Solution

It is given that $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$

$A$ and $B$ are independent events. Therefore,

$ P(A \cap B)=P(A) \cdot P(B)=\frac{3}{5} \cdot \frac{1}{5}=\frac{3}{25} $

Solution

There are 26 black cards in a deck of 52 cards.

Let $P(A)$ be the probability of getting a black card in the first draw.

$\therefore P(A)=\frac{26}{52}=\frac{1}{2}$

Let $P(B)$ be the probability of getting a black card on the second draw.

Since the card is not replaced,

$\therefore P(B)=\frac{25}{51}$

Thus, probability of getting both the cards black $=\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}$

Solution

Let $A, B$, and $C$ be the respective events that the first, second, and third drawn orange is good.

Therefore, probability that first drawn orange is good, $P(A)=\frac{12}{15}$

The oranges are not replaced.

Therefore, probability of getting second orange good, $P(B)=\frac{11}{14}$

Similarly, probability of getting third orange good, $P(C)=\frac{10}{13}$

The box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good $=\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}=\frac{44}{91}$

Therefore, the probability that the box is approved for sale is $\frac{44}{91}$.

Solution

If a fair coin and an unbiased die are tossed, then the sample space $S$ is given by,

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

Let A: Head appears on the coin

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

$\Rightarrow P(A)=\frac{6}{12}=\frac{1}{2}$

B: 3 on die $={(H, 3),(T, 3)}$

$P(B)=\frac{2}{12}=\frac{1}{6}$ $\therefore A \cap B={(H, 3)}$

$P(A \cap B)=\frac{1}{12}$

$P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{6}=P(A \cap B)$

Therefore, $A$ and $B$ are independent events.

Solution

When a die is thrown, the sample space (S) is

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

Let $A$ : the number is even $={2,4,6}$

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

$B:$ the number is red $={1,2,3}$

$\Rightarrow P(B)=\frac{3}{6}=\frac{1}{3}$

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

$P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \neq \frac{1}{6}$

$\Rightarrow P(A) \cdot P(B) \neq P(AB)$

Therefore, A and B are not independent.

Solution

It is given that $P(E)=\frac{3}{5}, P(F)=\frac{3}{10}$, and $P(EF)=P(E \cap F)=\frac{1}{5}$

$P(E) \cdot P(F)=\frac{3}{5} \cdot \frac{3}{10}=\frac{9}{50} \neq \frac{1}{5}$

$\Rightarrow P(E) \cdot P(F) \neq P(EF)$

Therefore, $E$ and $F$ are not independent.

Solution

It is given that $P(A)=\frac{1}{2}, P(A \cap B)=\frac{3}{5}$, and $P(B)=p$

(i) When $A$ and $B$ are mutually exclusive, $A \cap B=\Phi$

$\therefore P(A \cap B)=0$

It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$ \begin{aligned} & \Rightarrow \frac{3}{5}=\frac{1}{2}+p-0 \\ & \Rightarrow p=\frac{3}{5}-\frac{1}{2}=\frac{1}{10} \end{aligned} $

(ii) When $A$ and $B$ are independent, $P(A \cap B)=P(A) \cdot P(B)=\frac{1}{2} p$

It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{1}{2} p$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+\frac{p}{2}$

$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

$\Rightarrow p=\frac{2}{10}=\frac{1}{5}$

Solution

It is given that $P(A)=0.3$ and $P(B)=0.4$

(i) If $A$ and $B$ are independent events, then

$P(A \cap B)=P(A) \cdot P(B)=0.3 \times 0.4=0.12$

(ii) It is known that, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$\Rightarrow P(A \cup B)=0.3+0.4-0.12=0.58$ (iii) It is known that, $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

$\Rightarrow P(A \mid B)=\frac{0.12}{0.4}=0.3$

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

$\Rightarrow P(B \mid A)=\frac{0.12}{0.3}=0.4$

Solution

It is given that, $P(A)=\frac{1}{2}$ and $P(A \cap B)=\frac{1}{8}$

$P(not$ on $A$ and not on $B)=P(A^{\prime} \cap B^{\prime})$

$P(not$ on $A$ and not on $B)=P((A \cup B))^{\prime}$

$=1-P(A \cup B)$

$ [A^{\prime} \cap B^{\prime}=(A \cup B)^{\prime}] $

$=1-[P(A)+P(B)-P(A \cap B)]$

$=1-[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}]$

$=1-\frac{5}{8}$

$=\frac{3}{8}$

Solution

It is given that $P(A)=\frac{1}{2}, P(B)=\frac{7}{12}$, and $P(not A$ or $not B)=\frac{1}{4}$

$\Rightarrow P(A^{\prime} \cup B^{\prime})=\frac{1}{4}$

$\Rightarrow P((A \cap B)^{\prime})=\frac{1}{4} \quad[A^{\prime} \cup B^{\prime}=(A \cap B)^{\prime}]$

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

$\Rightarrow P(A \cap B)=\frac{3}{4}$

However, $P(A) \cdot P(B)=\frac{1}{2} \cdot \frac{7}{12}=\frac{7}{24}$

Here, $\frac{3}{4} \neq \frac{7}{24}$

$\therefore P(A \cap B) \neq P(A) \cdot P(B)$

Therefore, A and B are independent events.

