# 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          $\dpi{100} P(E|F)=\frac{Number\:of\:elementary\:events\:favourable\:to \:E\cap F}{Number\:of\:elementary\:events\:which\:are\:favourable\:to \:F}$

## 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

## Solutions of NCERT for class 12 maths chapter 13 probability-Exercise: 13.1 Question:1 Given that $\inline E$ and $\inline F$ are events such that $P(E)=0.6,P(F)=0.3$ and  $p(E\cap F)=0.2,$  find  $\inline P(E\mid F)$  and  $\inline P(F\mid E)$

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

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

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

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

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

(i)  $\inline P(A\cap B)$

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

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

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

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

$p(A\cap B) = 0.32$

(ii)  $\inline P(A\mid B)$

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

$P(A\cap B)=0.32$

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

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

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

(iii)  $P(A\cup B)$

It is given that  $P(A)=0.8,P(B)=0.5$

$P(A\cap B)=0.32$

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

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

$P(A\cup B)=1.3-0.32$

$P(A\cup B)=0.98$

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

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

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

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

(i)  $P(A\cap B)$

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:

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

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

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

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

(ii)  $\inline P(A\mid B)$

It is given that -$\dpi{80} P(A)=\frac{6}{11},P(B)=\frac{5}{11}$

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

We know that:

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

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

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

(iii)  $\inline P(B\mid A)$

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

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

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

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

$E\cap F =HHH$

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

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

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

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

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

Question:6 A coin is tossed three times, where

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

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

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

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

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

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

Total outcomes $=6^{3}=216$

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

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

$n(F)=6$

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

$n(E\cap F)=1$

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

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

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

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

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

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

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 $\inline 9$, given that the black die resulted in a $\inline 5.$

A black and a red dice are rolled.

Total outcomes $=6^{2}=36$

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

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

$n(A)=6$

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

$n(B)=6$

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

$n(A\cap B)=2$

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

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

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

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

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

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

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 $\inline 4$.

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

$n(A)=5$

Red dice is rolled after black dice.

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

$n(B)=18$

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

$n(A\cap B)=2$

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

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

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

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

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

A fair die is rolled.

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

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

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

$n(E\cap F)=1$

$n( F)=2$

$n( E)=3$

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

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

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

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

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

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

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

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

A fair die is rolled.

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

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

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

$n(E\cap G)=2$

$n( G)=4$

$n( E)=3$

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

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

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

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

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

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

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

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

A fair die is rolled.

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

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

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

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

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

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

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

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

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

$=\frac{3}{4}$

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

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

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

(i) the youngest is a girl,

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

(ii) at least one is a girl?

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 =$=\left \{(G1G2),(B1G2),(G1B2)\right \}$

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

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

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

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

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

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

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

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’.

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

$n(B)=7$

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

$A\cap B= \phi$

$n(A\cap B)= 0$

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

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

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

$P(A| B)=0$

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

(A)  $0$

(B)  $\frac{1}{2}$

(C) $\inline not\; defined$

(D)  $1$

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

Thus, correct option is C.

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

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

(B)  $\inline A=B$

(C)  $\inline A\cap B=\psi$

(D)  $\inline P(A)=P(B)$

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

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

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

Hence, option D is correct.

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

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

Given :   $A$  and $B$  are independent events.

So we have, $P(A\cap B)=P(A).P(B)$

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

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

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 $=P(A).P(B)$

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

$=\frac{25}{102}$

Total oranges = 15

Good oranges = 12

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 $=P(A).P(B).P(C)$

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

$=\frac{44}{91}$

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.

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

$\inline A$ is the event, ‘the number is even,’

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

$n(A)=3.$

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

$\inline B$ 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.

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.

