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# NCERT Solutions for Class 12 Maths Chapter 13 - Exercise Last updated date: 25th Nov 2023
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## NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.2) Exercise 13.2

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.2 (Ex 13.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 13 Probability Exercise 13.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

### Important Topics

The topics and sub-topics of NCERT Solutions for Class 12 Maths Chapter 13 Probability are given below.

 Section Name Topic Name 13 Probability 13.1 Introduction 13.2 Conditional Probability 13.3 Multiplication Theorem on Probability 13.4 Independent Events 13.5 Bayes’ Theorem 13.6 Random Variables and its Probability Distributions 13.7 Bernoulli Trials and Binomial Distribution

## Access NCERT Solutions for Class 12 Maths Chapter 13 – Probability

Exercise 13.2

1. Let $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$, then determine the probability $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)$ when $\mathbf{A}$ and $\mathbf{B}$ are independent events.

Ans. We are provided that,

$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{3}}{\text{5}}$ and $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{5}}$.

Then we have

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$, since $\text{A}$ and $\text{B}$ are independent.

$\text{=}\dfrac{3}{5}\cdot \dfrac{1}{5}$

$\text{=}\dfrac{3}{25}$.

Hence, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{3}}{\text{25}}$.

2. Let there are $\mathbf{52}$ playing cards in a packet. If two cards are needed to be drawn at random and without replacement, then what will be the probability that both the cards are black.

Ans. It is known that in the pack of $\text{52}$ cards, there are $\text{26}$ black cards.

Let $\text{P}\left( \text{A} \right)$ denotes the probability that a black card is drawn in the first draw.

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{26}}{\text{52}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Let suppose $\text{P}\left( \text{B} \right)$ denotes the probability that a black card is drawn in the second draw.

Therefore, $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{25}}{\text{51}}$.

Hence, the probability that both the cards drawn are black $\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{25}}{\text{51}}\text{=}\dfrac{\text{25}}{\text{102}}$.

3. Three oranges are chosen randomly without replacement from a box of oranges. The box is ready to go if all three oranges seem good, otherwise it is rejected. Determine the probability that a box having $\mathbf{15}$ oranges out of which $\mathbf{12}$ are good and $\mathbf{3}$ are bad ones will be ready to go for sale.

Ans. Let the events of choosing three oranges be $\text{A,}\,\,\text{B,}$ and $\text{C}$ respectively.

So, the probability that the orange drawn at first is good, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{12}}{\text{15}}$.

Since, that orange is not replaced so, there only $14$ oranges left.

Similarly, the probability that the second orange chosen is good, $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{11}}{\text{14}}$.

Also, the probability that the third orange drawn is good,

$\text{P}\left( \text{C} \right)\text{=}\dfrac{\text{10}}{\text{13}}$.

Now, it is described that, if all three oranges are good, then the whole box will be ready to go for sale.

Therefore, the probability that all the three oranges are good

$=\dfrac{12}{15}\times \dfrac{11}{14}\times \dfrac{10}{13}=\dfrac{44}{91}$.

Hence, the probability that the whole box of oranges will be ready to go for sale is $\dfrac{44}{91}$.

4. Let a fair coin and an unbiased coin are tossed together. If $\mathbf{A}$ and $\mathbf{B}$ denotes the following events, then decide whether the events $\mathbf{A}$ and $\mathbf{B}$ are independent.

$\mathbf{A}:$ head comes on the coin.

$\mathbf{B}:$ $\mathbf{3}$ appears in the die.

Ans. The sample space when a fair coin and an unbiased coin are tossed is

$\text{S=}\left\{ \left( \text{H,1} \right)\text{,}\left( \text{H,2} \right)\text{,}\left( \text{H,3} \right)\text{,}\left( \text{H,4} \right)\text{,}\left( \text{H,5} \right)\text{,}\left( \text{H,6} \right)\text{,}\left( \text{T,1} \right)\text{,}\left( \text{T,2} \right)\text{,}\left( \text{T,3} \right)\text{,}\left( \text{T,4} \right)\text{,}\left( \text{T,5} \right)\text{,}\left( \text{T,6} \right) \right\}$

Since, for the event $\text{A}$, head comes on the fair coin, so

$\text{A=}\left\{ \left( \text{H,1} \right)\text{,}\left( \text{H,2} \right)\text{,}\left( \text{H,3} \right)\text{,}\left( \text{H,4} \right)\text{,}\left( \text{H,5} \right)\text{,}\left( \text{H,6} \right) \right\}$

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{6}}{\text{12}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Now, $\text{B}$ suggests the event of appearing $\text{3}$ in the die.

