NCERT Solutions for Exercise 13.1 Class 12 Maths Chapter 13 - Probability

NCERT solutions for exercise 13.1 Class 12 Maths chapter 13 is very important for the students who wish to perform well in the 12th board exam. NCERT book exercise has covered topics like the probability of events, conditional probability, properties of conditional property, etc. Probability has a lot of real-life applications like forecasting the weather, sports, online gaming, insurance, politics lottery tickets, playing cards, statistics, etc. There area total of 17 questions covered in the exercise 13.1 Class 12 Maths which are mostly related to conditional probability. You are advised to solve all these NCERT syllabus problems on your own. You can go through NCERT solutions for Class 12 Maths chapter 13 exercise 13.1 If you are facing difficulties while solving these problems.

Also, see

  • Probability Exercise 13.2
  • Probability Exercise 13.3
  • Probability Exercise 13.4
  • Probability Exercise 13.5
  • Probability Miscellaneous Exercise

Probability Class 12 Chapter 13-Exercise: 13.1

Question:1

Given that E and F are events such that P(E)=0.6,P(F)=0.3 and p(E\cap F)=0.2, find P(E\mid F) and P(F\mid E)

Answer:

It is given that P(E)=0.6,P(F)=0.3 and p(E\cap F)=0.2,

P ( E | F ) = \frac{p(E\cap F)}{P(F)}= \frac{0.2}{0.3}= \frac{2}{3}

P ( F | E ) = \frac{p(E\cap F)}{P(E)}= \frac{0.2}{0.6}= \frac{1}{3}

Question:2 Compute P(A\mid B), if P(B)=0.5 and P(A\cap B)=0.32

Answer:

It is given that P(B)=0.5 and P(A\cap B)=0.32

P ( A | B ) = \frac{p(A\cap B)}{P(B)}= \frac{0.32}{0.5}=0.64

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(i) P(A\cap B)

Answer:

It is given that P(A)=0.8, P(B)=0.5 and P(B|A)=0.4

P ( B | A ) = \frac{p(A\cap B)}{P(A)}

0.4 = \frac{p(A\cap B)} {0.8}

p(A\cap B) = 0.4 \times 0.8

p(A\cap B) = 0.32

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(ii) P(A\mid B)

Answer:

It is given that P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4,

P(A\cap B)=0.32

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

P ( A | B ) = \frac{0.32}{0.5}

P ( A | B ) = \frac{32}{50}=0.64

Question:3 If P(A)=0.8,P(B)=0.5 and P(B\mid A)=0.4, find

(iii) P(A\cup B)

Answer:

It is given that P(A)=0.8,P(B)=0.5

P(A\cap B)=0.32

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

P(A\cup B)=0.8+0.5-0.32

P(A\cup B)=1.3-0.32

P(A\cup B)=0.98

Question:4 Evaluate P(A\cup B), if 2P(A)=P(B)=\frac{5}{13} and P(A\mid B)=\frac{2}{5}

Answer:

Given in the question 2P(A)=P(B)=\frac{5}{13} and P(A\mid B)=\frac{2}{5}

We know that:

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

\frac{2}{5} = \frac{p(A\cap B)}{\frac{5}{13}}

\frac{2\times 5}{5\times 13} = p(A\cap B)

p(A\cap B)=\frac{2}{ 13}

Use, p(A\cup B)=p(A)+p(B)-p(A\cap B)

p(A\cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}

p(A\cup B)=\frac{11}{26}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}. , find

(i) P(A\cap B)

Answer:

Given in the question

P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}.

