Solution

It is given that $P(E)=0.6, P(F)=0.3$, and $P(E \cap F)=0.2$

$\Rightarrow P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}$

$\Rightarrow P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}$

Solution

It is given that $P(B)=0.5$ and $P(A \cap B)=0.32$

$\Rightarrow P(\frac{A}{B})=\frac{P(A \cap B)}{P(B)}=\frac{0.32}{0.5}=\frac{16}{25}$

Solution

It is given that $P(A)=0.8, P(B)=0.5$, and $P(B \mid A)=0.4$

(i) $P(B \mid A)=0.4$

$\therefore \frac{P(A \cap B)}{P(A)}=0.4$

$\Rightarrow \frac{P(A \cap B)}{0.8}=0.4$

$\Rightarrow P(A \cap B)=0.32$

(ii) $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

$\Rightarrow P(A \mid B)=\frac{0.32}{0.5}=0.64$

(iii) $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow P(A \cup B)=0.5+0.5-0.32=0.98$

Solution

It is given that, $2 P(A)=P(B)=\frac{5}{13}$

$\Rightarrow P(A)=\frac{5}{26}$ and $P(B)=\frac{5}{13}$

$P(A \mid B)=\frac{2}{5}$

$\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{2}{5}$

$\Rightarrow P(A \cap B)=\frac{2}{5} \times P(B)=\frac{2}{5} \times \frac{5}{13}=\frac{2}{13}$

It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ $\Rightarrow P(A \cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$

$\Rightarrow P(A \cup B)=\frac{5+10-4}{26}$

$\Rightarrow P(A \cup B)=\frac{11}{26}$

Solution

It is given that $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$, and $P(A \cup B)=\frac{7}{11}$

(i) $P(A \cup B)=\frac{7}{11}$

$\therefore P(A)+P(B)-P(A \cap B)=\frac{7}{11}$

$\Rightarrow \frac{6}{11}+\frac{5}{11}-P(A \cap B)=\frac{7}{11}$

$\Rightarrow P(A \cap B)=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}$

(ii) It is known that, $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

$\Rightarrow P(A \mid B)=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}$

(iii) It is known that, $P(B \mid A)=\frac{P(A \cap B)}{P(A)}$

$ \Rightarrow P(B \mid A)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3} $

Solution

If a coin is tossed three times, then the sample space $S$ is

$S={H H H, H H T, H T H, H T T, T H H, T H T, T T H, ~ T T T}$

It can be seen that the sample space has 8 elements.

(i) $E={HHH, HTH, THH, TTH}$

$F={HHH, HHT}$

$\therefore E \cap F={HHH}$

$P(F)=\frac{2}{8}=\frac{1}{4}$ and $P(E \cap F)=\frac{1}{8}$

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2}$

(ii) $E={HHH, HHT, HTH, THH}$

$F={H H T, H T H, H T T, T H H, T H T, T T H, T T T}$

$\therefore E \cap F={HHT, HTH, THH}$

Clearly, $P(E \cap F)=\frac{3}{8}$ and $P(F)=\frac{7}{8}$

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}$

(iii) $E={HHH, HHT, HTT, HTH, THH, THT, TTH}$

$F={HHT, HTT, HTH, THH, THT, TTH, TTT}$ $\therefore E \cap F={HHT, HTT, HTH, THH, THT, TTH}$

$P(F)=\frac{7}{8}$ and $P(E \cap F)=\frac{6}{8}$

Therefore, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{6}{7}}{\frac{7}{8}}=\frac{6}{7}$

Solution

If two coins are tossed once, then the sample space $S$ is

$S={HH, HT, TH, TT}$

(i) $E={HT, TH}$

$F={HT, TH}$

$\therefore E \cap F={HT, TH}$

$P(F)=\frac{2}{8}=\frac{1}{4}$

$P(E \cap F)=\frac{2}{8}=\frac{1}{4}$

$\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{2}{2}=1$

(ii) $E={HH}$

$F={TT}$

$\therefore E \cap F=\Phi$

$P(F)=1$ and $P(E \cap F)=0$ $\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{0}{1}=0$

Solution

If a die is thrown three times, then the number of elements in the sample space will be 6 $\times 6 \times 6=216$

