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# NCERT Solutions for Class 12 Maths Chapter 13 - Probability Exercise 13.3

Last updated date: 11th Sep 2024
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## NCERT Solutions for Maths Class 12 Chapter 13 Probability Exercise 13.3 - FREE PDF Download

NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3 - Probability by Vedantu offers clear and detailed solutions to understanding complex probability topics. This exercise covers main concepts like Bayes' theorem, which plays an important role in various probability problems. Class 12 Maths NCERT Solutions are created to make complex topics easy to understand and apply.

Table of Content
1. NCERT Solutions for Maths Class 12 Chapter 13 Probability Exercise 13.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 13 Exercise 13.3 Class 12 | Vedantu
3. Formulas Used in Class 12 Chapter 13 Exercise 13.3
4. Access NCERT Solutions for Maths Class 12 Chapter 13 - Probability
4.1Exercise 13.3
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

Focus on mastering Bayes' theorem and its applications, as these are important for solving advanced probability questions. With the help of the detailed solutions, students were capable of creating a strong basis and facing the exams with confidence. Students can perform better and increase their analytical skills by working through the Class 12 Maths Syllabus.

## Glance on NCERT Solutions Maths Chapter 13 Exercise 13.3 Class 12 | Vedantu

• Class 12 Maths Chapter 13 Exercise 13.3 explains Baye’s theorem, a fundamental concept in probability which is crucial for solving problems where conditional probabilities are involved.

• Calculates the probability of an event by summing the probabilities of the event occurring under different conditions.

• Partition of a sample space: Divides the sample space into mutually exclusive and exhaustive events.

• Baye’s Theorem is used to Determine the probability of a cause given an event, using the event's probability given the cause and the cause's overall probability.

• Conditional probability measures the probability of an event given that another event has already occurred.

• Independent events are events that do not affect each other's occurrence, meaning the probability of one does not change the probability of the other.

• This article contains exercise notes, important questions, exemplar solutions, exercises, and video links for Class 12 Maths Chapter 13 Exercise 13.3 - Probability, which you can download as PDFs.

• There are 14 fully solved questions in Ex 13.3 Class 12 Maths NCERT Solutions.

## Formulas Used in Class 12 Chapter 13 Exercise 13.3

• Bayes' Theorem: $P\left ( \frac{A}{B} \right )=\frac{P\left ( \frac{B}{A}\right )\cdot P\left ( A \right ) }{P\left ( B \right )}$

• Conditional Probability: $P\left ( \frac{A}{B} \right )=\frac{P\left ( A\bigcap B \right )}{P\left ( B \right )}$

Competitive Exams after 12th Science

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

### Exercise 13.3

1. An urn contains 5 red and 5 black balls. A ball is drawn at random; its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Ans: Given the urn contains 5 red and 5 black balls.

Let a red ball be drawn in the first attempt.

Then, probability of drawing a red ball $=\dfrac{5}{10}=\dfrac{1}{2}$

Now, we add two more red balls. Therefore the urns contain 7 red and 5 black balls.

Probability of drawing the second red ball $=\dfrac{7}{12}$

Let a black ball be drawn in the first attempt.

Then, probability of drawing a black ball $=\dfrac{5}{10}=\dfrac{1}{2}$

Now, we add two more red balls. Therefore the urns contain 5 red and 7 black balls.

Probability of drawing the second red ball $=\dfrac{5}{12}$

Therefore, total probability of drawing second ball as red is

$\dfrac{1}{2}\times \dfrac{7}{12}+\dfrac{1}{2}\times \dfrac{5}{12}=\dfrac{1}{2}\left( \dfrac{7}{12}+\dfrac{5}{12} \right)$

$=\dfrac{1}{2}\times 1$

$=\dfrac{1}{2}$

2. A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Ans: Let A be the event of getting a red ball.

Let ${{E}_{1}}$ and ${{E}_{2}}$ be the events of selecting the first bag and second bag respectively.