So, $\text{B=}\left\{ \left( \text{H,3} \right)\text{,}\left( \text{T,3} \right) \right\}$.

So, $\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{2}}{\text{12}}\text{=}\dfrac{\text{1}}{\text{6}}$.

Thus, $\text{A}\cap \text{B=}\left\{ \left( \text{H,3} \right) \right\}$ and so,

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{1}{12}$.

Now,

$\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{12}}\text{=P}\left( \text{A}\cap \text{B} \right)$.

Hence, it is concluded that the events $\text{A}$ and $\text{B}$ are independent.

5. An unbiased die is tossed where $\mathbf{1,2,3}$ are colored as red and $\mathbf{4},\mathbf{5},\mathbf{6}$ are in green color. If $\mathbf{A}$ and $\mathbf{B}$ denotes the following events, then determine whether the events are independent.

$\mathbf{A}:$ the number is even.

$\mathbf{B}:$ the number is red.

Ans. The sample space when a die thrown is

$\text{S=}\left\{ \text{1,2,3,4,5,6} \right\}$.

Therefore, when the number is even, then

$\text{A=}\left\{ 2,4,6 \right\}$ and so

$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Also, when the number turns out red, then

$\text{B=}\left\{ 1,2,3 \right\}$ and therefore,

$\text{P}\left( \text{B} \right)\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Thus, $\text{A}\cap \text{B=}\left\{ 2 \right\}$ and so,

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{1}}{\text{6}}$.

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=}\dfrac{\text{1}}{\text{6}}$.

Now, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{4}}\ne \dfrac{1}{6}$

That is, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \left( \text{AB} \right)$.

Hence, the events $\text{A}$ and $\text{B}$ are not independent.

6. If the two events $\mathbf{E}$ and $\mathbf{F}$ be such that $\mathbf{P}\left( \mathbf{E} \right)=\dfrac{\mathbf{3}}{\mathbf{5}},\mathbf{P}\left( \mathbf{F} \right)=\dfrac{\mathbf{3}}{\mathbf{10}}$ and $\mathbf{P}\left( \mathbf{E}\cap \mathbf{F} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$, then determine whether the events $\mathbf{E}$ and $\mathbf{F}$ are independent.

Ans. We have, $\text{P}\left( \text{E} \right)\text{=}\dfrac{\text{3}}{\text{5}}\text{,P}\left( \text{F} \right)\text{=}\dfrac{\text{3}}{\text{10}}$ and $\text{P}\left( \text{E}\cap \text{F} \right)\text{=}\dfrac{\text{1}}{\text{5}}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{10}}\text{=}\dfrac{\text{9}}{\text{50}}\ne \dfrac{\text{1}}{\text{5}}$.

Therefore, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{E}\cap \text{F} \right)$.

That is, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{EF} \right)$.

Hence, the events $\text{E}$ and $\text{F}$ are not independent.

7. Let the two events $\mathbf{A}$ and $\mathbf{B}$ be such that $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{1}}{\mathbf{2}},\mathbf{P}\left( \mathbf{A}\cup \mathbf{B} \right)=\dfrac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}\left( \mathbf{B} \right)=\mathbf{p}$. Then answer the following questions.

(i) Determine the value of $\mathbf{p}$ if the events $\mathbf{A},\,\mathbf{B}$ are mutually exclusive.

Ans. It is known that if two events $\text{A}$ and $\text{B}$ are mutually exclusive, then

$\text{A}\cap \text{B=}\phi$.

Therefore, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}$.

Now, we also know that, $\text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

$\Rightarrow \dfrac{\text{3}}{\text{5}}\text{=}\dfrac{\text{1}}{\text{2}}\text{+p-0}$

$\Rightarrow \text{p=}\dfrac{\text{3}}{\text{5}}\text{-}\dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\text{1}}{\text{10}}$.