Solution

It is given that $P(A)=0.3$ and $P(B)=0.6$

Also, $A$ and $B$ are independent events.

(i) $\therefore P(A$ and $B)=P(A) \cdot P(B)$

$\Rightarrow P(A \cap B)=0.3 \times 0.6=0.18$

(ii) $P(A$ and not $B)=P(A \cap B^{\prime})$ $=P(A)-P(A \cap B)$

$=0.3-0.18$

$=0.12$

(iii) $P(A$ or $B)=P(A \cup B)$

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

$=0.3+0.6-0.18$

$=0.72$

(iv) $P($ neither $A$ nor $B)=P(A^{\prime} \cap B^{\prime})$

$=P((A \cup B)^{\prime})$

$=1-P(A \cup B)$

$=1-0.72$

$=0.28$

Solution

Probability of getting an odd number in a single throw of a die $=\frac{3}{6}=\frac{1}{2}$

Similarly, probability of getting an even number $=\frac{3}{6}=\frac{1}{2}$

Probability of getting an even number three times $=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{1}{8}$

Therefore, probability of getting an odd number at least once

$=1-$ Probability of getting an odd number in none of the throws

$=1-$ Probability of getting an even number thrice

$=1-\frac{1}{8}$

$=\frac{7}{8}$

Solution

Total number of balls $=18$

Number of red balls $=8$

Number of black balls $=10$

(i) Probability of getting a red ball in the first draw $=\frac{8}{18}=\frac{4}{9}$

The ball is replaced after the first draw.

$\therefore$ Probability of getting a red ball in the second draw $=\frac{8}{18}=\frac{4}{9}$

Therefore, probability of getting both the balls red $=\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}$

(ii) Probability of getting first ball black $=\frac{10}{18}=\frac{5}{9}$

The ball is replaced after the first draw.

Probability of getting second ball as red $=\frac{8}{18}=\frac{4}{9}$

Therefore, probability of getting first ball as black and second ball as red $=\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}$

(iii) Probability of getting first ball as red $=\frac{8}{18}=\frac{4}{9}$

The ball is replaced after the first draw.

Probability of getting second ball as black $=\frac{10}{18}=\frac{5}{9}$

Therefore, probability of getting first ball as black and second ball as red $=\frac{4}{9} \times \frac{5}{9}=\frac{20}{81}$ Therefore, probability that one of them is black and other is red

$=$ Probability of getting first ball black and second as red + Probability of getting first ball

red and second ball black $=\frac{20}{81}+\frac{20}{81}$

$ =\frac{40}{81} $

Solution

Probability of solving the problem by $A, P(A)=\frac{1}{2}$

Probability of solving the problem by $B, P(B)=\frac{1}{3}$

Since the problem is solved independently by $A$ and $B$,

$\therefore P(AB)=P(A) \cdot P(B)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$

$P(A^{\prime})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}$

$P(B^{\prime})=1-P(B)=1-\frac{1}{3}=\frac{2}{3}$ i. $\quad$ Probability that the problem is solved $=P(A \cup B)$

$=P(A)+P(B)-P(A B)$ $=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{4}{6}$

$=\frac{2}{3}$

(ii) Probability that exactly one of them solves the problem is given by,

$ \begin{aligned} & P(A) \cdot P(B^{\prime})+P(B) \cdot P(A^{\prime}) \\ & =\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3} \\ & =\frac{1}{3}+\frac{1}{6} \\ & =\frac{1}{2} \end{aligned} $

Solution

(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.

$\therefore P(E)=P($ the card drawn is a spade $)=\frac{13}{52}=\frac{1}{4}$ $\therefore P(F)=P($ the card drawn is an ace $)=\frac{4}{52}=\frac{1}{13}$

In the deck of cards, only 1 card is an ace of spades.

$P(E F)=P($ the card drawn is spade and an ace $)=\frac{1}{52}$

$P(E) \times P(F)=\frac{1}{4} \cdot \frac{1}{13}=\frac{1}{52}=P(E F)$

$\Rightarrow P(E) \times P(F)=P(E F)$

Therefore, the events $E$ and $F$ are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.

