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# NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.3) Exercise 13.3

## NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.3) Exercise 13.3

Last updated date: 28th Jan 2023
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Free PDF download of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.3 (Ex 13.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 13 Probability Exercise 13.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

Exercise 13.3 is based on the following topics:

• Bayes’ Theorem

• Partition of a sample space

• The theorem of total probability

Solve all of the sums covered in this exercise to have a thorough understanding of the concepts and topics discussed in the chapter on Probability

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## Access NCERT Solutions for Class 12 Maths 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,

\begin{align} 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.

Importance of Probability

Tossing a coin will result in either a head or a tail, and the outcome can be easily predicted. But what happens if you toss two coins at once? The end result could be a mix of head and tail. As the correct answer cannot be obtained in this case, only one may predict the probability of a result. Probability is the name given to this type of prediction. Sports, weather reports, blood tests, statics, and many more aspects of daily life use probability.

### NCERT Solution Class 12 Maths of Chapter 13 All Exercises

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

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

Opting for the NCERT solutions for Ex 13.3 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 13.3 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.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Subject Probability textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 13 Exercise 13.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.3, 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.

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.3) Exercise 13.3

1. What is conditional probability?

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 an event B is based on the knowledge that an event A has already occured. This probability is usually written as P (B/A) which is read as probability of B given A. In case both the events are not independent then the probability of intersection of both the events is equal to the product of the probability of previous event and conditional probability.

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

2. What are the conditions for bernoulli trials?

Bernoulli trials are the trials of random experiment 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 class 12 maths chapter 13?

The chapter 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 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 by Vedantu?

NCERT solutions 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 with an aim to make the learning process more convenient and fun.

5. Explain the partition of a sample space.

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, it is advisable that you download the NCERT solutions by Vedantu for Class 12 Maths that are sure to help you in clearing your concepts easily. The NCERT Solutions are available on the Vedantu website and the app.

6. What is the theorem of Total Probability?

The theorem 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 Exercise 13.3 Of Class 12 Maths?

A total of 14 questions have been included in Exercise 13.3 of 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?

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 Chapter 13?

• 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.