(ii) Determine the value of $\mathbf{p}$ if the events $\mathbf{A},\,\mathbf{B}$ are independent.

Ans. Since, the events $\text{A}$ and $\text{B}$ are independent, so

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\text{p}$.

Now, we know that, $\text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

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

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

$\Rightarrow \text{p=}\dfrac{\text{2}}{\text{10}}\text{=}\dfrac{\text{1}}{\text{5}}$.

8. Let the events $\mathbf{A}$ and $\mathbf{B}$ are independent such that $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}.\mathbf{3},\,\,\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{4}$. Then determine the values of the following probabilities.

(i) $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)$

Ans. We are provided that $\text{P}\left( \text{A} \right)\text{=0}\text{.3,}\,\,\text{P}\left( \text{B} \right)\text{=0}\text{.4}$.

Since the two events are independent, so

$\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=0}\text{.3}\times \text{0}\text{.4=0}\text{.12}$.

(ii) $\mathbf{P}\left( \mathbf{A}\cup \mathbf{B} \right)$

Ans. We know that, $\text{P}\left( \text{A}\cup \text{B} \right)=\text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

Since, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}\text{.12}$, so

$\text{P}\left( \text{A}\cup \text{B} \right)=\text{0}\text{.3+0}\text{.4-0}\text{.12=0}\text{.58}$.

(iii) $\mathbf{P}\left( \mathbf{A}|\mathbf{B} \right)$

Ans. We know that, the conditional probability

$\text{P}\left( \text{A }\!\!|\!\!\text{ B} \right)\text{=}\dfrac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{B} \right)}$

$\Rightarrow \text{P}\left( \text{A }\!\!|\!\!\text{ B} \right)\text{=}\dfrac{0.12}{0.4}\text{=0}\text{.3}$.

(iv) $\mathbf{P}\left( \mathbf{B}|\mathbf{A} \right)$

Ans. We know that, the conditional probability

$\text{P}\left( \text{B }\!\!|\!\!\text{ A} \right)\text{=}\dfrac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{A} \right)}$

Therefore, $\text{P}\left( \text{B }\!\!|\!\!\text{ A} \right)\text{=}\dfrac{0.12}{0.3}\text{=0}\text{.4}$.

9. Let the two events $\mathbf{A}$ and $\mathbf{B}$ be such that $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{1}}{\mathbf{4}},\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{2}}$ and $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{8}}$. Then determine the probability $\mathbf{P}\left( \mathbf{{A}'}\cap \mathbf{{B}'} \right)$.

Ans. We are provided that, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{1}}{\text{4}}\text{,P}\left( \text{B} \right)\text{=}\dfrac{\text{1}}{\text{2}}$ and $\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{\text{1}}{\text{8}}$.

Now, it is known that, $\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$.

Therefore,

$\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A}\cup \text{B} \right)$.

Again, it is known that,

$\text{P}\left( \text{A}\cup \text{B} \right)=\text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$.

Thus,

$\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A}\cup \text{B} \right)\text{=1-}\left[ \text{P}\left( \text{A} \right)\text{+P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right) \right]$

$\text{=1-}\left[ \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{8} \right]$

$\text{=1-}\dfrac{5}{8}$

$\text{=}\dfrac{3}{8}$.

Hence, $\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=}\dfrac{3}{8}$.

10. Let the two events $\mathbf{A}$ and $\mathbf{B}$ be such that $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{1}}{\mathbf{2}},\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{7}}{\mathbf{12}}$, and $\mathbf{P}\left( \mathbf{{A}'}\cap \mathbf{{B}'} \right)=\dfrac{\mathbf{1}}{\mathbf{4}}$. Then determine whether the events $\mathbf{A}$ and $\mathbf{B}$ are independent.

Ans. We are provided that, $\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{,P}\left( \text{B} \right)\text{=}\dfrac{\text{7}}{\text{12}}$, and $\text{P}\left( \text{{A}'}\cup \text{{B}'} \right)\text{=}\dfrac{\text{1}}{\text{4}}$.

It is known that, $\text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$.