$\therefore P(E)=P($ the card drawn is black $)=\frac{26}{52}=\frac{1}{2}$

$\therefore P(F)=P($ the card drawn is a king $)=\frac{4}{52}=\frac{1}{13}$

In the pack of 52 cards, 2 cards are black as well as kings.

$\therefore P(EF)=P($ the card drawn is a black king $)=\frac{2}{52}=\frac{1}{26}$ $P(E) \times P(F)=\frac{1}{2} \cdot \frac{1}{13}=\frac{1}{26}=P(EF)$

Therefore, the given events $E$ and $F$ are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.

$\therefore P(E)=P($ the card drawn is a king or a queen $)=\frac{8}{52}=\frac{2}{13}$

$\therefore P(F)=P($ the card drawn is a queen or a jack $)=\frac{8}{52}=\frac{2}{13}$

There are 4 cards which are king or queen and queen or jack.

$\therefore P(EF)=P$ (the card drawn is a king or a queen, or queen or a jack)

$=\frac{4}{52}=\frac{1}{13}$

$P(E) \times P(F)=\frac{2}{13} \cdot \frac{2}{13}=\frac{4}{169} \neq \frac{1}{13}$

$\Rightarrow P(E) \cdot P(F) \neq P(EF)$

Therefore, the given events $E$ and $F$ are not independent.

Solution

Let $H$ denote the students who read Hindi newspaper and $E$ denote the students who read English newspaper.

It is given that,

$ \begin{aligned} & P(H)=60 %=\frac{6}{10}=\frac{3}{5} \\ & P(E)=40 %=\frac{40}{100}=\frac{2}{5} \\ & P(H \cap E)=20 %=\frac{20}{100}=\frac{1}{5} \end{aligned} $

i. Probability that a student reads Hindi or English newspaper is,

$ \begin{aligned} (H \cup E)^{\prime} & =1-P(H \cup E) \\ & =1-{P(H)+P(E)-P(H \cap E)} \\ & =1-(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}) \\ & =1-\frac{4}{5} \\ & =\frac{1}{5} \end{aligned} $

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by $P(E \mid H)$.

$ \begin{aligned} P(E \mid H) & =\frac{P(E \cap H)}{P(H)} \\ & =\frac{\frac{1}{3}}{\frac{3}{5}} \\ & =\frac{1}{3} \end{aligned} $

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by $P(H \mid E)$.

$ \begin{aligned} P(H \mid E) & =\frac{P(H \cap E)}{P(E)} \\ & =\frac{\frac{1}{5}}{\frac{2}{5}} \\ & =\frac{1}{2} \end{aligned} $

Choose the correct answer in Exercises 17 and 18.

Solution

When two dice are rolled, the number of outcomes is 36 .

The only even prime number is 2 .

Let $E$ be the event of getting an even prime number on each die.

$\therefore E={(2,2)}$

$\Rightarrow P(E)=\frac{1}{36}$

Therefore, the correct answer is D.

Solution

Two events $A$ and $B$ are said to be independent, if $P(A B)=P(A) \times P(B)$

Consider the result given in alternative $\mathbf{B}$.

$P(A^{\prime} B^{\prime})=[1-P(A)][1-P(B)]$

$\Rightarrow P(A^{\prime} \cap B^{\prime})=1-P(A)-P(B)+P(A) \cdot P(B)$

$\Rightarrow 1-P(A \cup B)=1-P(A)-P(B)+P(A) \cdot P(B)$

$\Rightarrow P(A \cup B)=P(A)+P(B)-P(A) \cdot P(B)$

$\Rightarrow P(A)+P(B)-P(AB)=P(A)+P(B)-P(A) \cdot P(B)$

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

This implies that $A$ and $B$ are independent, if $P(A^{\prime} B^{\prime})=[1-P(A)][1-P(B)]$

Distracter Rationale

A. Let $P(A)=m, P(B)=n, 0<m, n<1$

$A$ and $B$ are mutually exclusive.

$\therefore A \cap B=\phi$

$\Rightarrow P(AB)=0$

However, $P(A) \cdot P(B)=m n \neq 0$

$\therefore P(A) \cdot P(B) \neq P(AB)$

C. Let A: Event of getting an odd number on throw of a die $={1,3,5}$

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

B: Event of getting an even number on throw of a die $={2,4,6}$

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

Here, $A \cap B=\phi$

$\therefore P(AB)=0$

$P(A) \cdot P(B)=\frac{1}{4} \neq 0$

$\Rightarrow P(A) \cdot P(B) \neq P(AB)$

D. From the above example, it can be seen that, $P(A)+P(B)=\frac{1}{2}+\frac{1}{2}=1$

However, it cannot be inferred that $A$ and $B$ are independent.

Thus, the correct answer is $B$.