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

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## NCERT Solutions for Maths Chapter 13 Probability Exercise 13.2 Class 12 - FREE PDF Download

In Exercise 13.2 Class 12, students will work on problems that help them understand and apply the principles of conditional probability and the multiplication theorem of probability. This exercise includes questions that require logical reasoning and a good grasp of basic probability concepts. By practicing problems in Ex 13.2 Class 12, students will develop a solid understanding of calculating probabilities in complex situations, which is vital for their academic and competitive exam preparations. Students can access the revised Class 12 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Maths Chapter 13 Probability Exercise 13.2 Class 12 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 13 Exercise 13.2 Class 12 | Vedantu
3. Formulas Used in Class 12 Chapter 13 Exercise 13.2
4. Access NCERT Solutions for Maths Class 12 Chapter 13 - Probability
4.1Exercise 13.2
5. Class 12 Maths Chapter 13: Exercises Breakdown
6. CBSE Class 12 Maths Chapter 13 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

These solutions are aligned with the updated Class 12 Maths Syllabus guidelines for Class 12, ensuring students are well-prepared for exams. Class 12 Chapter 13 Maths Exercise 13.2 Questions and Answers PDF provides accurate answers to textbook questions and assists in effective exam preparation and better performance.

## Glance on NCERT Solutions Maths Chapter 13 Exercise 13.2 Class 12 | Vedantu

• NCERT Solution for Class 12 Maths Chapter 13 Exercise 13.2 covers topics such as Multiplication Theorem on Probability and Independent Events.

• Multiplication Theorem on Probability - This theorem helps us calculate the probability of two independent events happening together in a single experiment.

• Imagine two events, A and B. Their independence means the occurrence (or non-occurrence) of one event doesn't affect the probability of the other event happening.

• The theorem states that the probability of both A and B occurring (denoted as P(A∩B)) is the product of the individual probabilities of A (P(A)) and B (P(B)).

• Independent Events - For the Multiplication Theorem to apply, events A and B need to be independent.

• Knowing whether event A happened doesn't give you any additional information about the likelihood of event B happening (and vice versa).

• For example, tossing a coin and picking a card from a deck are independent events. The outcome of the coin toss (heads or tails) doesn't influence what card you will pick from the deck.

• Ex 13.2 Class 12 covers 18 fully solved questions and solutions.

## Formulas Used in Class 12 Chapter 13 Exercise 13.2

• P(A∩B) = P(A) * P(B)

Competitive Exams after 12th Science

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

### Exercise 13.2

1. If $\mathbf{P}\left( \mathbf{A} \right)=\dfrac{\mathbf{3}}{\mathbf{5}}$ and $\mathbf{P}\left( \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$, find P $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)$ if $\mathbf{A}$ 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. Two cards are drawn at random and without replacement from a pack of $\mathbf{52}$ playing cards. Find 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 $\text{P}\left( \text{A} \right)$ denote 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. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $\mathbf{15}$ oranges out of which $\mathbf{12}$ are good and $\mathbf{3}$ are bad ones will be approved 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 are 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. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

$\mathbf{A}:$ 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. A die marked $\mathbf{1,2,3}$ in red and $\mathbf{4},\mathbf{5},\mathbf{6}$ in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

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

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

Ans. The sample space when a die is 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 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. Let $\mathbf{E}$ and $\mathbf{F}$ be events with $\mathbf{P}\left( \mathbf{E} \right)=\dfrac{\mathbf{3}}{\mathbf{5}},\mathbf{P}\left( \mathbf{F} \right)=\dfrac{\mathbf{3}}{\mathbf{10}}$ and $\mathbf{P}\left( \mathbf{E}\cap \mathbf{F} \right)=\dfrac{\mathbf{1}}{\mathbf{5}}$. Are $\mathbf{E}$ and $\mathbf{F}$ 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. Given that the events $\mathbf{A}$ and $\mathbf{B}$ are 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 $\mathbf{P}\left( \mathbf{B} \right)=\mathbf{p}$. Find p if they are (i) mutually exclusive (ii) independent.

Ans.

(i) 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)  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 $\mathbf{A}$ and $\mathbf{B}$ be independent events with $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}.\mathbf{3},\,\,\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{4}$. Find

(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. If $\mathbf{A}$ and $\mathbf{B}$ are two events 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 $\mathbf{P}\left( \mathbf{A}\cap \mathbf{B} \right)=\dfrac{\mathbf{1}}{\mathbf{8}}$. find P (not A and not B).