By using formula:

p(A\cup B)=p(A)+p(B)-p(A\cap B)

\frac{7}{11}=\frac{6}{11}+\frac{5}{11}-p(A\cap B)

p(A\cap B)=\frac{11}{11}-\frac{7}{11}

p(A\cap B)=\frac{4}{11}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}, find

(ii) P(A\mid B)

Answer:

It is given that - P(A)=\frac{6}{11},P(B)=\frac{5}{11}

p(A\cap B)=\frac{4}{11}

We know that:

P ( A | B ) = \frac{p(A\cap B)}{P(B)}

P ( A | B ) = \frac{\frac{4}{11}}{\frac{5}{11}}

P ( A | B ) = \frac{4}{5}

Question:5 If P(A)=\frac{6}{11},P(B)=\frac{5}{11} and P(A\cup B)=\frac{7}{11}, find

(iii) P(B\mid A)

Answer:

Given in the question-

P(A)=\frac{6}{11},P(B)=\frac{5}{11} and p(A\cap B)=\frac{4}{11}

Use formula

P ( B | A ) = \frac{p(A\cap B)}{P(A)}

P ( B | A ) = \frac{\frac{4}{11}}{\frac{6}{11}}

P ( B | A ) = \frac{4}{6}=\frac{2}{3}

Question:6 A coin is tossed three times, where

(i)E : head on third toss ,F : heads on first two tosses

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E: head on third toss, F: heads on first two tosses

E=\left \{ {HHH},{TTH},{HTH},{THH} \right \}

F=\left \{ {HHH},{HHT} \right \}

E\cap F =HHH

P(F)=\frac{2}{8}=\frac{1}{4}

P(E\cap F)=\frac{1}{8}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{8}}{\frac{1}{4}}

P(E| F)=\frac{4}{8}=\frac{1}{2}

Question:6 A coin is tossed three times, where

(ii)E : at least two heads ,F : at most two heads

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E : at least two heads , F : at most two heads

E=\left \{ {HHH},{HTH},{THH},{HHT}\right \}=4

F=\left \{ {HTH},{HHT},{THH},{TTT},{HTT},{TTH},{THT} \right \}=7

E\cap F =\left \{ {HTH},THH,HHT\right \}=3

P(F)=\frac{7}{8}

P(E\cap F)=\frac{3}{8}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{3}{8}}{\frac{7}{8}}

P(E| F)=\frac{3}{7}

Question:6 A coin is tossed three times, where

(iii)E : at most two tails ,F : at least one tail

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =2^{3}=8

According to question

E: at most two tails, F: at least one tail

E=\left \{ {HHH},{TTH},{HTH},{THH},THT,HTT,HHT \right \}=7

F=\left \{ {TTT},{TTH},{HTH},{THH},THT,HTT,HHT \right \}=7

E\cap F=\left \{ {TTH},{HTH},{THH},THT,HTT,HHT \right \}=6

P(F)=\frac{7}{8}

P(E\cap F)=\frac{6}{8}=\frac{3}{4}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{3}{4}}{\frac{7}{8}}

P(E| F)=\frac{6}{7}

Question:7 Two coins are tossed once, where

(i) E : tail appears on one coin, F : one coin shows head

Answer:

E : tail appears on one coin, F : one coin shows head

Total outcomes =4

E=\left \{ HT,TH \right \}=2

F=\left \{ HT,TH \right \}=2

E\cap F=\left \{ HT,TH \right \}=2

P(F)=\frac{2}{4}=\frac{1}{2}

P(E\cap F)=\frac{2}{4}=\frac{1}{2}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{2}}{\frac{1}{2}}

P(E| F)=1

Question:7 Two coins are tossed once, where

(ii)E : no tail appears,F : no head appears

Answer:

E : no tail appears, F : no head appears

Total outcomes =4

\\E={HH}\\F={TT}

E\cap F=\phi

n(E\cap F)=0

P(F)=1

P(E\cap F)=\frac{0}{4}=0

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{0}{1}=0

Question:8 A die is thrown three times,

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Answer:

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Total outcomes =6^{3}=216

E=\left \{ 114,124,134,144,154,164,214,224,234,244,254,264,314,324,334,344,354,364,414,424,434,454,464,514,524,534,544,554,564,614,624,634,644,654,664 \right \} n(E)=36

