NCERT Solutions for Class 12 Maths Chapter 13 Probability

 

NCERT Solutions for Class 12 Maths Chapter 13 Probability: You have already studied the basics of probability in our previous classes like probability as a measure of uncertainty of events in a random experiment, addition rule of probability etc. NCERT solutions for class 12 maths chapter 13 probability will build your base to study probability theory in higher studies. It includes concepts of discrete probability, computational probability and stochastic process. The important topics like conditional probability, Bayes' theorem, multiplication rule of probability, and independence of events are covered in this chapter. Questions from all these topics are covered in CBSE NCERT solutions for class 12 maths chapter 13 probability. Also, you will learn some important concepts of the random variable, probability distribution, and the mean and variance of a probability distribution in this chapter. Solutions of NCERT for class 12 maths chapter 13 probability will help you to learn the concept of probability distribution which will be required in higher study. Check all NCERT solutions from class 6 to 12 at a single place, which will help you get a better understanding of concepts in a much easy way.

Generally, two questions( 8 marks) are asked from this chapter in 12th board final examination. You can score this 8 marks very easily with the help of NCERT solutions for class 12 maths chapter 13 probability. In NCERT textbook there are 37 solved examples are given, so you can understand the concept easily. In this chapter, there is a total of 81 questions in 5 exercises. You should try to solve every question on your own. If you are not able to do, you can take help of these CBSE NCERT solutions for class 12 maths chapter 13 probability. These NCERT questions are solved and explained in a step-by-step manner, so it can be understood very easily. This chapter requires lots of practice to understand it better. So, you are advised to solve miscellaneous exercise too.

 
 

The conditional probability of an event E, given the occurrence of the event F is given by

         P(E|F)=\frac{Number\:of\:elementary\:events\:favourable\:to \:E\cap F}{Number\:of\:elementary\:events\:which\:are\:favourable\:to \:F}

Topics of NCERT class 12 maths chapter-13 Probability

13.1 Introduction

13.2 Conditional Probability

13.2.1 Properties of conditional probability

13.3 Multiplication Theorem on Probability

13.4 Independent Events

13.5 Bayes' Theorem

13.5.1 Partition of a sample space

3.5.2 Theorem of total probability

13.6 Random Variables and its Probability Distributions

13.6.1 Probability distribution of a random variable

13.6.2 Mean of a random variable

13.6.3 Variance of a random variable

13.7 Bernoulli Trials and Binomial Distribution

13.7.1 Bernoulli trials

13.7.2 Binomial distribution

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

Solutions of NCERT for class 12 maths chapter 13 probability-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

CBSE NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.2

Question:1 If  P(A)=\frac{3}{5}   and P(B)=\frac{1}{5},  find P(A\cap B)  if A  and B  are independent events.

Answer:

  P(A)=\frac{3}{5}   and P(B)=\frac{1}{5}, 

Given :   A  and B  are independent events.

So we have, P(A\cap B)=P(A).P(B)

     \Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{5}\times \frac{1}{5}

\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{25}

Question:2 Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer:

  Two cards are drawn at random and without replacement from a pack of 52 playing cards.

  There are 26 black cards in a pack of 52.

  Let  P(A) be the probability that first  cards is black.

Then, we have 

                           P(A)= \frac{26}{52}=\frac{1}{2}

  Let  P(B) be the probability that second  cards is black.

Then, we have 

                           P(B)= \frac{25}{51}

The probability that both the cards are black =P(A).P(B)

                                                                         =\frac{1}{2}\times \frac{25}{51}

                                                                       =\frac{25}{102}

Question: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 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer:

Total oranges = 15

Good oranges = 12

Bad oranges = 3

  Let  P(A) be the probability that first orange is good.

The, we have 

                      P(A)= \frac{12}{15}=\frac{4}{5}

  Let  P(B) be the probability that second orange is good.