(i) mutually exclusive

Given,

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

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

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

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

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

(ii) independent

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 $P(B)=p.$

$P(A\cap B) = P(A).P(B)=\frac{p}{2}$

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

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

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

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

(i)  $\inline P(A\cap B)$

$\inline P(A)=0.3$  and $\inline P(B)=0.4$

Given :  A and B be independent events

So, we have

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

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

(ii)  $\inline P(A\cup B)$

$\inline P(A)=0.3$  and $\inline P(B)=0.4$

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

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

(iii)  $\inline P(A\mid B)$

$\inline P(A)=0.3$  and $\inline P(B)=0.4$

Given :  A and B be independent events

So, we have  $P(A\cap B)=0.12$

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

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

(iv)  $P(B\mid A)$

$\inline P(A)=0.3$  and $\inline P(B)=0.4$

Given :  A and B be independent events

So, we have  $P(A\cap B)=0.12$

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

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

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,      $(P(A'\cap B')= P(A\cup B)')$

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

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

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

$= 1-\frac{5}{8}$

$= \frac{3}{8}$

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.

(i)  $\inline P(A \; and\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

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

$P(A\cap B)=0.3\times 0.6=0.18$

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

(ii)  $\inline P(A \; and \; not\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

Given two independent events $\inline A$ and $\inline B$.

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

$=0.3-0.18=0.12$

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

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

(iv)  $\inline P(neither\; A\; nor\; B)$

$\inline P(A)=0.3,P(B)=0.6,$

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

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

$= P((A\cup B)')$

$=1-P(A\cup B)$

$=1-0.72$

$=0.28$

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

$=1-\frac{1}{8}$

$=\frac{7}{8}$

(i) both balls are red.

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

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

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

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

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

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

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                                                                                       $=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}$

(i) the problem is solved

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

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

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

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

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

(ii) exactly one of them solves the problem

$P(A)=\frac{1}{2}$           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  $=P(A).P(B')+P(A')P(B)$

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

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

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

(i) E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

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

Total ace = 4

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 .

(ii) E : ‘the card drawn is

F : ‘the card drawn is a king’

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 .

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

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

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

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

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

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

$H \cap E :$  $\inline 20\; ^{o}/_{o}$ 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 $=1-P(H\cup E)$

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

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

$=1-\frac{4}{5}$

$=\frac{1}{5}$

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

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

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

$H \cap E :$  $\inline 20\; ^{o}/_{o}$ 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 $=P(E|H)$

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

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

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

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

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

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

$H \cap E :$  $\inline 20\; ^{o}/_{o}$ 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 $= P(H |E)$

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

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

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

(A)  $0$

(B)  $\frac{1}{3}$

(C)   $\frac{1}{12}$

(D)   $\frac{1}{36}$

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

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 =

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

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

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

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

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

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

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

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

$P(B1|R) = \frac{2}{3}$

H :   reside in hostel

D :  day scholars

A : students who attain grade A

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

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

$P(A|H)=\frac{30}{100}=\frac{3}{10}$

$P(A|D)=\frac{20}{100}=\frac{2}{10}= \frac{1}{5}$

By Bayes theorem :

$P(H|A)=\frac{P(H).P(A|H)}{P(H).P(A|H)+P(D).P(A|D)}$

$P(H|A)=\frac{\frac{3}{5}\times \frac{3}{10}}{\frac{3}{5}\times \frac{3}{10}+\frac{2}{5}\times \frac{1}{5}}$

$P(H|A)=\frac{\frac{9}{50}}{\frac{9}{50}+\frac{2}{25}}$

$P(H|A)=\frac{\frac{9}{50}}{\frac{13}{50}}$

$P(H|A)=\frac{9}{13}$

B : Student guess the answer

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

$P(C|A)=1$

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

By Bayes theorem :

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

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

$=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}$     $=\frac{\frac{3}{4}}{\frac{13}{16}}$

$P(A|C)=\frac{12}{13}$

A  : Person selected is  having the disease

B : Person selected is not having the disease.

C :Blood result is positive.