Therefore,

$\text{P}\left( \text{{A}'}\cup \text{{B}'} \right)\text{=}\dfrac{1}{4}$

$\text{P}{{\left( \text{A}\cap \text{B} \right)}^{\prime }}\text{=}\dfrac{1}{4}$

$\Rightarrow 1-\text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{1}{4}$

$\Rightarrow \text{P}\left( \text{A}\cap \text{B} \right)\text{=}\dfrac{3}{4}$.

But, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{7}{12}\text{=}\dfrac{7}{24}$.

Therefore, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \left( \text{A}\cap \text{B} \right)$.

Hence, the events $\text{A}$ and $\text{B}$ are not independent.

11. Let the two events $\mathbf{A}$ and $\mathbf{B}$ be such that $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}\mathbf{.3},\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{6}$. Then determine the values of the following probabilities.

(i) $\mathbf{P}$($\mathbf{A}$ and $\mathbf{B}$)

Ans. Since, the events $\text{A}$and $\text{B}$ are independent, so

$\text{P}$($\mathbf{A}$ and $\text{B}$)$\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$

$\text{=0}\text{.3}\times \text{0}\text{.6=0}\text{.18}$.

(ii) $\mathbf{P}$($\mathbf{A}$ and not $\mathbf{B}$)

Ans. Note that,

$\text{P}$($\text{A}$ and not $\text{B}$)$\text{=P}\left( \text{A}\cap \text{{B}'} \right)\text{=P}\left( \text{A} \right)\text{-P}\left( \text{A}\cap \text{B} \right)$

$\text{=0}\text{.3-0}\text{.18}$

$\text{=0}\text{.12}$.

(iii) $\mathbf{P}$($\mathbf{A}$ or $\mathbf{B}$)

Ans.

It is known that,

$\text{P}$($\text{A}$ or $\text{B}$)$\text{=P}\left( \text{A}\cup \text{B} \right)$

$\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)\text{-P}\left( \text{A}\cap \text{B} \right)$

$\text{=0}\text{.3+0}\text{.6-0}\text{.18}$

$\text{=0}\text{.72}$.

(iv) $\mathbf{P}$(neither $\mathbf{A}$ nor $\mathbf{B}$)

Ans. It is known that,

$\text{P}$(neither $\text{A}$ nor $\text{B}$)$\text{=P}{{\left( \text{A}\cup \text{B} \right)}^{\prime }}$

$\text{=1-P}\left( \text{A}\cup \text{B} \right)$

$\text{=1-0}\text{.72}$

$\text{=0}\text{.28}$

12. If an unbiased die thrown three times, then what will be the probability of obtaining an odd number at least one times.

Ans. The probability of having an odd number in tossing the die one times $\text{=}\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{1}}{\text{2}}$.

Again, the probability of having an even number $\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$.

Therefore, the probability of having an even number thrice $\text{=}\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\text{=}\dfrac{1}{8}$.

Thus, the probability of obtaining an od number at least one times

$\text{=1-}$probability of not having an odd number in any of the throws

$\text{=1-}$probability of having an even number three times

$\text{=1-}\dfrac{1}{8}$

$\text{=}\dfrac{7}{8}$.

13. Suppose a box contains $\mathbf{10}$ black and $\mathbf{8}$ red balls and two balls are chosen randomly with replacement from the box. Then determine the values of the following probabilities.

(i) When both balls are red.

Ans. Note that, there is total $\text{18}$ balls, among them $\text{8}$ are red and $\text{10}$ are black.

Therefore, the probability of having a red ball on drawing the first time $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

The total count of balls remains the same as the balls are replaced.

So, the probability of having a red ball on drawing second times $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

Thus, the probability of having both the balls is red $\text{=}\dfrac{4}{9}\times \dfrac{4}{9}\text{=}\dfrac{16}{81}$.

(ii) When the first ball is black and the other is red.

Ans. The probability of having a black ball in first draw $\text{=}\dfrac{10}{18}\text{=}\dfrac{5}{9}$.

The total count of balls remains the same as the balls are replaced.

Then, the probability of having a red ball in second draw $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

Thus, the probability of having the black and red balls in first and second draw respectively

$\text{=}\dfrac{5}{4}\times \dfrac{4}{9}\text{=}\dfrac{20}{81}$.

(iii) When one of the balls is black and the other is red.

