability of success must be the same for all trials.
 This is a requirement of a Binomial distribution because the probability of success must remain constant across all trials to ensure the trials are identically distributed.
84. If two cards are drawn from a well shuffled deck of 52 playing cards with replacement, then the probability that both cards are queens is
(a) $\frac{1}{13} \times \frac{1}{13}$
(b) $\frac{1}{13}+\frac{1}{13}$
(c) $\frac{1}{13} \times \frac{1}{17}$
(d) $\frac{1}{13} \times \frac{4}{51}$
Show Answer
Solution
Probability of getting Queen $=\frac{4}{52}$
So, the required probability
$ \begin{aligned} & =P(\text{ Queen }) \cdot P(\text{ Queen }) \\ & =\frac{4}{52} \cdot \frac{4}{52}=\frac{1}{13} \cdot \frac{1}{13} \quad \text{ (with replacement) } \end{aligned} $
Hence, the correct option is $(a)$.

Option (b) $\frac{1}{13}+\frac{1}{13}$: This option is incorrect because it represents the sum of probabilities rather than the product. When calculating the probability of two independent events both occurring, you multiply their probabilities, not add them.

Option (c) $\frac{1}{13} \times \frac{1}{17}$: This option is incorrect because it uses an incorrect probability for the second card. The probability of drawing a queen remains $\frac{1}{13}$ for both draws since the cards are drawn with replacement, meaning the deck is reset to its original state after each draw.

Option (d) $\frac{1}{13} \times \frac{4}{51}$: This option is incorrect because it assumes the cards are drawn without replacement. If the cards were drawn without replacement, the probability of drawing a queen on the second draw would be $\frac{3}{51}$ after drawing a queen on the first draw. However, since the problem states the cards are drawn with replacement, the probability for each draw remains $\frac{1}{13}$.
85. The probability of guessing correctly atleast 8 out of 10 answers on a true false type examination is
(a) $\frac{7}{64}$
(b) $\frac{7}{128}$
(c) $\frac{45}{1024}$
(d) $\frac{7}{41}$
Show Answer
Solution
Here, $n=10, p=\frac{1}{2}$ and $q=\frac{1}{2} \quad$ (for true/false questions) and $r \geq 8$ i.e. $8,9,10$
$ \begin{aligned} \therefore P(X \geq 8) & =P(x=8)+P(x=9)+P(x=10) \\ & ={ }^{10} C_8(\frac{1}{2})^{8}(\frac{1}{2})^{2}+{ }^{10} C_9(\frac{1}{2})^{9}(\frac{1}{2})+{ }^{10} C _{10}(\frac{1}{2})^{10}(\frac{1}{2})^{0} \\ & =45 \cdot(\frac{1}{2})^{10}+10 \cdot(\frac{1}{2})^{10}+(\frac{1}{2})^{10}=(\frac{1}{2})^{10}(45+10+1) \\ & =56 \times \frac{1}{1024}=\frac{7}{128} \end{aligned} $
Hence, the correct option is (b).

Option (a) $\frac{7}{64}$: This option is incorrect because the calculated probability of guessing correctly at least 8 out of 10 answers is $\frac{7}{128}$, not $\frac{7}{64}$. The value $\frac{7}{64}$ is twice the correct probability, indicating a miscalculation.

Option (c) $\frac{45}{1024}$: This option is incorrect because it represents only the probability of guessing exactly 8 out of 10 answers correctly. The problem requires the probability of guessing at least 8 answers correctly, which includes the probabilities of guessing 8, 9, and 10 answers correctly.

Option (d) $\frac{7}{41}$: This option is incorrect because it does not match the calculated probability of $\frac{7}{128}$. The value $\frac{7}{41}$ is not derived from the binomial probability formula used in the solution and is not a simplified form of the correct probability.
86. The probability that a person is not a swimmer is 0.3 . The probability that out of 5 persons 4 are swimmers is
(a) ${ }^{5} C_4(0.7)^{4}(0.3)$
(b) ${ }^{5} C_1(0.7)(0.3)^{4}$
(c) ${ }^{5} C_4(0.7)(0.3)^{4}$
(d) $(0.7)^{4}(0.3)$
Show Answer
Solution
Given that: $\overline{P}=0.3 \therefore p=0.7$ and $q=10.7=0.3$ $n=5$ and $r=4$
We know that
$ \begin{aligned} & P(X=r)={ }^{n} C_r(p)^{r} \cdot(q)^{nr} \\ \therefore \quad & P(x=4)={ }^{5} C_4(0.7)^{4}(0.3)^{54}={ }^{5} C_4(0.7)^{4}(0.3) \end{aligned} $
Hence, the correct option is (a).