$E=\begin{bmatrix} (1,1,4),(1,2,4), \ldots(1,6,4) \\ (2,1,4),(2,2,4), \ldots(2,6,4) \\ (3,1,4),(3,2,4), \ldots(3,6,4) \\ (4,1,4),(4,2,4), \ldots(4,6,4) \\ (5,1,4),(5,2,4), \ldots(5,6,4) \\ (6,1,4),(6,2,4), \ldots(6,6,4) \end{bmatrix} $

$F={(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}$

$\therefore E \cap F={(6,5,4)}$

$P(F)=\frac{6}{216}$ and $P(E \cap F)=\frac{1}{216}$

$\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{216}}{\frac{6}{216}}=\frac{1}{6}$

Solution

If mother (M), father (F), and son (S) line up for the family picture, then the sample space will be

$S={M F S, M S F, F M S, F S M, S M F, S F M}$

$\Rightarrow E={MFS, FMS, SMF, SFM}$

$F={M F S, S F M}$

$\therefore E \cap F={MFS, SFM}$

$P(E \cap F)=\frac{2}{6}=\frac{1}{3}$

$P(F)=\frac{2}{6}=\frac{1}{3}$

$\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$

Solution

Let the first observation be from the black die and second from the red die.

When two dice (one black and another red) are rolled, the sample space $S$ has $6 \times 6=$ 36 number of elements.

1. Let

A: Obtaining a sum greater than 9

$={(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}$

B: Black die results in a 5 .

$={(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}$

$\therefore A \cap B={(5,5),(5,6)}$

The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5 , is given by $P(A \mid B)$.

$\therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{2}{36}}{\frac{6}{35}}=\frac{2}{6}=\frac{1}{3}$

(b) E: Sum of the observations is 8 .

$={(2,6),(3,5),(4,4),(5,3),(6,2)}$

$F$ : Red die resulted in a number less than 4.

$$=\begin{bmatrix}(1,1),(1,2),(1,3),(2,1),(2,2),(2,3), \\ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3), \\ (5,1),(5,2),(5,3),(6,1),(6,2),(6,3)\end{bmatrix}$$

$\begin{aligned} & \therefore E \cap F=\{(5,3),(6,2)\} \\ & P(F)=\frac{18}{36} \text { and } P(E \cap F)=\frac{2}{36}\end{aligned}$

The conditional probability of obtaining the sum equal to 8 , given that the red die resulted in a number less than 4 , is given by $P$ (E|F).

Therefore, $P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}$

Solution

When a fair die is rolled, the sample space $S$ will be

$S={1,2,3,4,5,6}$

It is given that $E={1,3,5}, F={2,3}$, and $G={2,3,4,5}$

$\therefore P(E)=\frac{3}{6}=\frac{1}{2}$

$P(F)=\frac{2}{6}=\frac{1}{3}$

$P(G)=\frac{4}{6}=\frac{2}{3}$

(i) $E \cap F={3}$

$\therefore P(E \cap F)=\frac{1}{6}$

$\therefore P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}$

$P(F \mid E)=\frac{P(E \cap F)}{P(E)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{3}$

(ii) $E \cap G={3,5}$ $\therefore P(E \cap G)=\frac{2}{6}=\frac{1}{3}$

$\therefore P(E \mid G)=\frac{P(E \cap G)}{P(G)}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$

$P(G \mid E)=\frac{P(E \cap G)}{P(E)}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$

(iii) $E \cup F={1,2,3,5}$

$(E \cup F) \cap G={1,2,3,5} \cap{2,3,4,5}={2,3,5}$

$E \cap F={3}$

$(E \cap F) \cap G={3} \cap{2,3,4,5}={3}$ $\therefore P(E \cup G)=\frac{4}{6}=\frac{2}{3}$

$P((E \cup F) \cap G)=\frac{3}{6}=\frac{1}{2}$

$P(E \cap F)=\frac{1}{6}$

$P((E \cap F) \cap G)=\frac{1}{6}$

$\therefore P((E \cup F) \mid G)=\frac{P((E \cup F) \cap G)}{P(G)}$

$=\frac{\frac{1}{2}}{\frac{2}{3}}=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}$

$P((E \cap F) \mid G)=\frac{P((E \cap G) \cap G)}{P(G)}$

$=\frac{\frac{1}{6}}{\frac{2}{3}}=\frac{1}{6} \times \frac{3}{2}=\frac{1}{4}$

Solution

Let $b$ and $g$ represent the boy and the girl child respectively. If a family has two children, the sample space will be

$S={(b, b),(b, g),(g, b),(g, g)}$

Let $A$ be the event that both children are girls.