                    P(B)=\frac{11}{14}

  Let  P(C) be the probability that third orange is good.

                 P(C)=\frac{10}{13}

The probability that a box  will be approved for sale =P(A).P(B).P(C)

                                                                                   =\frac{4}{5}.\frac{11}{14}.\frac{10}{13}

                                                                                   =\frac{44}{91}

Question: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.

Answer:

 A fair coin and an unbiased die are tossed,then total outputs are:

                       = \left \{ (H1),(H2),(H3),(H4),(H5),(H6),(T1),(T2),(T3),(T4),(T5),(T6) \right \}

                      =12                                         

A is the event ‘head appears on the coin’ .

Total outcomes of A are : = \left \{ (H1),(H2),(H3),(H4),(H5),(H6) \right \}

                                                                              P(A)=\frac{6}{12}=\frac{1}{2}

 B is the event ‘3 on the die’.

Total outcomes of B are : = \left \{ (T3),(H3)\right \}

                                             P(B)=\frac{2}{12}=\frac{1}{6}

                       \therefore A\cap B = (H3)

                           P (A\cap B) = \frac{1}{12}

  Also,                P (A\cap B) = P(A).P(B) 

                         P (A\cap B) = \frac{1}{2}\times \frac{1}{6}=\frac{1}{12}

Hence, A and B are independent events.

Question:5 A die marked 1,2,3  in red and 4,5,6 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?

Answer:

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

A is the event, ‘the number is even,’

Outcomes of A  =\left \{ 2,4,6 \right \}

                  n(A)=3.

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

B is the event, ‘the number is red’.

Outcomes of B  =\left \{ 1,2,3 \right \}

                  n(B)=3.

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

           \therefore (A\cap B)=\left \{ 2 \right \}

             n(A\cap B)=1

         P(A\cap B)=\frac{1}{6}

Also,

             P(A\cap B)=P(A).P(B)

           P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\neq \frac{1}{6}

Thus, both the events A and B are not independent.

Question:6 Let E  and F be events with P(E)=\frac{3}{5},P(F)=\frac{3}{10}   and P(E\cap F)=\frac{1}{5}.  Are E and F independent?

Answer:

Given :

               P(E)=\frac{3}{5},P(F)=\frac{3}{10}   and P(E\cap F)=\frac{1}{5}.

For events E and F to be independent , we need 

                                                         P(E\cap F)=P(E).P(F)

                                                         P(E\cap F)=\frac{3}{5}\times \frac{3}{10}=\frac{9}{50}\neq \frac{1}{5}

Hence, E and F are not indepent events.

Question:7  Given that the events A and B are such that P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}   and P(B)=p. Find p  if they are

(i) mutually exclusive

Answer:

Given, 

          P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}

Also, A and B  are mutually exclusive means A\cap B=\phi.

     P(A\cup B)=P(A)+P(B)-P(A\cap B)

                        \frac{3}{5}=\frac{1}{2}+P(B)-0

                       P(B)=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}

Question:7 Given that the eventsA and B are such that P(A)=12,P(A\cup B)=\frac{3}{5}   and P(B)=p. Find p if they are

 (ii) independent

Answer:

Given, 

          P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}

Also, A and B  are independent events means

 P(A\cap B) = P(A).P(B).   Also P(B)=p.

 P(A\cap B) = P(A).P(B)=\frac{p}{2}

     P(A\cup B)=P(A)+P(B)-P(A\cap B)

                        \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}

                       \frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}

                     p=\frac{2}{10}=\frac{1}{5}

Question:8 Let A and B be independent events with P(A)=0.3  and P(B)=0.4  Find

 (i)  P(A\cap B)

Answer:

P(A)=0.3  and P(B)=0.4

Given :  A and B be independent events

So, we have 

                       P(A\cap B)=P(A).P(B)

                        P(A\cap B)=0.3\times 0.4=0.12   

Question:8 LetA and B be independent events with P(A)=0.3 and P(B)=0.4  Find

(ii)  P(A\cup B)