Option (b) is incorrect because it incorrectly places the power of 4 on the probability of not being a swimmer (0.3) instead of the probability of being a swimmer (0.7). The correct formula should have the power of 4 on the probability of being a swimmer.

Option (c) is incorrect because it incorrectly places the power of 4 on the probability of not being a swimmer (0.3) instead of the probability of being a swimmer (0.7). The correct formula should have the power of 4 on the probability of being a swimmer.

Option (d) is incorrect because it does not include the binomial coefficient ${ }^{5} C_4$, which is necessary to account for the different ways 4 swimmers can be chosen out of 5 people.
87. The probability distribution of a discrete random variable $X$ is given below:
$\begin{array}{lllll} \hline X & 2 & 3 & 4 & 5 \\ \hline P(X) & \frac{5}{k} & \frac{7}{k} & \frac{9}{k} & \frac{11}{k} \\ \hline \end{array}$
The value of $k$ is
(a) 8
(b) 16
(c) 32
(d) 48
Show Answer
Solution
We know that $\sum _{i=1}^{n} P(X_i)=1$
$ \begin{aligned} \therefore \quad \frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k} & =1 \\ \frac{32}{k} & =1 \Rightarrow k=32 . \end{aligned} $
Hence, the correct option is (c).

Option (a) 8 is incorrect because if ( k = 8 ), then the sum of the probabilities would be (\frac{5}{8} + \frac{7}{8} + \frac{9}{8} + \frac{11}{8} = \frac{32}{8} = 4 ), which is not equal to 1.

Option (b) 16 is incorrect because if ( k = 16 ), then the sum of the probabilities would be (\frac{5}{16} + \frac{7}{16} + \frac{9}{16} + \frac{11}{16} = \frac{32}{16} = 2 ), which is not equal to 1.

Option (d) 48 is incorrect because if ( k = 48 ), then the sum of the probabilities would be (\frac{5}{48} + \frac{7}{48} + \frac{9}{48} + \frac{11}{48} = \frac{32}{48} = \frac{2}{3} ), which is not equal to 1.
88. For the following probability distribution:
$\begin{array}{llllll} \hline X & 4 & 3 & 2 & 1 & 0\\ \hline P(X) & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \\ \hline \end{array}$
$E(X)$ is equal to:
(a) 0
(b) 1
(c) 2
(d) 1.8
Show Answer
Solution
We know that:
$ \begin{aligned} E(X) & =\sum _{i=1}^{n} X_i P_i \\ & =(4)(0.1)+(3)(0.2)+(2)(0.3)+(1)(0.2)+0(0.2) \\ & =0.40.60.60.2=1.8 \end{aligned} $
Hence, the correct option is $(d)$.

Option (a) 0 is incorrect because the expected value ( E(X) ) is calculated as a weighted average of the values of ( X ) with their respective probabilities. The calculation shows that the sum is (1.8), not 0.

Option (b) 1 is incorrect because the expected value ( E(X) ) is calculated as (1.8), not (1). The sum of the products of each value of ( X ) and its probability does not equal (1).

Option (c) 2 is incorrect because the expected value ( E(X) ) is calculated as (1.8), not (2). The sum of the products of each value of ( X ) and its probability does not equal (2).
89. For the following probability distribution
$\begin{array}{llllll} \hline X & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{1}{10} & \frac{3}{10} & \frac{3}{10} & \frac{2}{10} \\ \hline \end{array}$
$E(X^{2})$ is equal to
(a) 3
(b) 5
(c) 7
(d) 10
Show Answer
Solution
We know that
$ \begin{aligned} E(X^{2}) & =\sum _{i=1}^{n} P_i X_i^{2} \\ & =1 \times \frac{1}{10}+4 \times \frac{1}{5}+9 \times \frac{3}{10}+16 \times \frac{2}{5} \end{aligned} $
$ =\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}=\frac{28}{10}+\frac{36}{5}=\frac{100}{10}=10 $
Hence, the correct option is (d).