$P({{E}_{1}})=P({{E}_{2}})=\dfrac{1}{2}$

$\Rightarrow P(A|{{E}_{1}})=$P(drawing a red ball from first bag) $=\dfrac{4}{8}=\dfrac{1}{2}$

$\Rightarrow P(A|{{E}_{2}})=$P(drawing a red ball from second bag) $=\dfrac{2}{8}=\dfrac{1}{4}$

The probability of drawing a ball from the first bag,

given that it is red, is given by $P({{E}_{1}}|A)$

By using bayes theorem, we obtain

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{\dfrac{1}{2}\times \dfrac{1}{2}}{\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{1}{2}\times \dfrac{1}{4}}$

$=\dfrac{\dfrac{1}{4}}{\dfrac{1}{4}+\dfrac{1}{8}}$

$=\dfrac{2}{3}$

3. Of the students in a college, it is known that 60% reside in hostel and 40 %are day scholars (not residing in hostel). Previous year results report that 30 %of all students who reside in hostel attain A grade and 20 %of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is hostler?

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the events where the student is a hostler and a day scholar respectively.

Let A be the event that the chosen students get a grade A.

$P({{E}_{1}})=60%=0.6$

$P({{E}_{2}})=40%=0.4$

$\Rightarrow P(A|{{E}_{1}})=$P(student getting an A grade is a hostler) $=30%=0.3$

$\Rightarrow P(A|{{E}_{2}})=$P(student getting an A grade is a day scholar) $=20%=0.2$

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}$

$=\dfrac{0.18}{0.26}$

$=\dfrac{9}{13}$

4. In answering a question on a multiple choice test, a student either knows the answer or guesses.  Let $\dfrac{3}{4}$ be the probability that he knows the answer and $\dfrac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\dfrac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly?

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the events in which the student knows the answer and the student guesses the answer respectively.

Let A be the event that the answer is correct.

$P({{E}_{1}})=\dfrac{3}{4}$

$P({{E}_{2}})=\dfrac{1}{4}$

$\Rightarrow P(A|{{E}_{1}})=$P(student answer correctly, given he knows the answer) $=1$

$\Rightarrow P(A|{{E}_{2}})=$P(student answer correctly, given that he guessed) $=\dfrac{1}{4}$

The probability that the student knows the answer, given that the answer is correct, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{\dfrac{3}{4}\times 1}{\dfrac{3}{4}\times 1+\dfrac{1}{4}\times \dfrac{1}{4}}$

$=\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}+\dfrac{1}{16}}$

$=\dfrac{3}{4}\div \dfrac{13}{16}$

$=\dfrac{12}{13}$

5. A laboratory blood test is 99%effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5%of the healthy person tested (that is, if a healthy person is tested, then, with probability0.005, the test will imply he has the disease). If 0.1percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the respective events that a person has a disease and a person has no disease.

$P({{E}_{1}})=0.1%=0.001$

Since ${{E}_{1}}$ and ${{E}_{2}}$ are events complementary to each other

$P({{E}_{2}})=1-P({{E}_{1}})=1-0.001=0.999$

Let A be the event that the answer is correct.

$\Rightarrow P(A|{{E}_{1}})=$P(result is positive given the person has disease) $=99%=0.99$

$\Rightarrow P(A|{{E}_{2}})=$P(result is positive given that the person has no disease) $=0.5%=0.005$

The probability that a person has a disease, given that his test result is positive, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}$

$=\dfrac{0.00099}{0.00099+0.004995}$

$=\dfrac{0.00099}{0.005985}$

$=\dfrac{22}{133}$

6. There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Ans: Let ${{E}_{1}}$, ${{E}_{2}}$ and ${{E}_{3}}$be the events of choosing a two headed coin, a biased coin, and an unbiased coin respectively.

Let A be the event that the coin shows heads.

$\therefore P({{E}_{1}})=P({{E}_{2}})=P({{E}_{3}})=\dfrac{1}{3}$

$\Rightarrow P(A|{{E}_{1}})=$P(coin showing heads, given that it is two-headed coin) $=1$

Probability of getting heads, given that the coin is biased $=75%$

$\Rightarrow P(A|{{E}_{2}})=$P(coin showing heads, given that the coin is biased) $=\dfrac{75}{100}=\dfrac{3}{4}$

The third coin is unbiased hence the probability of getting heads is $=\dfrac{1}{2}$

$\Rightarrow P(A|{{E}_{3}})=$P(coin showing heads, given that the coin is unbiased) $=\dfrac{1}{2}$

The probability that the coin is two-headed, given that it shows heads, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})+P({{E}_{3}})P(A|{{E}_{3}})}$

$=\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times 1+\dfrac{1}{3}\times \dfrac{3}{4}+\dfrac{1}{3}\times \dfrac{1}{2}}$

$=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}\left( 1+\dfrac{3}{4}+\dfrac{1}{2} \right)}$

$=\dfrac{1}{9}\div 4$

$=\dfrac{4}{9}$

7. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Ans: Let ${{E}_{1}}$, ${{E}_{2}}$ and ${{E}_{3}}$be the events that the driver is a scooter driver, a car driver, and a truck driver respectively.