Answer:

P(A)=0.3  and P(B)=0.4

Given :  A and B be independent events

So, we have 

                       P(A\cap B)=P(A).P(B)

                        P(A\cap B)=0.3\times 0.4=0.12

   We have,    P(A\cup B)=P(A)+P(B)-P(A\cap B)

                    P(A\cup B)=0.3+0.4-0.12=0.58

Question:8 Let A and B be independent events with P(A)=0.3  and P(B)=0.4 Find

 (iii)  P(A\mid B)

Answer:

P(A)=0.3  and P(B)=0.4

Given :  A and B be independent events

So, we have  P(A\cap B)=0.12

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

P(A|B)=\frac{0.12}{0.4}= 0.3

Question:8 Let A and B be independent events with P(A)=0.3  and P(B)=0.4  Find 

 (iv)  P(B\mid A)

Answer:

P(A)=0.3  and P(B)=0.4

Given :  A and B be independent events

So, we have  P(A\cap B)=0.12

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

P(B|A)=\frac{0.12}{0.3}= 0.4

Question:9 If A  and B are two events such that P(A)=\frac{1}{4},P(B)=\frac{1}{2} and P(A\cap B)=\frac{1}{8},   find  P(not\; A\; and\; not\; B).

Answer:

 If A  and B are two events such that P(A)=\frac{1}{4},P(B)=\frac{1}{2} and P(A\cap B)=\frac{1}{8},  

       P(not\; A\; and\; not\; B)= P(A'\cap B')

       P(not\; A\; and\; not\; B)= P(A\cup B)'                         use,      (P(A'\cap B')= P(A\cup B)')

                                                    = 1-(P(A)+P(B)-P(A\cap B))

                                                     = 1-(\frac{1}{4}+\frac{1}{2}-\frac{1}{8})

                                                     = 1-(\frac{6}{8}-\frac{1}{8})

                                                     = 1-\frac{5}{8}

                                                    = \frac{3}{8}

Question:10 Events A and B are such that P(A)=\frac{1}{2},P(B)=\frac{7}{12}  and P(not \; A \; or\; not\; B)=\frac{1}{4}.  State whether A and B are independent ?

Answer:

 If A  and B are two events such that P(A)=\frac{1}{2},P(B)=\frac{7}{12} and P(not \; A \; or\; not\; B)=\frac{1}{4}.  

                    P(A'\cup B')=\frac{1}{4}

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

                   \Rightarrow \, \, 1-P(A\cap B)=\frac{1}{4}

                 \Rightarrow \, \, \, P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}

  Also \, \, \, P(A\cap B)=P(A).P(B)                                       

                P(A\cap B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}       

 As we can see \frac{3}{4}\neq \frac{7}{24}

Hence, A and B are not independent.                                          

Question:11 Given two independent events A and B such that P(A)=0.3,P(B)=0.6,  Find

 (i)  P(A \; and\; B)

Answer:

P(A)=0.3,P(B)=0.6,

Given two independent events A and B.

    P(A\cap B)=P(A).P(B)

  P(A\cap B)=0.3\times 0.6=0.18

Also , we know P(A \, and \, B)=P(A\cap B)=0.18

Question:11 Given two independent events A and B such that P(A)=0.3,P(B)=0.6,  Find

(ii)  P(A \; and \; not\; B)

Answer:

P(A)=0.3,P(B)=0.6,

Given two independent events A and B.