Option (a) 3: This option is incorrect because the calculation of (E(X^2)) involves squaring each value of (X) and multiplying by its corresponding probability. The sum of these products does not equal 3. Specifically, the correct calculation yields a value of 10, not 3.

Option (b) 5: This option is incorrect because, similar to option (a), the correct calculation of (E(X^2)) involves summing the products of each squared value of (X) and its corresponding probability. The sum of these products is 10, not 5.

Option (c) 7: This option is incorrect because the correct calculation of (E(X^2)) results in a value of 10. The sum of the products of each squared value of (X) and its corresponding probability does not equal 7.
90. Suppose a random variable $X$ follows the Binomial distribution with parameters $n$ and $p$, where $0<p<1$. If $\frac{P(x=r)}{P(x=nr)}$ is independent of $n$ and $r$, then $p$ equals
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{5}$
(d) $\frac{1}{7}$
Show Answer
Solution
$\quad P(X=r)={ }^{n} C_r p^{r} q^{nr}=\frac{n !}{r !(nr) !} p^{r} \cdot(1p)^{nr}$
$ \begin{aligned} P(X=nr) & ={ }^{n} C _{nr} p^{nr} \cdot(q)^{n(nr)}={ }^{n} C _{nr} p^{nr} \cdot q^{r} \\ & =\frac{n !}{(nr) !(nn+r) !} p^{nr} q^{r}=\frac{n !}{(nr) ! r !} \cdot p^{nr} \cdot q^{r} \end{aligned} $
Now $\frac{P(x=r)}{P(x=nr)}=\frac{\frac{n !}{r !(nr) !} \cdot p^{r} \cdot(1p)^{nr}}{\frac{n !}{r !(nr) !} \cdot p^{nr} \cdot(1p)^{r}}=\frac{(\frac{1p}{p})^{nr}}{(\frac{1p}{p})^{r}}$
The above expression will be independent of $n$ and $r$ if
$ (\frac{1p}{p})=1 \Rightarrow \frac{1}{p}=2 \Rightarrow p=\frac{1}{2} $
Hence, the correct option is (a).

Option (b) $\frac{1}{3}$: If ( p = \frac{1}{3} ), then ( 1  p = \frac{2}{3} ). The ratio (\frac{P(X=r)}{P(X=nr)}) would be (\left(\frac{\frac{2}{3}}{\frac{1}{3}}\right)^{n2r} = 2^{n2r}). This expression is dependent on ( n ) and ( r ), which contradicts the condition that the ratio should be independent of ( n ) and ( r ).

Option (c) $\frac{1}{5}$: If ( p = \frac{1}{5} ), then ( 1  p = \frac{4}{5} ). The ratio (\frac{P(X=r)}{P(X=nr)}) would be (\left(\frac{\frac{4}{5}}{\frac{1}{5}}\right)^{n2r} = 4^{n2r}). This expression is dependent on ( n ) and ( r ), which contradicts the condition that the ratio should be independent of ( n ) and ( r ).

Option (d) $\frac{1}{7}$: If ( p = \frac{1}{7} ), then ( 1  p = \frac{6}{7} ). The ratio (\frac{P(X=r)}{P(X=nr)}) would be (\left(\frac{\frac{6}{7}}{\frac{1}{7}}\right)^{n2r} = 6^{n2r}). This expression is dependent on ( n ) and ( r ), which contradicts the condition that the ratio should be independent of ( n ) and ( r ).
91. In a college, $30 %$ students fail in Physics, $25 %$ fail in Mathematics and $10 %$ fail in both. One student is chosen at random. The probability that she fails in Physics if she has failed in Mathematics is
(a) $\frac{1}{10}$
(b) $\frac{2}{5}$
(c) $\frac{9}{20}$
(d) $\frac{1}{3}$
Show Answer
Solution
Let $E_1$ be the event that the student fails in Physics and $E_2$ be the event that she fails in Mathematics.
$\therefore \quad P(E_1)=\frac{30}{100}, P(E_2)=\frac{25}{100}$ and $P(E_1 \cap E_2)=\frac{10}{100}$
$\therefore P(E_1 / E_2)=\frac{P(E_1 \cap E_2)}{P(E_2)}=\frac{10 / 100}{25 / 100}=\frac{2}{5}$
Hence, the correct option is (b).

Option (a) $\frac{1}{10}$ is incorrect because it does not match the calculated conditional probability $\frac{2}{5}$ for the event that a student fails in Physics given that they have failed in Mathematics. The correct calculation is $\frac{10/100}{25/100} = \frac{2}{5}$, not $\frac{1}{10}$.