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. Events $\mathbf{A}$ and $\mathbf{B}$ are 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 $\mathbf{P}\left( \mathbf{{A}'}\cap \mathbf{{B}'} \right)=\dfrac{\mathbf{1}}{\mathbf{4}}$. State whether $\mathbf{A}$ 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.

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find (i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B)

11. Given two independent events $\mathbf{A}$ and $\mathbf{B}$ such that $\mathbf{P}\left( \mathbf{A} \right)=\mathbf{0}\mathbf{.3},\mathbf{P}\left( \mathbf{B} \right)=\mathbf{0}.\mathbf{6}$. Find

(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. A die is tossed thrice. Find the probability of getting an odd number at least once

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. Two balls are drawn at random with replacement from a box containing $\mathbf{10}$ black and $\mathbf{8}$ red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red.

Ans.

(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 the second draw $\text{=}\dfrac{8}{18}\text{=}\dfrac{4}{9}$.

Thus, the probability of having the black and red balls in the 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 the second draw $\text{=}\dfrac{10}{18}\text{=}\dfrac{5}{9}$.

Thus, the probability of having the red and black balls in the 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 other 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. Probability of solving specific problem independently by $\mathbf{A}$ and $\mathbf{B}$ are $\dfrac{\mathbf{1}}{\mathbf{2}}$ and $\dfrac{\mathbf{1}}{\mathbf{3}}$ respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved

(ii) exactly one of them solves the problem.

Ans.

(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. One card is drawn at random from a well-shuffled deck of $\mathbf{52}$ cards. In which of the following cases are the events E and F independent?

(i) E : ‘The card drawn is a spade’

F : ‘the card drawn is an ace’

(ii) E : ‘the card drawn is black’

F : ‘the card drawn is a king’

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

Ans:

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

$\mathbf{F}:$ the card drawn is 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 a 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 drawn is black.

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

$\mathbf{F}:$ the card drawn is 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 a 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. In a hostel, $\mathbf{60}%$of the students read Hindi newspaper, $\mathbf{40}%$ read English newspaper and $\mathbf{20}%$ read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Ans:

(a) Find the probability that she reads neither Hindi nor English newspapers.

Ans. Let $\text{H}$ and $\text{E}$ denote the students who read Hindi newspapers and the students who read English newspapers 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}. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. Ans. The probability that a student who reads an English newspaper also reads Hindi newspapers, \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}. (c) If she reads English newspaper, find the probability that she reads Hindi newspapers. Ans. The probability that a student reads Hindi newspaper also reads English newspapers, \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. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is (A) 0 (B) \frac{1}{3} (C) \frac{1}{12} (D) \frac{1}{36} Ans. The sample space contains a total of 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. Two events \mathbf{A} and \mathbf{B} will be independent, 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.

## Conclusion

This exercise provides a variety of problems that enhance your analytical and problem-solving skills. By understanding these concepts is not only essential for your exams but also for applying mathematical reasoning in real-life situations. By thoroughly practicing the problems in Class 12 13.2 Exercise, you are better prepared to tackle more advanced topics in probability and succeed in your academic journey.

## Class 12 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 17 Questions & Solutions Exercise 13.3 14 Questions & Solutions Miscellaneous Exercise 13 Questions & Soluitions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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

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

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?

Class 12 Maths Chapter 13 deals with Probability. It is a very important chapter from the perspective of the board exam as well as for competitive exams, such as NEET and JEE. Therefore it's critical to understand it thoroughly. The NCERT solutions offered by Vedantu are your best choice because they have been carefully curated to ensure accuracy and effectiveness for these exams. Similarly, it is also simplified to guarantee that students understand the concepts and theorems clearly.

10. Can you give an example of conditional probability?

Suppose we have a deck of 52 cards, and we draw one card. The probability of drawing an Ace is P(Ace) = 4/52. If we know that the drawn card is a spade, then the conditional probability of the card being an Ace given that it is a spade is P(Ace|Spade) = 1/13. Also understand more about Multiplication Theorem on Probability and Independent Events in Class 12 13.2 Execise.