             P(A \; and \; not\; B)=P(A)-P(A\cap B)

                                                =0.3-0.18=0.12

Question:11 Given two independent events A and B such that P(A)=0.3,P(B)0.6,  Find

(iii) P(A\; or \; B)

Answer:

P(A)=0.3,P(B)=0.6,

P(A\cap B)=0.18

P(A\; or \; B)=P(A\cup B)

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

                      =0.3+0.6-0.18

                     =0.9-0.18

                      =0.72  

Question:11 Given two independent events A  and B  such that P(A)=0.3,P(B)=0.6,  Find 

 (iv)  P(neither\; A\; nor\; B)

Answer:

P(A)=0.3,P(B)=0.6,

P(A\cap B)=0.18

P(A\; or \; B)=P(A\cup B)

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

                      =0.3+0.6-0.18

                      =0.9-0.18

                      =0.72  

P(neither\; A\; nor\; B) =P(A'\cap B')

                                         = P((A\cup B)')

                                        =1-P(A\cup B)

                                         =1-0.72

                                         =0.28

Question:12 A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer:

A die is tossed thrice.

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

Odd numbers =\left \{ 1,3,5 \right \}

The probability of getting an odd number at first throw 

                                                                                       =\frac{3}{6}=\frac{1}{2}

The probability of getting an even number 

                                                                          =\frac{3}{6}=\frac{1}{2}

Probability of getting even number three times 

                                                                         =\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}

The probability of getting an odd number at least once = 1 - the probability of getting an odd number in none of throw

                                                                                       = 1 - probability of getting even number three times 

                                                                                       =1-\frac{1}{8}

                                                                                       =\frac{7}{8}

Question:13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a red ball in first draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

The ball is repleced after drawing first ball.

The probability of getting a red ball in second draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

the probability that both balls are red

                                                            =\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}

Question:13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

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

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a black ball in the first draw 

                                                                            =\frac{10}{18}=\frac{5}{9}

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

the probability that the first  ball is black and the second is red

                                                                                     =\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}

Question:13 Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

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

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw 

                                                                            =\frac{10}{18}=\frac{5}{9}

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

the probability that the first  ball is black and the second is red

                                                                                     =\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}    ...........................1

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw 

                                                                            =\frac{8}{18}=\frac{4}{9}

The probability of getting a black ball in the second  draw 

                                                                            =\frac{10}{18}=\frac{5}{9}

the probability that the first  ball is red and the second is black

                                                                                     =\frac{4}{9}\times \frac{5}{9}=\frac{20}{81}    ...........................2

Thus, 

The probability that one of them is black and the other is red = the probability that the first  ball is black and the second is red  + the probability that the first  ball is red and the second is black                                                                                       =\frac{20}{81}+\frac{20}{81}=\frac{40}{81}

Question:14 Probability of solving specific problem independently by A and B are \frac{1}{2} and  \frac{1}{3} respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved

Answer:

P(A)=\frac{1}{2}        and         P(B)=\frac{1}{3}

Since, problem is solved independently by A and B,

\therefore           P(A\cap B)=P(A).P(B)

               P(A\cap B)=\frac{1}{2}\times \frac{1}{3}

                 P(A\cap B)=\frac{1}{6}

  probability that the problem is solved = P(A\cup B)

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

        P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}

        P(A\cup B)=\frac{5}{6}-\frac{1}{6}

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

Question:14 Probability of solving specific problem independently by A and B are \frac{1}{2}  and \frac{1}{3} respectively. If both try to solve the problem independently, find the probability that

(ii) exactly one of them solves the problem

Answer:

  P(A)=\frac{1}{2}           and        P(B)=\frac{1}{3}

P(A')=1-P(A),      P(B')=1-P(B)

 P(A')=1-\frac{1}{2}=\frac{1}{2}  ,           P(B')=1-\frac{1}{3}=\frac{2}{3}

   probability that exactly one of them solves the problem =P(A\cap B') + P(A'\cap B)

    probability that exactly one of them solves the problem  =P(A).P(B')+P(A')P(B)

                                                                                              =\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}

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

                                                                                            = \frac{3}{6}=\frac{1}{2}

Question:15 One card is drawn at random from a well shuffled deck of 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’

Answer:

One card is drawn at random from a well shuffled deck of 52 cards

Total ace = 4

total spades =13

E : ‘the card drawn is a spade

 F : ‘the card drawn is an ace’

P(E)=\frac{13}{52}=\frac{1}{4}

P(F)=\frac{4}{52}=\frac{1}{13}

E\cap F : a card which is spade and ace = 1

P(E\cap F)=\frac{1}{52}

P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}

\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{52}

Hence, E and F are indepentdent events .