$\therefore A={(g, g)}$

(i) Let $B$ be the event that the youngest child is a girl. $\therefore B=[(b, g),(g, g)]$

$\Rightarrow A \cap B={(g, g)}$

$\therefore P(B)=\frac{2}{4}=\frac{1}{2}$

$P(A \cap B)=\frac{1}{4}$

The conditional probability that both are girls, given that the youngest child is a girl, is given by $P(A \mid B)$.

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$

Therefore, the required probability is $\frac{1}{2}$.

(ii) Let $C$ be the event that at least one child is a girl.

$\therefore C={(b, g),(g, b),(g, g)}$

$\Rightarrow A \cap C={g, g}$

$\Rightarrow P(C)=\frac{3}{4}$

$P(A \cap C)=\frac{1}{4}$

The conditional probability that both are girls, given that at least one child is a girl, is given by $P(A \mid C)$.

Therefore, $P(A \mid C)=\frac{P(A \cap C)}{P(C)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Solution

The given data can be tabulated as

Let us denote $E=$ easy questions, $M=$ multiple choice questions, $D=$ difficult questions, and $T=$ True/False questions

Total number of questions $=1400$

Total number of multiple choice questions $=900$

Therefore, probability of selecting an easy multiple choice question is

$P(E \cap M)=\frac{500}{1400}=\frac{5}{14}$

Probability of selecting a multiple choice question, $P(M)$, is $\frac{900}{1400}=\frac{9}{14}$

$P(E \mid M)$ represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

$\therefore P(E \mid M)=\frac{P(E \cap M)}{P(M)}=\frac{\frac{5}{14}}{\frac{9}{14}}=\frac{5}{9}$

Therefore, the required probability is $\frac{5}{9}$.

Solution

When dice is thrown, number of observations in the sample space $=6 \times 6=36$

Let $A$ be the event that the sum of the numbers on the dice is 4 and $B$ be the event that the two numbers appearing on throwing the two dice are different.

$ \begin{aligned} & \therefore A={(1,3),(2,2),(3,1)} \\ & B={\begin{matrix} (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) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{matrix} } \end{aligned} $

$A \cap B={(1,3),(3,1)}$

$\therefore P(B)=\frac{30}{36}=\frac{5}{6}$ and $P(A \cap B)=\frac{2}{36}=\frac{1}{18}$

Let $P(A \mid B)$ represent the probability that the sum of the numbers on the dice is 4 , given that the two numbers appearing on throwing the two dice are different.

$ \therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{18}}{\frac{5}{6}}=\frac{1}{15} $

Therefore, the required probability is $\frac{1}{15}$.

Solution

The outcomes of the given experiment can be represented by the following tree diagram.

The sample space of the experiment is,

$ S={\begin{matrix} (1, H),(1, T),(2, H),(2, T),(3,1)(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4, H),(4, T),(5, H),(5, T),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{matrix} } $

Let $A$ be the event that the coin shows a tail and $B$ be the event that at least one die shows 3.

$ \begin{aligned} & \therefore A={(1, T),(2, T),(4, T),(5, T)} \\ & B={(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(6,3)} \\ & \Rightarrow A \cap B=\phi \\ & \therefore P(A \cap B)=0 \end{aligned} $

Then, $P(B)=P({3,1})+P({3,2})+P({3,3})+P({3,4})+P({3,5})+P({3,6})+P({6,3})$

$ \begin{aligned} & =\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36} \\ & =\frac{7}{36} \end{aligned} $

Probability of the event that the coin shows a tail, given that at least one die shows 3, is given by $P(A \mid B)$.

Therefore,

$ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0}{\frac{7}{36}}=0 $

Solution

It is given that $P(A)=\frac{1}{2}$ and $P(B)=0$ $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{0}$

Therefore, $P(A \mid B)$ is not defined.

Thus, the correct answer is $C$.

Solution

It is given that, $P(A \mid B)=P(B \mid A)$

$\Rightarrow \frac{P(A \cap B)}{P(B)}=\frac{P(A \cap B)}{P(A)}$

$\Rightarrow P(A)=P(B)$

Thus, the correct answer is D.