Option (c) $\frac{9}{20}$ is incorrect because it is not the result of the conditional probability calculation. The correct probability is $\frac{2}{5}$, which simplifies to $\frac{4}{10}$, not $\frac{9}{20}$.

Option (d) $\frac{1}{3}$ is incorrect because it does not represent the correct conditional probability. The correct calculation yields $\frac{2}{5}$, not $\frac{1}{3}$.
92. A and B are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$ respectively. If the probability of their making a common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is
(a) $\frac{1}{12}$
(b) $\frac{1}{40}$
(c) $\frac{13}{120}$
(d) $\frac{10}{13}$
Show Answer
Solution
Let $E_1$ be the event that both of them solve the problem.
$\therefore \quad P(E_1)=\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}$
and $E_2$ be the event that both of them same incorrectly the problem.
$\therefore \quad P(E_2)=(1\frac{1}{3}) \times(1\frac{1}{4})=\frac{2}{3} \times \frac{3}{4}=\frac{1}{2}$
Let $H$ be the event that both of them get the same answer.
Here, $P(H / E_1)=1, P(H / E_2)=\frac{1}{20}$
$\therefore P(E_1 / H)=\frac{P(E_1) \cdot P(H / E_1)}{P(E_1) \cdot P(H / E_1)+P(E_2) \cdot P(H / E_2)}$
$=\frac{\frac{1}{12} \times 1}{\frac{1}{12} \times 1+\frac{1}{2} \times \frac{1}{20}}=\frac{\frac{1}{12}}{\frac{1}{12}+\frac{1}{40}}=\frac{\frac{1}{12}}{\frac{10+3}{120}}=\frac{1 / 12}{13 / 120}=\frac{10}{13}$
Hence, the correct option is (d).

Option (a) $\frac{1}{12}$: This option is incorrect because it represents the probability that both A and B solve the problem correctly independently, not the conditional probability given that they obtained the same answer.

Option (b) $\frac{1}{40}$: This option is incorrect because it represents the probability that both A and B make a common error independently, not the conditional probability given that they obtained the same answer.

Option (c) $\frac{13}{120}$: This option is incorrect because it is derived from the sum of the probabilities of both solving the problem correctly and both making a common error, but it does not account for the conditional probability given that they obtained the same answer.
93. A box has 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement atmost one is defective?
(a) $(\frac{9}{10})^{5}$
(b) $\frac{1}{2}(\frac{9}{10})^{4}$
(c) $\frac{1}{2}(\frac{9}{10})^{5}$
(d) $(\frac{9}{10})^{5}+\frac{1}{2}(\frac{9}{10})^{4}$
Show Answer
Solution
Here, $n=5, p=\frac{10}{100}=\frac{1}{10}$ and $q=1\frac{1}{10}=\frac{9}{10}$ and $r \leq 1$
We know that
$ \text{ So } \quad \begin{aligned} P(X=r) & ={ }^{n} C_r p^{r} q^{nr} \\ P(x \leq 1) & =P(x=0)+P(x=1) \\ & ={ }^{5} C_0(\frac{1}{10})^{0}(\frac{9}{10})^{5}+{ }^{5} C_1(\frac{1}{10})(\frac{9}{10})^{4} \end{aligned} $
$ =(\frac{9}{10})^{5}+\frac{5}{10}(\frac{9}{10})^{4}=(\frac{9}{10})^{5}+\frac{1}{2}(\frac{9}{10})^{4} $
Hence, the correct option is (d).

Option (a) $(\frac{9}{10})^{5}$ is incorrect because it only accounts for the probability that all 5 pens drawn are nondefective. It does not include the scenario where exactly one pen is defective.

Option (b) $\frac{1}{2}(\frac{9}{10})^{4}$ is incorrect because it only considers the probability of drawing exactly one defective pen and does not include the probability of drawing no defective pens at all.