Question:15 One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?

 (ii) E : ‘the card drawn is

 F : ‘the card drawn is a king’

Answer:

One card is drawn at random from a well shuffled deck of 52 cards

Total black card = 26

total king =4

E : ‘the card drawn is black’

 F : ‘the card drawn is a king’

P(E)=\frac{26}{52}=\frac{1}{2}

P(F)=\frac{4}{52}=\frac{1}{13}

E\cap F : a card which is black and king = 2

P(E\cap F)=\frac{2}{52}=\frac{1}{26}

P(E).P(F)=\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}

\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{26}

Hence, E and F are indepentdent events .

Question:15 One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?

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

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

Answer:

One card is drawn at random from a well shuffled deck of 52 cards

Total king or queen = 8

total queen or jack = 8

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

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

P(E)=\frac{8}{52}=\frac{2}{13}

P(F)=\frac{8}{52}=\frac{2}{13}

E\cap F : a card which is queen  = 4

P(E\cap F)=\frac{4}{52}=\frac{1}{13}

P(E).P(F)=\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}

\Rightarrow P(E\cap F)\neq P(E).P(F)

Hence, E and F are not indepentdent  events 

Question:16 In a hostel, 60\; ^{o}/_{o}  of the students read Hindi newspaper,40\; ^{o}/_{o} read English newspaper and 20\; ^{o}/_{o} read both Hindi and English newspapers. A student  is selected at random.

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

Answer:

H : 60\; ^{o}/_{o}  of the students read Hindi newspaper,

E : 40\; ^{o}/_{o} read English newspaper       and  

 H \cap E :  20\; ^{o}/_{o} read both Hindi and English newspapers.

P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}

P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}

P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}

the probability that she reads neither Hindi nor English newspapers =1-P(H\cup E) 

                                                                                                             =1-(P(H)+P(E)-P(H\cap E))

                                                                                                            =1-(\frac{3}{5}+\frac{2}{5}-\frac{1}{5})    

                                                                                                             =1-\frac{4}{5}

                                                                                                              =\frac{1}{5}

Question:16 In a hostel, 60\; ^{o}/_{o} of the students read Hindi newspaper, 40\; ^{o}/_{o}  read English newspaper and 20\; ^{o}/_{o}  read both Hindi and English newspapers. A student is selected at random.

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

Answer:

H : 60\; ^{o}/_{o}  of the students read Hindi newspaper,

E : 40\; ^{o}/_{o} read English newspaper       and  

 H \cap E :  20\; ^{o}/_{o} read both Hindi and English newspapers.

P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}

P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}

P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}

The probability that she reads English newspape if she reads Hindi newspaper =P(E|H)

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

                                                                                  P(E|H)=\frac{\frac{1}{5}}{\frac{3}{5}}

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

Question:16 In a hostel, 60^{o}/_{o} of the students read Hindi newspaper, 40^{o}/_{o}  read English newspaper and 20^{o}/_{o} read both Hindi and English newspapers. A student is selected at random.

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

Answer:

H : 60\; ^{o}/_{o}  of the students read Hindi newspaper,

E : 40\; ^{o}/_{o} read English newspaper       and  

 H \cap E :  20\; ^{o}/_{o} read both Hindi and English newspapers.