Ans. The probability of having a red ball in first draw $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

The total count of balls remains the same as the balls are replaced.

Then, the probability of having a black ball in second draw $\text{=}\dfrac{10}{18}\text{=}\dfrac{5}{9}$.

Thus, the probability of having the red and black balls in first and second draw respectively

$\text{=}\dfrac{4}{9}\times \dfrac{5}{4}\text{=}\dfrac{20}{81}$.

Therefore, the probability of having one ball as red and the another as black

$\text{=}$ probability of obtaining the first ball is black and the other is red

$+$ probability of obtaining the first ball is red and the other is black

$\text{=}\dfrac{20}{81}+\dfrac{20}{81}$

$=\dfrac{40}{81}$.

14. The probability that a specific problem can be solved independently by $\mathbf{A}$ and $\mathbf{B}$ are $\dfrac{\mathbf{1}}{\mathbf{2}}$ and $\dfrac{\mathbf{1}}{\mathbf{3}}$ respectively. Let both will try to solve the problem independently. Then determine the probabilities for the following two cases.

(i) the problem is solved.

Ans. Probability that $\text{A}$ can solve the problem, $\text{P}\left( \text{A} \right)\text{=}\dfrac{1}{2}$,

Probability that $\text{B}$ can solve the problem, $\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{3}$.

It is given that the problem is solved by them independently.

Therefore, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{3}\text{=}\dfrac{1}{6}$.

Also, $\text{P}\left( {\text{{A}'}} \right)\text{=1-P}\left( \text{A} \right)\text{=1-}\dfrac{1}{2}\text{=}\dfrac{1}{2}$ and

$\text{P}\left( {\text{{B}'}} \right)\text{=1-P}\left( \text{B} \right)\text{=1-}\dfrac{1}{3}\text{=}\dfrac{2}{3}$.

Thus, the probability that $\text{A}$ or $\text{B}$ can solve the problem

$\text{=P}\left( \text{A}\cup \text{B} \right)$

$\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{A}\cap \text{B} \right)$

$\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}$

$\text{=}\dfrac{4}{6}$

That is, the probability that the problem is solved $\text{=}\dfrac{2}{3}$.

(ii) exactly one of them solves that specific problem.

Ans. The probability that exactly one of $\text{A}$ and $\text{B}$ solves the problem,

$\text{P}\left( \text{A} \right)\cdot \text{P}\left( {\text{{B}'}} \right)+\text{P}\left( \text{B} \right)\cdot \text{P}\left( {\text{{A}'}} \right)$

$\text{=}\dfrac{1}{2}\times \dfrac{2}{3}+\dfrac{1}{2}\times \dfrac{1}{3}$

$\text{=}\dfrac{1}{3}+\dfrac{1}{6}$

$\text{=}\dfrac{1}{2}$.

15. Let there are $\mathbf{52}$ well shuffled cards and one card is chosen randomly from them. Then determine whether in the following cases events $\mathbf{E}$ and $\mathbf{F}$ are independent.

(i) $\mathbf{E}:$ the card chosen is showing a spade.

$\mathbf{F}:$ the card chosen is showing an ace.

Ans. It is known that there are $13$ spade and $4$ ace cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is a spade,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{13}{52}\text{=}\dfrac{1}{4}$.

Also, the probability that the card chosen is an ace,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Since only one card is an ace as well as spade, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{1}{52}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{1}{4}\times \dfrac{1}{13}\text{=}\dfrac{1}{52}\text{=P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are independent events.

(ii) $\mathbf{E}:$ the card chosen is showing black.

$\mathbf{F}:$ the card chosen is showing a king.

Ans. It is known that there are $26$ black and $4$ king cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is black,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{26}{52}\text{=}\dfrac{1}{2}$.

Also, the probability that the card chosen is a king,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Since only two cards are black as well as king, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{2}{52}\text{=}\dfrac{1}{26}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{13}\text{=}\dfrac{1}{26}\text{=P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are independent events

(iii) $\mathbf{E}:$ the card drawn is showing a king or queen.

$\mathbf{F}:$ the card drawn is showing a queen or jack.

Ans. It is known that there are $4$ king, $4$ queen, and $4$ jack cards in a deck of $\text{52}$ cards.