$P(A)= 0.1 \%=\frac{1}{1000}=0.001$

$P(B)= 1 -P(A)=1-0.001=0.999$

$P(C|A)=99\%=0.99$

$P(C|B)=0.5\%=0.005$

By Bayes theorem :

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

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

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

$=\frac{0.00099}{0.005985}$ $=\frac{990}{5985}$

$=\frac{22}{133}$

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 $\inline 75^{o}/_{o}$ of the time.

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

$P(D|C)=\frac{1}{2}$

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

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

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

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

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

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

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.

$P(D|A)= 0.01$

$P(D|B)= 0.03$

$P(D|C)= 0.15$

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

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

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

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

$P(A|D)= 0.019$

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 $\inline B$ =

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

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 :

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

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 =

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

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 $\inline A$ =

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

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 :

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

(A)  $\frac{4}{5}$

(B)  $\frac{1}{2}$

C)  $\frac{1}{5}$

(D)  $\frac{2}{5}$

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.

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

(B) $\inline P(A\mid B)< P(A)$

(C) $\inline P(A\mid B)\geq P(A)$

(D) None of these

If  $\inline A\subset B$  and $\inline P(B)\neq 0,$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

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

Sum of probabilities $=0.4+0.4+0.2=1$

$\therefore$ The given table is the probability distributions of a random variable.

As we know  probabilities cannot be negative for  a probability distribution .

$P(3) = -0.1$

$\therefore$ The given table is not a the probability distributions of a random variable.

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

Sum of probablities $=0.6+0.1+0.2=0.9\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

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

Sum of probablities $=0.3+0.2+0.4+0.1+0.05=1.05\neq 1$

$\therefore$ The given table is not a the probability distributions of a random variable because sum of probabilities is not 1.

B = black balls

R = red balls

The two balls can be selected as BR,BB,RB,RR.

X = number of black balls.

$\therefore$     $X(BB)=2$

$X(RB)=1$

$X(BR)=1$

$X(RR)=0$

Hence, possible values of X can be 0, 1 and 2.

Yes, X is a random variable.

The difference between the number of heads and the number of tails obtained when a coin is tossed $6$ times are :

$\therefore$   $X(6H,0T)=\left | 6-0 \right |=6$

$\Rightarrow \, \, X(5H,1T)=\left | 5-1 \right |=4$

$\Rightarrow \, \, X(4H,2T)=\left | 4-2 \right |=2$

$\Rightarrow \, \, X(3H,3T)=\left | 3-3 \right |=0$

$\Rightarrow \, \, X(2H,4T)=\left | 2-4\right |=2$

$\Rightarrow \, \, X(1H,5T)=\left | 1-5\right |=4$

$\Rightarrow \, \, X(0H,6T)=\left |0-6\right |=6$

Thus, possible values of X are 0, 2, 4 and 6.

Question:4(i) Find the probability distribution of

number of heads in two tosses of a coin.

When coin is tossed twice then sample space $=\left \{ HH,HT,TH,TT \right \}$

Let X be number of heads.

$\therefore$  $X(HH)=2$

$X(HT)=1$

$X(TH)=1$

$X(TT)=0$

X can take values of 0,1,2.

$P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}$

$P(X=0)=P(TT)=\frac{1}{4}$

$P(X=2)=P(HH)=\frac{1}{4}$

$P(X=1)=P(HT)+P(TH)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

Table is as shown :

 X 0 1 2 P(X) $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$

Question:4(ii) Find the probability distribution of

number of tails in the simultaneous tosses of three coins.

When 3  coins are  simultaneous tossed  then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

Let X be number of tails.

$\therefore$  X can be 0,1,2,3

X can take values of 0,1,2.

$P(X=0)=P(HHH)=\frac{1}{8}$

$P(X=1)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(THT)+P(HTT)+P(TTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(TTT)=\frac{1}{8}$

Table is as shown :

 X 0 1 2 3 P(X) $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

Question:4(iii) Find the probability distribution of

number of heads in four tosses of a coin.