P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}

P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}

P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}

 the probability that she reads Hindi newspaper if she reads English newspaper = P(H |E)

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

                                                                                                                 P(H |E)=\frac{\frac{1}{5}}{\frac{2}{5}}

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

Question: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}

Answer:

when a pair of dice is rolled, total outcomes =6^2=36

Even prime number =\left \{ 2 \right \}

n(even \, \, prime\, \, number)=1

 The probability of obtaining an even prime number on each die =P(E)

                                                                                        P(E)=\frac{1}{36}

Option D is correct.

Solutions of NCERT for class 12 maths chapter 13 probability-Exercise: 13.3

Question: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?

Answer:

Black balls = 5

Red balls = 5

Total balls = 10

CASE 1  Let red ball be drawn in  first attempt.

P(drawing\, red\, ball)=\frac{5}{10}=\frac{1}{2}

Now two red balls are added in urn .

Now  red balls = 7, black balls = 5

Total balls = 12

P(drawing\, red\, ball)=\frac{7}{12}

CASE 2 

Let black ball be drawn in  first attempt.

P(drawing\, black\, ball)=\frac{5}{10}=\frac{1}{2}

Now two black balls are added in urn .

Now  red balls = 5, black balls = 7

Total balls = 12

P(drawing\, red\, ball)=\frac{5}{12}

the probability that the second ball is red =

=\frac{1}{2}\times \frac{7}{12}+\frac{1}{2}\times \frac{5}{12}

= \frac{7}{24}+ \frac{5}{24}

= \frac{12}{24}=\frac{1}{2}

Question: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 frome bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer:

BAG 1 : Red balls =4        Black balls=4     Total balls = 8                                                                                                          

BAG 2 : Red balls = 2       Black balls = 6     Total balls = 8                                                                              

B1 : selecting bag 1

B2 : selecting bag 2

P(B1)=P(B2)=\frac{1}{2}

Let R be a event of getting red ball

P(R|B1) = P(drawing\, \, red\, \, ball\, \, from \, first \, \, bag)= \frac{4}{8}=\frac{1}{2}

P(R|B2) = P(drawing\, \, red\, \, ball\, \, from \, second \, \, bag)= \frac{2}{8}=\frac{1}{4}

probability that the ball is drawn from the first bag,

given that it is red is P(B1|R).

Using Baye's theorem, we have

             P(B1|R) = \frac{P(B1).P(R|B1)}{P(B1).P(R|B1)+P(B2).P(R|B2)}

           P(B1|R) = \frac{\frac{1}{2}\times \frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{4}}

            P(B1|R) =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}}

         P(B1|R) =\frac{\frac{1}{4}}{\frac{3}{8}}

        P(B1|R) = \frac{2}{3}

Question:6 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75^{o}/_{o} 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?

Answer:

Given :   A :  chossing a two headed coin

              B :  chossing a biased coin

              C : chossing a unbiased coin 

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

             D : event that coin tossed show head.

         P(D|A)=1

     Biased coin that comes up heads 75^{o}/_{o} of the time.

       P(D|B)=\frac{75}{100}=\frac{3}{4}

     P(D|C)=\frac{1}{2}

P(B|D)=\frac{P(B).P(D|B)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

P(B|D)=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{3}\times \frac{1}{2}}

P(B|D)=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}

P(B|D)=\frac{\frac{1}{3}}{\frac{9}{12}}

P(B|D)={\frac{1\times 12}{3\times 9}}

P(B|D)={\frac{4}{9}}

Question:7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an 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?

Answer:

Let  A : scooter drivers = 2000

       B : car drivers = 4000

       C : truck drivers = 6000

       Total drivers = 12000

P(A)=\frac{2000}{12000}=\frac{1}{6}=0.16

P(B)=\frac{4000}{12000}=\frac{1}{3}=0.33

P(C)=\frac{6000}{12000}=\frac{1}{2}=0.5

  D : the event that person meets with an accident.