Option (c) $\frac{1}{2}(\frac{9}{10})^{5}$ is incorrect because it incorrectly scales the probability of drawing no defective pens by $\frac{1}{2}$, which is not part of the correct calculation.
True/False
94. Let $P(A)>0$ and $P(B)>0$. Then $A$ and $B$ can be both mutually exclusive and independent.
Show Answer
Solution
False
 If events ( A ) and ( B ) are mutually exclusive, it means that ( P(A \cap B) = 0 ). However, for events to be independent, the probability of their intersection must equal the product of their individual probabilities, i.e., ( P(A \cap B) = P(A) \cdot P(B) ). Since ( P(A) > 0 ) and ( P(B) > 0 ), ( P(A) \cdot P(B) > 0 ). Therefore, ( P(A \cap B) ) cannot be both 0 and greater than 0 simultaneously. Hence, events ( A ) and ( B ) cannot be both mutually exclusive and independent.
95. If $A$ and $B$ are independent events, then $A^{\prime}$ and $B^{\prime}$ are also independent.
Show Answer
Solution
True
 The statement is true because if ( A ) and ( B ) are independent events, then the probability of their complements ( A’ ) and ( B’ ) occurring together can be shown to be the product of their individual probabilities. Specifically, ( P(A’ \cap B’) = P(A’) \cdot P(B’) ). This follows from the fact that ( P(A’) = 1  P(A) ) and ( P(B’) = 1  P(B) ), and the independence of ( A ) and ( B ) implies ( P(A \cap B) = P(A) \cdot P(B) ). Therefore, the complements ( A’ ) and ( B’ ) are also independent.
96. If A and B are mutually exclusive events, then they will be independent also.
Show Answer
Solution
False

Reason 1: Mutually exclusive events cannot be independent because if one event occurs, the probability of the other event occurring is zero. For two events to be independent, the occurrence of one event should not affect the probability of the other event occurring.

Reason 2: For mutually exclusive events A and B, P(A ∩ B) = 0. However, for independent events, P(A ∩ B) should equal P(A) * P(B). Since P(A ∩ B) = 0 for mutually exclusive events, and P(A) * P(B) is generally not zero unless one of the events has a probability of zero, mutually exclusive events cannot be independent.

Reason 3: Independence implies that the occurrence of one event does not provide any information about the occurrence of the other event. In contrast, if events are mutually exclusive, the occurrence of one event completely rules out the occurrence of the other event, which means they are not independent.
97. Two independent events are always mutually exclusive.
Show Answer
Solution
False
 Independent events are events where the occurrence of one event does not affect the probability of the other event occurring. Mutually exclusive events, on the other hand, are events that cannot occur at the same time. If two events are mutually exclusive, the occurrence of one event means the other event cannot occur, which directly contradicts the definition of independent events. Therefore, independent events cannot be mutually exclusive.
98. If $A$ and $B$ are two independent events, then
$ P(A \text{ and } B)=P(A) \cdot P(B) \text{. } $
Show Answer
Solution
True
 The statement is true because, by definition, if two events (A) and (B) are independent, the probability of both events occurring together (i.e., (P(A \text{ and } B))) is the product of their individual probabilities, (P(A) \cdot P(B)). This is a fundamental property of independent events in probability theory.
99. Another name for the mean of a probability distribution is expected value.
Show Answer
Solution
True $\qquad [\because E(X)=\sum X_i P(X_i)]$
 The statement is true because the mean of a probability distribution is indeed referred to as the expected value. The expected value, denoted as ( E(X) ), is calculated as the sum of each possible value of the random variable ( X ) multiplied by its corresponding probability ( P(X_i) ). This is mathematically represented as ( E(X) = \sum X_i P(X_i) ).
100. If $A$ and $B^{\prime}$ are independent events, then $P(A^{\prime} \cup B)=1P(A) \cdot P(B^{\prime})$
Show Answer
Solution
True $\quad[\because P(A^{\prime} \cup B)=1P(A \cap B^{\prime})=1P(A) \cdot P(B^{\prime})]$
To provide reasons why each of the other options is incorrect, we need to consider the context of the problem and the given answer. The problem states that if ( A ) and ( B’ ) are independent events, then ( P(A’ \cup B) = 1  P(A) \cdot P(B’) ). The given answer is “True” with the explanation provided.
Here are the reasons why other options (i.e., “False”) would be incorrect:
 Incorrect Reasoning for “False” Option:
 Reason 1: The given statement is mathematically correct based on the properties of probability and independence of events. Specifically, for independent events ( A ) and ( B’ ), the probability of their intersection is ( P(A \cap B’) = P(A) \cdot P(B’) ). Using the complement rule and the union rule, we get ( P(A’ \cup B) = 1  P(A \cap B’) = 1  P(A) \cdot P(B’) ). Therefore, stating “False” would contradict these established probability rules.
 Reason 2: The derivation provided in the answer is accurate. It correctly applies the complement rule ( P(A’) = 1  P(A) ) and the fact that ( P(A \cup B) = 1  P(A’ \cap B’) ) for independent events. Hence, any claim that the statement is “False” would ignore these correct applications of probability laws.
 Reason 3: The independence of ( A ) and ( B’ ) is crucial in the given problem. If someone claims “False,” they might be misunderstanding or misapplying the concept of independence. The independence ensures that ( P(A \cap B’) = P(A) \cdot P(B’) ), which is a key step in the provided solution. Ignoring this would lead to an incorrect conclusion.
In summary, the reasons why the “False” option is incorrect are rooted in the proper application of probability rules and the correct understanding of event independence. The given statement is mathematically sound, and any claim to the contrary would be based on a misunderstanding of these principles.
101. If $A$ and $B$ are independent events, then
$P($ exactly one of $A, B$ occurs $)=P(A) \cdot P(B^{\prime})+P(B) \cdot P(A^{\prime})$
Show Answer
Solution
True
To determine why each of the other options is incorrect, we need to understand the given statement and the nature of independent events.
Given:
 ( A ) and ( B ) are independent events.
 The statement to verify is: ( P(\text{exactly one of } A \text{ or } B \text{ occurs}) = P(A) \cdot P(B’) + P(B) \cdot P(A’) ).
Reasoning:

Incorrect Option: ( P(A \cup B) = P(A) + P(B) )
 This is incorrect because ( P(A \cup B) ) represents the probability that either event ( A ) or event ( B ) or both occur. For independent events, the correct formula is: [ P(A \cup B) = P(A) + P(B)  P(A \cap B) ]
 Since ( A ) and ( B ) are independent, ( P(A \cap B) = P(A) \cdot P(B) ). Therefore: [ P(A \cup B) = P(A) + P(B)  P(A) \cdot P(B) ]
 The given option does not account for the intersection term ( P(A \cap B) ), making it incorrect.

Incorrect Option: ( P(A \cap B) = P(A) + P(B) )
 This is incorrect because ( P(A \cap B) ) represents the probability that both events ( A ) and ( B ) occur simultaneously. For independent events, the correct formula is: [ P(A \cap B) = P(A) \cdot P(B) ]
 The given option incorrectly suggests that the probability of both events occurring is the sum of their individual probabilities, which is not true for any events, independent or not.

Incorrect Option: ( P(A \cap B’) = P(A) \cdot P(B) )
 This is incorrect because ( P(A \cap B’) ) represents the probability that event ( A ) occurs and event ( B ) does not occur. For independent events, the correct formula is: [ P(A \cap B’) = P(A) \cdot P(B’) ]
 Since ( B’ ) (the complement of ( B )) is independent of ( A ), we have: [ P(B’) = 1  P(B) ]
 Therefore: [ P(A \cap B’) = P(A) \cdot (1  P(B)) ]
 The given option incorrectly uses ( P(B) ) instead of ( P(B’) ).

Incorrect Option: ( P(A \cup B’) = P(A) \cdot P(B’) )
 This is incorrect because ( P(A \cup B’) ) represents the probability that either event ( A ) occurs or event ( B ) does not occur. For independent events, the correct formula is: [ P(A \cup B’) = P(A) + P(B’)  P(A \cap B’) ]
 Since ( A ) and ( B’ ) are independent, ( P(A \cap B’) = P(A) \cdot P(B’) ). Therefore: [ P(A \cup B’) = P(A) + P(B’)  P(A) \cdot P(B’) ]
 The given option incorrectly suggests that the probability of the union of ( A ) and ( B’ ) is the product of their probabilities, which is not true.
Conclusion:
The given statement ( P(\text{exactly one of } A \text{ or } B \text{ occurs}) = P(A) \cdot P(B’) + P(B) \cdot P(A’) ) is true, and the reasons for the incorrectness of each of the other options are provided above.
102. If $A$ and $B$ are two events such that $P(A)>0$ and $P(A)+P(B)>1$,
then $P(B / A) \geq 1\frac{P(B^{\prime})}{P(A)}$
Show Answer
Solution
False
$ \begin{matrix} {[\because P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{P(A)+P(B)P(A \cup B)}{P(A)}}>\frac{1P(A \cup B)}{P(A)}] \end{matrix} $

The given inequality ( P(B / A) \geq 1  \frac{P(B’)}{P(A)} ) is incorrect because the correct expression for ( P(B / A) ) is ( \frac{P(A \cap B)}{P(A)} ). The inequality ( P(B / A) > \frac{1  P(A \cup B)}{P(A)} ) does not necessarily imply ( P(B / A) \geq 1  \frac{P(B’)}{P(A)} ).