Total number of drivers = 2000 scooter drivers + 4000 car drivers + 6000 truck drivers = 12000 drivers

$\therefore P({{E}_{1}})=\dfrac{2000}{12000}=\dfrac{1}{6}$

$P({{E}_{2}})=\dfrac{4000}{12000}=\dfrac{1}{3}$

$PP({{E}_{3}})=\dfrac{6000}{12000}=\dfrac{1}{2}$

Let A be the event that the person meets with an accident.

$\Rightarrow P(A|{{E}_{1}})=$P(scooter driver met with an accident) $=0.01=\dfrac{1}{100}$

$\Rightarrow P(A|{{E}_{2}})=$P(car driver met with an accident) $=0.03=\dfrac{3}{100}$

$\Rightarrow P(A|{{E}_{3}})=$P(truck driver met with an accident) $=0.15=\dfrac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})+P({{E}_{3}})P(A|{{E}_{3}})}$

$=\dfrac{\dfrac{1}{6}\times \dfrac{1}{100}}{\dfrac{1}{6}\times \dfrac{1}{100}+\dfrac{1}{3}\times \dfrac{3}{100}+\dfrac{1}{2}\times \dfrac{15}{100}}$

$=\dfrac{\dfrac{1}{6}\times \dfrac{1}{100}}{\dfrac{1}{100}\left( \dfrac{1}{6}+1+\dfrac{15}{2} \right)}$

$=\dfrac{1}{6}\div \dfrac{104}{12}$

$=\dfrac{1}{52}$

8. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Future, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen random from this and is found to be defective. What is the probability that was produced by machine B?

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the respective events of items produced by machines A and B.

$\therefore P({{E}_{1}})=60%=\dfrac{3}{5}$

$P({{E}_{2}})=40%=\dfrac{2}{5}$

Let A be the event that the produced items were found to be defective.

$\Rightarrow P(A|{{E}_{1}})=$P(product is defective, given that machine A produced) $=2%=\dfrac{2}{100}$

$\Rightarrow P(A|{{E}_{2}})=$P(product is defective, given that machine B produced) $=1%=\dfrac{1}{100}$

The probability that the randomly selected items was from B, given that it is defective, is given by $P({{E}_{2}}|A)$

By using Bayes theorem, we get,

$P({{E}_{2}}|A)=\dfrac{P({{E}_{2}})P(A|{{E}_{2}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{\dfrac{2}{5}\times \dfrac{1}{100}}{\dfrac{3}{5}\times \dfrac{2}{100}+\dfrac{2}{5}\times \dfrac{1}{100}}$

$=\dfrac{\dfrac{2}{500}}{\dfrac{6}{500}+\dfrac{2}{500}}$

$=\dfrac{2}{8}$

$=\dfrac{1}{4}$

9. Two groups are competing for the position on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the respective events in which the first group and the second group win the competition.

$\therefore P({{E}_{1}})=0.6$

$P({{E}_{2}})=0.4$

Let A be the event of introducing a new product.

$\Rightarrow P(A|{{E}_{1}})=$P(introducing a new product, given that first group wins) $=0.7$

$\Rightarrow P(A|{{E}_{2}})=$P(introducing a new product, given that second group wins) $=0.3$

The probability that the new product is introduced by the second group, is given by $P({{E}_{2}}|A)$

By using Bayes theorem, we get,

$P({{E}_{2}}|A)=\dfrac{P({{E}_{2}})P(A|{{E}_{2}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3}$

$=\dfrac{0.12}{0.42+0.12}$

$=\dfrac{0.12}{0.54}$

$=\dfrac{12}{54}$

$=\dfrac{2}{9}$

10. Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Ans: Let ${{E}_{1}}$ be the event that the outcome on the die is 5 or 6 and ${{E}_{2}}$ be the event that the outcome on the die is 1,2,3, or 4.