  P(D|A)= 0.01

P(D|B)= 0.03

P(D|C)= 0.15

P(A|D)=\frac{P(A).P(D|A)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

 P(A|D)= \frac{0.16\times 0.01}{0.16\times 0.01+0.33\times 0.03+0.5\times 0.15}

P(A|D)= \frac{0.0016}{0.0016+0.0099+0.075}

P(A|D)= \frac{0.0016}{0.0865}

P(A|D)= 0.019

Question:8 A factory has two machines A and B. Past record shows that machine A produced 60^{o}/_{o} of the items of output and machine B produced 40^{o}/_{o}  of the items. Further, 2^{o}/_{o}  of the items produced by machine A and 1^{o}/_{o} produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Answer:

A  : Items produced by machine A =60\%

B : Items produced by machine B=40\%

P(A)= \frac{60}{100}=\frac{3}{5}

P(B)= \frac{40}{100}=\frac{2}{5}

X : Produced item found to be defective.

P(X|A)= \frac{2}{100}=\frac{1}{50}

P(X|B)= \frac{1}{100}

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{2}{5}\times \frac{1}{100}}{\frac{2}{5}\times \frac{1}{100}+\frac{3}{5}\times \frac{1}{50}}

P(B|X)= \frac{\frac{1}{250}}{\frac{1}{250}+\frac{3}{250}}

P(B|X)= \frac{\frac{1}{250}}{\frac{4}{250}}

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

Hence, the probability that defective item was produced by machine B =    

                                                                                                                     P(B|X)= \frac{1}{4}.

Question: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 are0.6 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.

Answer:

A: the first groups will win

B:  the second groups will win

P(A)=0.6

P(B)=0.4

X: Event of introducing a new product.

Probability of introducing a new product if the first group wins : P(X|A)=0.7

Probability of introducing a new product if the second group wins : P(X|B)=0.3

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

p(B|X) = \frac{0.4\times 0.3}{0.4\times 0.3+0.6\times 0.7}

p(B|X) = \frac{0.12}{0.12+0.42}

p(B|X) = \frac{0.12}{0.54}

p(B|X) = \frac{12}{54}

p(B|X) = \frac{2}{9}

Hence,  the probability that the new product introduced was by the second group : 

                                                                                                                                  p(B|X) = \frac{2}{9} 

Question: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 4with the die?

Answer:

Let, A: Outcome on die is 5 or 6.

       B: Outcome on die is 1,2,3,4

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

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

X: Event of getting exactly one head.

Probability of  getting exactly one head when she tosses a coin three times : P(X|A)=\frac{3}{8}

Probability of  getting exactly one head when she tosses a coin one time :  P(X|B)=\frac{1}{2}

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{2}{3}\times \frac{1}{2}}{\frac{2}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{3}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{8}}

P(B|X)= \frac{\frac{1}{3}}{\frac{11}{24}}

P(B|X)= \frac{1\times 24}{3\times 11}=\frac{8}{11}

Hence, the probability that she threw 1,2,3 or 4 with the die = 

                                                                                                      P(B|X)=\frac{8}{11}

Question:11  A manufacturer has three machine operators A,B and C. The first operator A produces 1^{o}/_{o} defective items, where as the other two operators B and C produce 5^{o}/_{o} and 7^{o}/_{o} defective items respectively. A is on the job for 50^{o}/_{o} of the time, B is on the job for 30^{o}/_{o} of the time and C is on the job for 20^{o}/_{o}  of the time. A defective item is produced, what is the probability that it was produced by A?