F=\left \{ 651,652,653,654,655,656 \right \}

n(F)=6

E\cap F=\left \{ 654 \right \}

n(E\cap F)=1

P(E\cap F)=\frac{1}{216}

P( F)=\frac{6}{216}=\frac{1}{36}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{216}}{\frac{1}{36}}

P(E| F)=\frac{1}{6}

Question:9 Mother, father and son line up at random for a family picture

E : son on one end, F : father in middle

Answer:

E : son on one end, F : father in middle

Total outcomes =3!=3\times 2=6

Let S be son, M be mother and F be father.

Then we have,

E= \left \{ SMF,SFM,FMS,MFS \right \}

n(E)=4

F=\left \{ SFM,MFS \right \}

n(F)=2

E\cap F=\left \{ SFM,MFS \right \}

n(E\cap F)=2

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

P(E\cap F)=\frac{2}{6}=\frac{1}{3}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{3}}{\frac{1}{3}}

P(E| F)=1

Question:10 A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5.

Answer:

A black and a red dice are rolled.

Total outcomes =6^{2}=36

Let the A be event obtaining a sum greater than 9 and B be a event that the black die resulted in a 5.

A=\left \{ 46,55,56,64,65,66 \right \}

n(A)=6

B=\left \{ 51,52,53,54,55,56 \right \}

n(B)=6

A\cap B=\left \{ 55,56 \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P( B)=\frac{6}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{6}{36}}=\frac{2}{6}=\frac{1}{3}

Question:10 A black and a red dice are rolled.

(b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4 .

Answer:

A black and a red dice are rolled.

Total outcomes =6^{2}=36

Let the A be event obtaining a sum 8 and B be a event thatthat the red die resulted in a number less than 4 .

A=\left \{ 26,35,53,44,62, \right \}

n(A)=5

Red dice is rolled after black dice.

B=\left \{ 11,12,13,21,22,23,31,32,33,41,42,43,51,52,53,61,62,63 \right \}

n(B)=18

A\cap B=\left \{ 53,62 \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P( B)=\frac{18}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{18}{36}}=\frac{2}{18}=\frac{1}{9}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(i) P(E\mid F) and P(F\mid E)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \}

E\cap F=\left \{ 3\right \}

n(E\cap F)=1

n( F)=2

n( E)=3

P( E)=\frac{3}{6} P( F)=\frac{2}{6} and P(E\cap F)=\frac{1}{6}

P(E| F)=\frac{P(E\cap F)}{P(F)}

P(E| F)=\frac{\frac{1}{6}}{\frac{2}{6}}

P(E| F)=\frac{1}{2}

P(F| E)=\frac{P(F\cap E)}{P(E)}

P(F| E)=\frac{\frac{1}{6}}{\frac{3}{6}}

P(F| E)=\frac{1}{3}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F=\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(ii) P(E\mid G) and P(G\mid E)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \} , G=\left \{ 2,3,4,5 \right \}

E\cap G=\left \{ 3,5\right \}

n(E\cap G)=2

n( G)=4

n( E)=3

P( E)=\frac{3}{6} P( G)=\frac{4}{6} P(E\cap F)=\frac{2}{6}

P(E| G)=\frac{P(E\cap G)}{P(G)}

P(E| G)=\frac{\frac{2}{6}}{\frac{4}{6}}

P(E| G)=\frac{2}{4}=\frac{1}{2}

P(G| E)=\frac{P(G\cap E)}{P(E)}

P(G| E)=\frac{\frac{2}{6}}{\frac{3}{6}}

P(G| E)=\frac{2}{3}

Question:11 A fair die is rolled. Consider events E=\left \{ 1,3,5 \right \},F=\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \} Find

(iii) P((E\cup F)\mid G) and P((E\cap F)\mid G)

Answer:

A fair die is rolled.