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

$P({{E}_{2}})=\dfrac{4}{6}=\dfrac{2}{3}$

Let A be the event of getting exactly one head.

$P({{E}_{1}})$ is same as tossing the coin 3 times, similarly $P({{E}_{2}})$ is same as tossing the coin exactly once

$\Rightarrow P(A|{{E}_{1}})=$P(getting exactly one head, given the coin is tossed 3 times) $=\dfrac{3}{8}$(TTH,THT,HTT)

$\Rightarrow P(A|{{E}_{2}})=$P(getting exactly one head, given the coin is tossed only once) $=\dfrac{1}{2}$

The probability that the girl threw 1,2,3, or 4 with die, if she obtained exactly one head, is given by $P({{E}_{2}}|A)$

By using Bayes theorem, we get,

$P({{E}_{2}}|A)=\dfrac{P({{E}_{2}})P(A|{{E}_{2}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{\dfrac{2}{3}\times \dfrac{1}{2}}{\dfrac{1}{3}\times \dfrac{3}{8}+\dfrac{2}{3}\times \dfrac{1}{2}}$

$=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}\left( \dfrac{3}{8}+1 \right)}$

$=\dfrac{1}{11}\div 8$

$=\dfrac{8}{11}$

11. A manufacture has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30%of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?

Ans: Let ${{E}_{1}}$, ${{E}_{2}}$ and ${{E}_{3}}$be the events that the time consumed by machine operator A, B and C on the job.

$\therefore P({{E}_{1}})=50%=\dfrac{50}{100}=\dfrac{1}{2}$

$P({{E}_{2}})=30%=\dfrac{30}{100}=\dfrac{3}{10}$

$P({{E}_{3}})=20%=\dfrac{20}{100}=\dfrac{1}{5}$

Let A be the event of producing defective items.

$\Rightarrow P(A|{{E}_{1}})=$P(item is defective, given that it is produced by A) $=1%=\dfrac{1}{100}$

$\Rightarrow P(A|{{E}_{2}})=$P(item is defective, given that it is produced by B) $=5%=\dfrac{5}{100}$

$\Rightarrow P(A|{{E}_{3}})=$P(item is defective, given that it is produced by C) $=7%=\dfrac{7}{100}$

The probability that the defective item was produced by A is given by, is given by $P({{E}_{1}}|A)$

By using Bayes theorem, we get,

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})+P({{E}_{3}})P(A|{{E}_{3}})}$

$=\dfrac{\dfrac{1}{2}\times \dfrac{1}{100}}{\dfrac{1}{2}\times \dfrac{1}{100}+\dfrac{3}{10}\times \dfrac{5}{100}+\dfrac{1}{5}\times \dfrac{7}{100}}$

$=\dfrac{\dfrac{1}{100}\times \dfrac{1}{2}}{\dfrac{1}{100}\left( \dfrac{1}{2}+\dfrac{3}{2}+\dfrac{7}{2} \right)}$

$=\dfrac{1}{2}\div \dfrac{17}{5}$

$=\dfrac{5}{34}$

12. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the respective events of choosing a diamond card and a card which is not diamond.

Let A be the event that denote the lost card

$\therefore P({{E}_{1}})=\dfrac{13}{52}=\dfrac{1}{4}$

$P({{E}_{2}})=\dfrac{39}{52}=\dfrac{3}{4}$

When the card lost is a diamond card, then there are 12 diamond cards out of 51 cards.

From 12 diamond cards 2 cards can be drawn in ${}^{12}{{C}_{2}}$ ways.

Therefore the probability of getting two diamond cards, when one diamond card is lost, is given by $P(A|{{E}_{1}})$.

$P(A|{{E}_{1}})=\dfrac{{}^{12}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\dfrac{12!}{2!\times 10!}\times \dfrac{2!\times 49!}{5!}=\dfrac{22}{425}$

When the card lost is not a diamond card, then there are 13 diamond cards out of 51 cards.

From 13 diamond cards 2 cards can be drawn in ${}^{13}{{C}_{2}}$ ways.

Therefore the probability of getting two diamond cards, when the card lost is not a diamond, is given by $P(A|{{E}_{2}})$.