Answer:

Let A: time consumed by machine A =50\%

       B: time consumed by machine B=30\%

       C: time consumed by machine C =20\%

       Total drivers = 12000

P(A)=\frac{50}{100}=\frac{1}{2}

P(B)=\frac{30}{100}=\frac{3}{10}

P(C)=\frac{20}{100}=\frac{1}{5}

  D: Event of producing defective items 

  P(D|A)= \frac{1}{100}

P(D|B)= \frac{5}{100}

P(D|C)= \frac{7}{100}

P(A|D)=\frac{P(A).P(D|A)}{P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)}

 P(A|D)=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{2}\times \frac{1}{100}+\frac{3}{10}\times \frac{5}{100}+\frac{1}{5}\times \frac{7}{100}}

P(A|D)=\frac{\frac{1}{2}\times \frac{1}{100}}{\frac{1}{100} (\frac{1}{2}+\frac{3}{2}+\frac{7}{5})}

P(A|D)=\frac{\frac{1}{2}}{ (\frac{17}{5})}

P(A|D)= \frac{5}{34}

Hence, the probability that defective item was produced by A = 

                                                                                                  P(A|D)= \frac{5}{34}

Question: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.

Answer:

Let   A : Event of choosing a diamond card.

        B :  Event of not choosing a diamond card.

P(A)=\frac{13}{52}=\frac{1}{4}

P(B)=\frac{39}{52}=\frac{3}{4}

X : The lost card.

If lost card is diamond then 12 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 12 diamond cards in ^{12}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is lost : P(X|A)= \frac{^{12}\textrm{C}_2}{^{51}\textrm{C}_2}

                                                                                                         P(X|A)=\frac{12!}{10!\times 2!}\times \frac{49!\times 2!}{51!}

                                                                                                          P(X|A)=\frac{11\times 12}{50\times 51}

                                                                                                            P(X|A)=\frac{22}{425}

If lost card is not diamond then 13 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 13 diamond cards in ^{13}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is not lost : P(X|B)= \frac{^{13}\textrm{C}_2}{^{51}\textrm{C}_2}

                                                                                                         P(X|B)=\frac{13!}{11!\times 2!}\times \frac{49!\times 2!}{51!}

                                                                                                          P(X|B)=\frac{13\times 12}{50\times 51}

                                                                                                            P(X|B)=\frac{26}{425}

The probability of the lost card being a diamond : P(B|X)

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}

P(B|X)= \frac{\frac{11}{2}}{25}

P(B|X)= \frac{11}{50}

Hence, the probability of the lost card being a diamond : 

                                                                                          P(B|X)= \frac{11}{50}

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

               (A)  \frac{4}{5}

               (B)  \frac{1}{2}

                C)  \frac{1}{5}

               (D)  \frac{2}{5}

Answer:

Let   A : A speaks truth

        B : A  speaks false

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

P(B)=1-\frac{4}{5}=\frac{1}{5}

X : Event that head appears.

A coin is tossed , outcomes are head or tail.

Probability of getting head whether A speaks thruth or not is \frac{1}{2}

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

P(A|X)= \frac{P(A).P(X|A)}{P(B).P(X|B)+P(A).P(X|A)}

P(A|X)= \frac{\frac{4}{5}\times \frac{1}{2}}{\frac{4}{5}\times \frac{1}{2}+\frac{1}{5}\times \frac{1}{2}}

P(A|X)= \frac{\frac{4}{5}}{\frac{4}{5}+\frac{1}{5}}

P(A|X)= \frac{\frac{4}{5}}{\frac{1}{1}}

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

The probability that actually there was head is P(A|X)={\frac{4}{5}}

Hence, option A is correct.

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

                (A) P(A\mid B)=\frac{P(B)}{P(A)}

                (B) P(A\mid B)< P(A)

                (C) P(A\mid B)\geq P(A)

                (D) None of these

Answer:

If  A\subset B  and P(B)\neq 0,then

                \Rightarrow \, \, \, (A\cap B) = A

 Also, P(A)< P(B)

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

We know that P(B)\leq 1

                       1\leq \frac{1}{P(B)}

                    P(A)\leq \frac{P(A)}{P(B)}

                   P(A)\leq P(A|B)

Hence, we can see option C is correct.

NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.4

Question:1(i) State which the following are not the probability distributions of a random variable. Give reasons for your answer.

   

Answer:

As we know the sum of probabilities of a probability distribution is 1.