Therefore, the probability that the card chosen is a king or queen,

$\text{P}\left( \text{E} \right)\text{=}\dfrac{8}{52}\text{=}\dfrac{2}{13}$.

Also, the probability that the card chosen is a queen or jack,

$\text{P}\left( \text{F} \right)\text{=}\dfrac{8}{52}\text{=}\dfrac{2}{13}$.

Since only $4$ cards are king or a queen, or queen or a jack, so

$\text{P}\left( \text{EF} \right)\text{=}\dfrac{4}{52}\text{=}\dfrac{1}{13}$.

Now, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\text{=}\dfrac{2}{13}\times \dfrac{2}{13}\text{=}\dfrac{4}{169}\ne \dfrac{1}{13}$.

Therefore, $\text{P}\left( \text{E} \right)\cdot \text{P}\left( \text{F} \right)\ne \text{P}\left( \text{EF} \right)$.

Hence, $\text{E}$ and $\text{F}$ are not independent events.

16. Let $\mathbf{60}%$ of students in a hotel read Hindi newspaper, $\mathbf{40}%$ of students read English and $\mathbf{20}%$ of students read both Hindi and English papers. If a student is chosen randomly, then determine the probabilities for the following three cases.

(i) he reads neither Hindi nor English newspapers.

Ans. Let $\text{H}$ and $\text{E}$ denote the students read Hindi newspaper and the students read English newspaper respectively.

It is provided that,

$\text{P}\left( \text{H} \right)\text{=60 }\!\!%\!\!\text{ =}\dfrac{60}{100}$$\text{=}\dfrac{3}{5}, \text{P}\left( \text{E} \right)\text{=40 }\!\!%\!\!\text{ =}\dfrac{40}{100}\text{=}\dfrac{2}{5}, and \text{P}\left( \text{H}\cap \text{E} \right)\text{=20 }\!\!%\!\!\text{ =}\dfrac{20}{100}\text{=}\dfrac{1}{5}. The probability that a student reads neither Hindi nor English newspaper is, \text{P}{{\left( \text{H}\cup \text{E} \right)}^{\prime }}\text{=1-P}\left( \text{H}\cup \text{E} \right) \text{=1-}\left\{ \text{P}\left( \text{H} \right)+\text{P}\left( \text{E} \right)-\text{P}\left( \text{H}\cap \text{E} \right) \right\} \text{=1-}\left( \dfrac{3}{5}+\dfrac{2}{5}-\dfrac{1}{5} \right) \text{=1-}\dfrac{4}{5} \text{=}\dfrac{1}{5}. (ii) if he reads Hindi newspapers, then determine the probability of reading English newspapers. Ans. The probability that a student reads English newspaper, also reads Hindi newspaper, \text{P}\left( \text{E }\!\!|\!\!\text{ H} \right)\text{=}\dfrac{\text{P}\left( \text{E}\cap \text{H} \right)}{\text{P}\left( \text{H} \right)} \text{=}\dfrac{\dfrac{1}{5}}{\dfrac{3}{5}} \text{=}\dfrac{1}{3}. (iii) if he reads English newspapers, then determine the probability of reading Hindi newspaper. Ans. The probability that a student reads Hindi newspaper, also reads English newspaper, \text{P}\left( \text{H }\!\!|\!\!\text{ E} \right)\text{=}\dfrac{\text{P}\left( \text{H}\cap \text{E} \right)}{\text{P}\left( \text{E} \right)} \text{=}\dfrac{\dfrac{1}{5}}{\dfrac{2}{5}} =\dfrac{1}{2}. 17. Let a pair of dice is thrown. Then determine the probability of getting an even prime number on each die. Ans. The sample space contains total 36 outcomes when two dice are thrown. 2 is the only even number which is prime. Therefore, if \text{E} be the event of obtaining an even prime number on each die, then \text{E=}\left\{ \left( 2,2 \right) \right\} and so, \text{P}\left( \text{E} \right)\text{=}\dfrac{1}{36}. 18. Select the correct answer(s). \mathbf{A} and \mathbf{B} will be two independent events, if (a) \mathbf{A} and \mathbf{B} are mutually exclusive. (b) \mathbf{P}\left( \mathbf{{A}'{B}'} \right)=\left[ 1-\mathbf{P}\left( \mathbf{A} \right) \right]\left[ \mathbf{1}-\mathbf{P}\left( \mathbf{B} \right) \right] (c) \mathbf{P}\left( \mathbf{A} \right)=\mathbf{P}\left( \mathbf{B} \right) (d) \mathbf{P}\left( \mathbf{A} \right)+\mathbf{P}\left( \mathbf{B} \right)=\mathbf{1} Ans.Correct Answer: (b) It is known that two events \text{A} and \text{B} are independent, if $\text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)\times \text{P}\left( \text{B} \right)$. Now, \text{P}\left( \text{{A}'{B}'} \right)\text{=}\left[ \text{1-P}\left( \text{A} \right) \right]\left[ \text{1-P}\left( \text{B} \right) \right]$$\Rightarrow \text{P}\left( \text{{A}'}\cap \text{{B}'} \right)\text{=1-P}\left( \text{A} \right)-\text{P}\left( \text{B} \right)+\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$