$P(A|{{E}_{2}})=\dfrac{{}^{13}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\dfrac{13!}{2!\times 11!}\times \dfrac{2!\times 49!}{5!}=\dfrac{26}{425}$

The probability that the lost card is diamond is given by $P({{E}_{1}}|A)$.

$P({{E}_{1}}|A)=\dfrac{P({{E}_{1}})P(A|{{E}_{1}})}{P({{E}_{1}})P(A|{{E}_{1}})+P({{E}_{2}})P(A|{{E}_{2}})}$

$=\dfrac{\dfrac{1}{4}\times \dfrac{22}{425}}{\dfrac{1}{4}\times \dfrac{22}{425}+\dfrac{3}{4}\times \dfrac{26}{425}}$

$=\dfrac{11}{2}\div 25$

$=\dfrac{11}{50}$

13. Probability that A speaks truth is $\dfrac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is

A) $\dfrac{4}{5}$

B) $\dfrac{1}{2}$

C) $\dfrac{1}{5}$

D) $\dfrac{2}{5}$

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the events such that

${{E}_{1}}$ : A speaks truth

${{E}_{2}}$ : A speaks false

Let X be the event that a head appears.

$P({{E}_{1}})=\dfrac{4}{5}$

$\therefore P({{E}_{2}})=1-P({{E}_{1}})=1-\dfrac{4}{5}=\dfrac{1}{5}$

If a coin is tossed, then it may result in either head(H) or tail(T).

The probability of getting a head is $\dfrac{1}{2}$ whether A speaks truth or not.

$\therefore P(X|{{E}_{1}})=P(X|{{E}_{2}})=\dfrac{1}{2}$

The probability that there actually a head is given by $P({{E}_{1}}|X)$.

$P({{E}_{1}}|X)=\dfrac{P({{E}_{1}})P(X|{{E}_{1}})}{P({{E}_{1}})P(X|{{E}_{1}})+P({{E}_{2}})P(X|{{E}_{2}})}$

$=\dfrac{\dfrac{4}{5}\times \dfrac{1}{2}}{\dfrac{4}{5}\times \dfrac{1}{2}+\dfrac{1}{5}\times \dfrac{1}{2}}$

$=\dfrac{\dfrac{4}{5}\times \dfrac{1}{2}}{\dfrac{1}{2}\left( \dfrac{4}{5}+\dfrac{1}{5} \right)}$

$=\dfrac{4}{5}\div 1$

$=\dfrac{4}{5}$

14. If A and B are two events such that $A\subset B$ and $P(B)\ne 0$, then which of the following is correct?

A) $P(A|B)=\dfrac{P(B)}{P(A)}$

B) $P(A|B)<P(A)$

C) $P(A|B)\ge P(A)$

D) None of these

Ans: If $A\subset B$ then $A\cap B=A$ ,

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

Also, $P(A)<P(B)$

Consider $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A)}{P(B)}\ne \dfrac{P(B)}{P(A)}$     - (Eq 1)

Consider $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(A)}{P(B)}$              - (Eq 2)

It is known that, $P(B)\le 1.$

$\Rightarrow \dfrac{1}{P(B)}\ge$

$\Rightarrow \dfrac{P(A)}{P(B)}\ge P(A)$

From (2), we obtain

$\Rightarrow P(A|B)\ge P(A)$          - (Eq 3)

$\therefore P(A|B)$ is not less than $P(A)$.

Thus, from (3), it can be concluded that the relation given in alternate C is correct.

## Conclusion

NCERT Solutions for Maths Exercise 13.3 Class 12 Chapter 13 - Probability by Vedantu help you understand Bayes' theorem clearly. Focus on how to use this theorem to solve tricky probability questions. The step-by-step solutions make learning easier. Practicing these problems will boost your understanding and help you do better in exams.

## Class 12 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 17 Questions & Solutions Exercise 13.2 18 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 13.3

1. What is conditional probability in Ex 13.3 Class 12 Maths NCERT Solutions?

Ex 13.3 Class 12 Maths NCERT Solutions, Conditional probability of an event is defined as the likelihood of an event or outcome which is based on the occurrence of the previous event or outcome. For example, the conditional probability of event B is based on the knowledge that event A has already occurred. This probability is usually written as P (B/A) which is read as the probability of B given A. In case both events are not independent then the probability of the intersection of both events is equal to the product of the probability of a previous event and conditional probability.

P(A&B) = P(A)P(B/A).