The expression ( 1  \frac{P(B’)}{P(A)} ) simplifies to ( 1  \frac{1  P(B)}{P(A)} = \frac{P(A) + P(B)  1}{P(A)} ). This does not directly relate to the probability ( P(B / A) ) in a straightforward manner.

The inequality ( P(A) + P(B) > 1 ) implies that ( P(A \cup B) ) is at least ( P(A) + P(B)  1 ), but this does not directly lead to the given inequality involving conditional probability.

The correct approach to conditional probability involves understanding the relationship between ( P(A \cap B) ) and ( P(A) ), not the complement of ( B ).
103. If $A, B$ and $C$ are three independent events such that
$P(A)=P(B)=P(C)=p$,
then P (atleast two of A, B and C occur) $=3 p^{2}2 p^{3}$
Show Answer
Solution
True
Since $P$ (atleast two of $A, B$ and $C$ occur)
$ \begin{aligned} & =p \times p \times(1p)+(1p) \cdot p \cdot p+p(1p) \cdot p+p \cdot p \cdot p \\ & =3 p^{2}(1p)+p^{3}=3 p^{2}3 p^{3}+p^{3}=3 p^{2}2 p^{3} \end{aligned} $
To determine why other options are incorrect, let’s analyze the given problem and the provided solution in detail.
Analysis of the Problem
We are given three independent events ( A ), ( B ), and ( C ) with equal probabilities ( P(A) = P(B) = P(C) = p ). We need to find the probability that at least two of these events occur.
Correct Solution
The correct solution is given as: [ P(\text{at least two of } A, B, \text{ and } C \text{ occur}) = 3p^2  2p^3 ]
Detailed Calculation
To find the probability that at least two of the events occur, we can use the principle of inclusionexclusion:

Calculate the probability of exactly two events occurring:
 ( P(A \cap B \cap \neg C) = p \cdot p \cdot (1  p) = p^2(1  p) )
 ( P(A \cap \neg B \cap C) = p \cdot (1  p) \cdot p = p^2(1  p) )
 ( P(\neg A \cap B \cap C) = (1  p) \cdot p \cdot p = p^2(1  p) )
Summing these probabilities: [ 3p^2(1  p) ]

Calculate the probability of all three events occurring:
 ( P(A \cap B \cap C) = p \cdot p \cdot p = p^3 )

Sum the probabilities of exactly two events and all three events: [ 3p^2(1  p) + p^3 = 3p^2  3p^3 + p^3 = 3p^2  2p^3 ]
Conclusion
The provided solution is correct. Now, let’s consider why other options might be incorrect.
Reasons for Incorrect Options

Option: ( 3p^2 )
 This option does not account for the overlap where all three events occur. It only considers the probability of exactly two events occurring, which is ( 3p^2(1  p) ). It misses the ( 2p^3 ) term that adjusts for the overcounting of the scenario where all three events occur.

Option: ( 2p^2  p^3 )
 This option underestimates the probability. It does not correctly account for the number of ways two events can occur and the adjustment for all three events occurring. The correct formula should be ( 3p^2  2p^3 ), not ( 2p^2  p^3 ).

Option: ( p^2(3  2p) )
 While this option looks similar to the correct answer, it is not in the simplified form. The correct simplified form is ( 3p^2  2p^3 ). This option might be a result of incorrect algebraic manipulation or misunderstanding of the inclusionexclusion principle.