Total oucomes =\left \{ 1,2,3,4,5,6 \right \}=6

E=\left \{ 1,3,5 \right \},F\left \{ 2,3 \right \} and G=\left \{ 2,3,4,5 \right \}

E\cap G=\left \{ 3,5 \right \} , F\cap G=\left \{ 2,3\right \}

(E\cap G)\cap G =\left \{ 3 \right \}

P[(E\cap G)\cap G] =\frac{1}{6} P(E\cap G) =\frac{2}{6} P(F\cap G) =\frac{2}{6}

P((E\cup F)|G) = P(E|G)+P(F|G) - P[(E\cap F)|G]

=\frac{P(E\cap G)}{P(G)}+\frac{P(F\cap G)}{P(G)}-\frac{P((E\cap F)\cap G)}{P(G)}

=\frac{\frac{2}{6}}{\frac{4}{6}}+\frac{\frac{2}{6}}{\frac{4}{6}}-\frac{\frac{1}{6}}{\frac{4}{6}}

=\frac{2}{4}+\frac{2}{4}-\frac{1}{4}

=\frac{3}{4}

P((E\cap F)|G)=\frac{P((E\cap F)\cap G)}{P(G)}

P((E\cap F)|G)=\frac{\frac{1}{6}}{\frac{4}{6}}

P((E\cap F)|G)=\frac{1}{4}

Question:12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(i) the youngest is a girl,

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1\, \, \, and \, \, \,G2 respectively also first and second boy are denoted by B1\, \, \, and \, \, \,B2

If a family has two children, then total outcomes =2^{2}=4 =\left \{ (B1B2),(G1G2),(G1B2),(G2B1)\right \}

Let A= both are girls =\left \{(G1G2)\right \}

and B= the youngest is a girl = =\left \{(G1G2),(B1G2)\right \}

A\cap B=\left \{(G1G2)\right \}

P(A\cap B)=\frac{1}{4} P( B)=\frac{2}{4}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{1}{4}}{\frac{2}{4}}

P(A| B)=\frac{1}{2}

Therefore, the required probability is 1/2

Question:12 Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(ii) at least one is a girl?

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1\, \, \, and \, \, \,G2 respectively also first and second boy are denoted by B1\, \, \, and \, \, \,B2

If a family has two children, then total outcomes =2^{2}=4 =\left \{ (B1B2),(G1G2),(G1B2),(G2B1)\right \}

Let A= both are girls =\left \{(G1G2)\right \}

and C= at least one is a girl = =\left \{(G1G2),(B1G2),(G1B2)\right \}

A\cap B=\left \{(G1G2)\right \}

P(A\cap B)=\frac{1}{4} P( C)=\frac{3}{4}

P(A| C)=\frac{P(A\cap C)}{P(C)}

P(A| C)=\frac{\frac{1}{4}}{\frac{3}{4}}

P(A| C)=\frac{1}{3}

Question:13 An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer:

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.

Total number of questions =300+200+500+400=1400

Let A = question be easy.

n(A)= 300+500=800

P(A)=\frac{800}{1400}=\frac{8}{14}

Let B = multiple choice question

n(B)=500+400=900

P(B)=\frac{900}{1400}=\frac{9}{14}

A\cap B = easy multiple questions

n(A\cap B) =500

P(A\cap B) =\frac{500}{1400}=\frac{5}{14}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{5}{14}}{\frac{9}{14}}

P(A| B)=\frac{5}{9}

Therefore, the required probability is 5/9

Question:14 Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer:

Two dice are thrown.

Total outcomes =6^2=36

Let A be the event ‘the sum of numbers on the dice is 4.

A=\left \{ (13),\left ( 22 \right ),(31) \right \}

Let B be the event that two numbers appearing on throwing two dice are different.