$\Rightarrow 1-\text{P}\left( \text{A}\cup \text{B} \right)\text{=1-P}\left( \text{A} \right)-\text{P}\left( \text{B} \right)+\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$

$\Rightarrow \text{P}\left( \text{A}\cup \text{B} \right)\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$

$\Rightarrow \text{P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)-\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)$

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)$.

Hence, this concludes that $\text{A}$ and $\text{B}$ are independent events.

Thus, option (b) is correct.

(a)

Now, let $\text{P}\left( \text{A} \right)\text{=m}$ and $\text{P}\left( \text{B} \right)\text{=n}$, where $0<\text{m,}\,\text{n1}$.

If possible, let $\text{A}$ and $\text{B}$ are two mutually exclusive events.

Therefore, $\text{A}\cap \text{B=}\phi$.

$\Rightarrow \text{P}\left( \text{AB} \right)\text{=0}$.

But, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\text{=mn}\ne \text{0}$.

Thus, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \text{P}\left( \text{AB} \right)$,

that is, $\text{A}$ and $\text{B}$ are not independent events.

Hence, option (a) is incorrect.

(c)

Let consider two events $\text{A}$ and $\text{B}$ such that

$\text{A=}\left\{ 1,3,5 \right\}$, $\text{B=}\left\{ 2,4,6 \right\}$ and the sample space $\text{S=}\left\{ 1,2,3,4,5,6 \right\}$.

Therefore, $\text{P}\left( \text{A} \right)\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$ and $\text{P}\left( \text{B} \right)\text{=}\dfrac{3}{6}\text{=}\dfrac{1}{2}$,

that is, $\text{P}\left( \text{A} \right)\text{=P}\left( \text{B} \right)$.

Also, $\text{A}\cap \text{B=}\left\{ \phi \right\}$ and so, $\text{P}\left( \text{A}\cap \text{B} \right)\text{=0}$.

That is, $\text{P}\left( \text{AB} \right)\text{=0}$.

Now, $\text{P}\left( \text{A} \right)\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}\times \dfrac{1}{2}\text{=}\dfrac{1}{4}\ne 0$

Thus, $\text{P}\left( \text{A} \right)\cdot \text{P}\left( \text{B} \right)\ne \text{P}\left( \text{AB} \right)$,

that is, $\text{A}$ and $\text{B}$ are not independent events.

Hence, option (c) is incorrect.

(d)

Considering the previous example, it can be seen that,

$\text{P}\left( \text{A} \right)+\text{P}\left( \text{B} \right)\text{=}\dfrac{1}{2}+\dfrac{1}{2}\text{=1}$, but it has been proved that the events $\text{A}$ and $\text{B}$ are not independent.

Hence, the option (d) is incorrect.

## NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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Besides these NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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 Chapter 13 - Probability Exercises in PDF Format Exercise 13.1 17 Questions & Solutions (4 Short Answers, 13 Long Answers) Exercise 13.2 18 Questions & Solutions Exercise 13.3 14 Questions & Solutions Exercise 13.4 17 Questions & Solutions Exercise 13.5 15 Questions & Solutions

## FAQs on NCERT Solutions for Class 12 Maths Chapter 13 - Exercise

1. What are the takeaways from Class 12 Maths Chapter 13 Probability Exercise 13.2?

Class 12 Maths Chapter 13 Probability Ex 13.2 deals with the following topics:

Multiplication Theorem on Probability: As per the Multiplication Theorem of Probability, If A and B are two independent events in a probability experiment, the probability that both events occur simultaneously is:

P(A and B)=P(A)⋅P(B)

Independent Events: If E and F are two events such that the probability of occurrence of one event is not affected by the occurrence of another event, it is called independent events.