Option: ( p^3 + 3p^2(1  p) )
 This option is actually correct in its expanded form but is not simplified. It represents the same calculation as the correct answer but is not presented in the simplified form ( 3p^2  2p^3 ). Therefore, while not incorrect in calculation, it is not the simplified version required.
Summary
 Option ( 3p^2 ): Incorrect because it does not account for the overlap of all three events occurring.
 Option ( 2p^2  p^3 ): Incorrect because it underestimates the probability.
 Option ( p^2(3  2p) ): Incorrect because it is not in the simplified form.
 Option ( p^3 + 3p^2(1  p) ): Correct in calculation but not in the simplified form.
These are the reasons why each of the other options is incorrect.
Fillers
104. If $A$ and $B$ are two events such that $P(A / B)=p, P(A)=p$,
$ P(B)=\frac{1}{3} \text{ and } P(A \cup B)=\frac{5}{9} \text{, then } p \text{ is equal to } $ ……
Show Answer
Solution
Given that: $P(A)=p, P(B)=\frac{1}{3}$ and $P(A \cup B)=\frac{5}{9}$
$ \begin{aligned} P(A / B) & =\frac{P(A \cap B)}{P(B)}=p \Rightarrow P(A \cap B)=p \cdot P(B)=p \cdot \frac{1}{3} \\ \text{ and } P(A \cup B) & =P(A)+P(B)P(A \cap B) \\ \frac{5}{9}=p+\frac{1}{3} & \frac{p}{3} \Rightarrow \frac{5}{9}\frac{1}{3}=\frac{2 p}{3} \Rightarrow \frac{2}{9}=\frac{2 p}{3} \Rightarrow p=\frac{1}{3} \end{aligned} $
Hence, $p$ is equal to $\frac{1}{3}$.
105. If $A$ and $B$ are such that $P(A^{\prime} \cup B^{\prime})=\frac{2}{3}$ and $P(A \cup B)=\frac{5}{9}$, then $P(A^{\prime})+P(B^{\prime})$ is equal to ……
Show Answer
Solution
Here,
$ P(A^{\prime} \cup B^{\prime})=\frac{2}{3} \text{ and } P(A \cup B)=\frac{5}{9} $
$ \begin{aligned} & \therefore & 1P(A \cap B) & =\frac{2}{3} \\ & \Rightarrow & P(A \cap B) & =1\frac{2}{3}=\frac{1}{3} \end{aligned} $
Now $\quad P(A^{\prime})+P(B^{\prime})=1P(A)+1P(B)=2[P(A)+P(B)]$
$ \begin{aligned} & =2[P(A \cup B)+P(A \cap B)] \\ & =2[\frac{5}{9}+\frac{1}{3}]=2\frac{8}{9}=\frac{10}{9} \end{aligned} $
Hence, the value of the filler is $\frac{10}{9}$.
106. If $X$ follows Binomial distribution with parameters $n=5, p$ and $P(X=2)=9 P(X=3)$, then $p$ is equal to ……
Show Answer
Solution
Given that: $P(X=2)=9 P(X=3)$
$ \Rightarrow{ }^{5} C_2 p^{2} q^{3}=9 .{ }^{5} C_3 p^{3} q^{2} $
$ \begin{matrix} \Rightarrow \frac{1}{9} =\frac{{ }^{5} C_3 p^{3} q^{2}}{{ }^{5} C_2 p^{2} q^{3}} \Rightarrow \frac{1}{9}=\frac{p}{q} \quad[\because{ }^{5} C_3={ }^{5} C_2] \\ \Rightarrow 9 p =q \\ \Rightarrow 9 p =1p \Rightarrow 9 p+p=1 \Rightarrow 10 p=1 \therefore p=\frac{1}{10} \end{matrix} $
Hence, the value of the filler is $\frac{1}{10}$.
107. Let $X$ be a random variable taking values $x_1, x_2, x_3, \ldots, x_n$ with probabilities $p_1, p_2, p_3, \ldots, p_n$ respectively. Then $Var(X)$ is equal to ……
Show Answer
Solution
$\quad Var(X)=E(X^{2})[E(X)]^{2}$
$=\sum X^{2} P(X)[\sum X . P(X)]^{2}=\sum p_i x_i^{2}(\sum p_i x_i)^{2}$
Hence, $Var(X)$ is equal to $\sum p_i x_i^{2}(\sum p_i x_i)^{2}$.
108. Let $A$ and $B$ be two events. If $P(A / B)=P(A)$, then $A$ is of B ……
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Solution
$ \begin{matrix} \because \qquad P(A / B) =\frac{P(A \cap B)}{P(B)} \\ \Rightarrow P(A) =\frac{P(A \cap B)}{P(B)} \\ \Rightarrow P(A \cap B) =P(A) \cdot P(B) \end{matrix} $
So, $A$ is independent of $B$.