B=\left \{ (12),(13),(14),(15),(16),(21)\left ( 23 \right ),(24),(25),(26,)(31),(32),(34),(35),(36),(41),(42),(43),(45),(46),(51),(52),(53),(54),(56) ,(61),(62),(63),(64),(65)\right \} n(B)=30

P(B)=\frac{30}{36}

A\cap B=\left \{ (13),(31) \right \}

n(A\cap B)=2

P(A\cap B)=\frac{2}{36}

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{\frac{2}{36}}{\frac{30}{36}}

P(A| B)=\frac{2}{30}=\frac{1}{15}

Therefore, the required probability is 1/15

Question:15 Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin.

Total outcomes

=\left \{ (1H),(1T),(2H),(2T),(31),(32),(33),(34),(35),(36),(4H),(4T),(5H),(5T),(61),(62),(63),(64),(65),(66)\right \}

Total number of outcomes =20

Let A be a event when coin shows a tail.

A=\left \{ ((1T),(2T),(4T),(5T)\right \}

Let B be a event that ‘at least one die shows a 3’.

B=\left \{ (31),(32),(33),(34),(35),(36),(63)\right \}

n(B)=7

P(B)=\frac{7}{20}

A\cap B= \phi

n(A\cap B)= 0

P(A\cap B)= \frac{0}{20}=0

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{0}{\frac{7}{20}}

P(A| B)=0

Question:16 In the following Exercise 16 choose the correct answer:

If P(A)=\frac{1}{2},P(B)=0, then P(A\mid B) is

(A) 0

(B) \frac{1}{2}

(C) not\; defined

(D) 1

Answer:

It is given that

P(A)=\frac{1}{2},P(B)=0,

P(A| B)=\frac{P(A\cap B)}{P(B)}

P(A| B)=\frac{P(A\cap B)}{0}

Hence, P(A\mid B) is not defined .

Thus, correct option is C.

Question:17 In the following Exercise 17 choose the correct answer:

If A and B are events such that P(A\mid B)=P(B\mid A), then

(A) A\subset B but A\neq B

(B) A=B

(C) A\cap B=\psi

(D) P(A)=P(B)

Answer:

It is given that P(A\mid B)=P(B\mid A),

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

\Rightarrow P(A)=P(B)

Hence, option D is correct.

More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1:-

Exercise 13.1 Class 12 Maths solutions are consist of questions related to conditional probability and different properties of conditional probability. There are 15 short answer type questions and 2 multiple choice types questions are given in this exercise. There are 7 solved examples given in the NCERT textbook before this exercise that is related to conditional probability. You can solve these examples before solving exercise problems, so it will be easy for you to solve exercise problems.

Also Read| Probability Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.1:-

  • Class 12th Maths chapter 13 exercise 13.1 solutions are helpful for the students to get conceptual clarity about conditional probability and its applications.
  • These exercise 13.1 Class 12 Maths solutions are designed by the subject matter experts who know how best to write the answer in the board exams in order to get good marks.
  • These Class 12 Maths chapter 13 exercise 13.1 solutions could be be used for revision before the exam.
  • These are some examples given in the Class 12 NCERT book before exercise 13.1 which could be solved before solving exercise 13.1.
  • As most of the questions in the 12th board exams are directly asked from NCERT textbooks, you are advised to be thorough with the NCERT textbook.

Also see-

  • NCERT Solutions for Class 12 Maths Chapter 13

  • NCERT Exemplar Solutions Class 12 Maths Chapter 13

NCERT Solutions of Class 12 Subject Wise

  • NCERT Solutions for Class 12 Maths

  • NCERT Solutions for Class 12 Physics

  • NCERT Solutions for Class 12 Chemistry

  • NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

  • NCERT Exemplar Solutions for Class 12 Maths

  • NCERT Exemplar Solutions for Class 12 Physics

  • NCERT Exemplar Solutions for Class 12 Chemistry

  • NCERT Exemplar Solutions for Class 12 Biology

Happy learning!!!