2. What are the conditions for Bernoulli trials in Ex 13.3 Class 12 Maths NCERT Solutions?

Bernoulli trials are the trials of random experiments if they satisfy the following conditions:

• The number of trials should be finite.

• All the trials done should be independent.

• Each trial must have only two outcomes. (i.e, success or failure).

• The probability of success should be the same for each trial.

3. What are the topics covered in Ex 13.3 Class 12 Maths NCERT Solutions?

The Class 12 Ex 13.3 starts with the introduction of probability where the discovery of probability is discussed. Later, we get to know what conditional probability is and what are its properties. This is followed by the multiplication theorem on probability which leads to the discussion of independent events. We learn about Bayes theorem which is divided into smaller sections like partition of sample space and theorem of total probability. The chapter ends with the probability distribution of random variables, mean of random variables, variance of random variables, Bernoulli's Trials and Binomial Distribution.

4. What are the benefits of solving NCERT solutions Class 12 Ex 13.3 by Vedantu?

NCERT Solutions Class 12 Ex 13.3 by Vedantu is prepared strictly according to the latest CBSE syllabus and it is in the most comprehensible form of the chapter containing all the important points and formulas. The subject matter experts prepared the NCERT solutions intending to make the learning process more convenient and fun.

5. Explain the partition of a sample space in Class 12 Ex 13.3.

A given group of events (denoted by E1, E2, ..., En) is considered as representing the partition of a sample space (denoted by S) when they are disjoint pairwise, and exhaustive while having nonzero probabilities. To understand better the partition of a sample space, you should download the NCERT solutions by Vedantu for Class 12 Maths that are sure to help you clear your concepts easily. The NCERT Solutions Class 12 Ex 13.3 are available on the Vedantu website and the app.

6. What is the theorem of Total Probability in Ex13.3 Class 12?

The theorem Ex13.3 Class 12 of total probability explains that when A1, A2, …… A denotes the partition of a given sample space(S) so that the probability of all the events is nonzero, in that case, the probability of the occurrence of event ‘E’ in this sample space can be obtained by summing up the probabilities of all other events in this sample space.

7. How many questions are included in Ex13.3 Class 12 Maths?

A total of 14 questions have been included in Ex13.3 Class 12 Maths Chapter 13 Probability. The problems to be solved here focus majorly on providing proper knowledge of the practical application of the concepts and theorems discussed above. If you require any further help regarding solving these questions, it's a good idea to download the NCERT solutions by Vedantu for free to help you properly understand and effectively solve all the questions both in the exercise as well as the exams.

8. What are the real-time applications of Probability in Class 12 Maths 13.3?

Class 12 Maths 13.3, Probability refers to the study of the chance that an event will occur or not. It is used practically in several cases without knowing. All things from the toss before a game of cricket to whether we'll study at a University are checked using probability. Some common Practical applications of Probability are:

• Weather forecasting

• Medical Decisions

• Politics

• Sports and games

• Choosing Insurance Policies

• Flipping a Coin or Dice.

9. What are the Independent event and Bayes theorem important formulas that come in Class 12 Maths 13.3?

• Independent events:

2 events E and F are independent, if

P(FIE) = P (F) provided P (E) = 0

OR

P (EIF) = P (E) provided P (F) = 0

OR

P(EF) = P(E). P (F)

• Bayes Theorem

E₁, E₂,..., En can represent the partition of sample space S when

En E, = 0, i is not equal to j, i, j = 1, 2, 3,..., n

OR

E1 E2 E1= S

OR

P(E) > 0 for all i = 1, 2, ..., n.

10. What is Bayes' theorem and its role in Class 12 Maths 13.3?

Baye’s theorem is used to determine the probability of an event based on prior knowledge of related conditions. It is central to Class 12 Maths 13.3 for solving advanced probability problems.

11. How do NCERT Solutions for Class 12 Maths Ex 13.3 help with Bayes' theorem?

The NCERT Solutions for Class 12 Maths Ex 13.3 provide detailed explanations and examples, making it easier to understand and apply Bayes' theorem in various probability scenarios.

12. What should I concentrate on when working on Class 12 Maths Ex 13.3?

Focus on learning Bayes' theorem and the conditional probability formula. Regular practice will help you understand these concepts and enhance your problem-solving abilities in Class 12 Maths Ex 13.3.