2. From where can I avail NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2?

Class 12 CBSE students can avail the free PDF of NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.2 by registering on Vedantu. Exercise-wise NCERT Solutions for Chapter 13 Probability are designed by Maths experts at Vedantu. The material is designed to help students revise all the exercises and score well in exams. NCERT Solutions for Class 12 Maths Chapter 13 are prepared as per the latest NCERT guidelines and exam pattern. The stepwise solutions to Ex 13.2 provide students with a thorough knowledge of the chapter and simple explanations to the problems. It is designed to help students in doubt clearance.

3. What are the benefits of downloading exercise-wise NCERT Solutions for Class 12 Maths Chapter 13 Probability?

CBSE Class 12 Maths Chapter 13 Probability is an important chapter from board examination point of view. The solutions, designed by Vedantu, explains the concepts related to Probability. This helps students to solve the problems on their own. It is important to solve each question of Ex 13.2 and other exercises to score well in exams. Vedantu’s NCERT Solutions for CBSE Class 12 Maths Chapter 13 are the most reliable online resource for students facing any doubt in the chapter. These solutions are provided by experts and provide doubt resolution. The experts have years of experience in the field of teaching. NCERT Solutions for Class 12 Chapter 13 Probability for Exercise 13.2 will help you score better in exams.

4. What are the key features of Vedantu’s NCERT Solutions for Class 12 Maths Chapter 13 Ex 13.2?

Vedantu’s NCERT Solutions for Class 12 Maths Chapter 13 Probability are available on its site. Students can find the free PDF of the solutions by registering on Vedantu. Following are some of the key features of NCERT Solutions for Class 12 Maths Chapter 13 offered by Vedantu:

• Designed in a stepwise manner.

• Prepared by subject matter experts.

• Available in the free PDF format.

• Provides complete coverage of the NCERT questions.

• Designed as per the latest syllabus and the exam pattern.

5. How many questions are there in Class 12 Maths NCERT Solutions Chapter 13 Probability Exercise 13.2?

There are 18 questions from the NCERT book that have been solved carefully and accurately by the subject matter experts at Vedantu for the Class 12 Maths NCERT solutions Chapter 13 Probability Exercise 13.2. NCERT Solutions provide accurate and effective methods of solving these problems so that you can have a clear picture of how to work for the exams. This way, you can understand the concepts well while solving the problems as well.

6. How many Exercises are there for Class 12 Maths Chapter 13?

There are five exercises for Class 12 Maths Chapter 13, which are 13.1, 13.2, 13.3, 13.4, 13.5. These exercises are solved in the NCERT solutions by Vedantu and all the questions have been effectively answered so that the students can follow along for exams. NCERT solutions by Vedantu will provide you with a clear picture of the concepts related to probability and help you to understand the exercise questions well. These solutions can also be downloaded free of cost.

7. Can you please brief about the Class 12 Maths Chapter 13 Probability?

Class 12 Maths Chapter 13 Probability covers topics including conditional probability, multiplication theorem, random variables, Bayes theorem, and others. Probability refers to the extent to which any given event may occur. It can be calculated by the number of favourite things to occur out of the whole number of events. Note here that the probability of any event P always has to be:

0 ≤ Pe ≤ 1.

8. What’s the underlying concept of Exercise 13.2 of Class 12 Maths?

Exercise 13.2 of Class 12 Maths Chapter 13, Probability includes the following underlying concepts and problems based on them:

• Multiplication Theorem on Probability

• Independent Events

You will be able to learn these concepts properly with the help of authentic guidance as provided by Vedantu's NCERT solutions. These solutions are available both on the app and the website. These are curated by experts in the field and are available to download for free to help students to excel in their studies.

9. Which is the best Solution book for NCERT Class 12 Maths Chapter 13?

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