When coin is tossed 4 times  then sample space $=\left \{ HHHH,HHHT,TTTH,HTHH,THHT,TTHH,HHTT ,TTTT,HTTH,THTH,HTHT,HTTT,THHH,THTT,HHTH,TTHT\right \}$

Let X be number of heads.

$\therefore$  X can be 0,1,2,3,4

$P(X=0)=P(TTTT)=\frac{1}{16}$

$P(X=1)=P(HTTT)+P(TTTH)+P(THTT)=(TTHT)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=2)=P(THHT)+P(HHTT)+P(HTTH)+P(TTHH)+P(HTHT)+P(THTH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}$

$P(X=3)=P(HHHT)+P(THHH)+P(HHTH)+P(HTHH)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$

$P(X=4)=P(HHHH)=\frac{1}{16}$

Table is as shown :

 X 0 1 2 3 4 P(X) $\frac{1}{16}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{4}$ $\frac{1}{16}$

number greater than 4

When a die is tossed twice , total outcomes = 36

Number less than or equal to 4 in both toss  :  $P(X=0)=\frac{4}{6} \times \frac{4}{6}=\frac{4}{9}$

Number less than or equal to 4 in first toss and number more than or equal to 4 in second toss + Number less than or equal to 4 in second toss and number more than or equal to 4 in first toss: $P(X=1)=\frac{4}{6} \times \frac{2}{6}+ \frac{2}{6}\times \frac{4}{6}=\frac{4}{9}$

Number less than 4 in both tosses : $P(X=2)=\frac{2}{6} \times \frac{2}{6}= \frac{1}{9}$

Probability distribution is as :

 X 0 1 2 P(X) $\frac{4}{9}$ $\frac{4}{9}$ $\frac{1}{9}$

When a die is tossed twice , total outcomes = 36

Six does not appear on any of the die  :  $P(X=0)=\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}$

Six appear on atleast one die  : $P(X=1)=\frac{1}{6} \times \frac{5}{6}+ \frac{1}{6}\times \frac{5}{6}=\frac{5}{18}$

Probability distribution is as :

 X 0 1 P(X) $\frac{25}{36}$ $\frac{5}{18}$

Total bulbs = 30

defective bulbs = 6

Non defective bulbs $=30-6=24$

$P(defective \, \, \, \, \, bulbs)=\frac{6}{30}=\frac{1}{5}$

$P(Non \, \, \, \, defective \, \, \, \, \, bulbs)=\frac{24}{30}=\frac{4}{5}$

$\inline 4$ bulbs is drawn at random with replacement.

Let X : number of defective bulbs

4 Non defective bulbs and 0  defective bulbs : $P(X=0)={4}\textrm{C}_0\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

3 Non defective bulbs and 1  defective bulbs : $P(X=1)={4}\textrm{C}_1\frac{1}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}$

2 Non defective bulbs and 2  defective bulbs : $P(X=2)={4}\textrm{C}_2\frac{1}{5}.\frac{1}{5}.\frac{4}{5}.\frac{4}{5}=\frac{96}{625}$

1 Non defective bulbs and 3  defective bulbs : $P(X=3)={4}\textrm{C}_3\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{4}{5}=\frac{16}{625}$

0 Non defective bulbs and 4  defective bulbs : $P(X=4)={4}\textrm{C}_4\frac{1}{5}.\frac{1}{5}.\frac{1}{5}.\frac{1}{5}=\frac{1}{625}$

the probability distribution of the number of defective bulbs is as :

 X 0 1 2 3 4 P(X) $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$

the coin is tossed twice, total outcomes =4 $=\left \{ HH,TT,HT,TH \right \}$

probability of getting a tail be x.

i.e. $P(T)=x$

Then $P(H)=3x$

$P(T)+P(H)=x+3x=1$

$4x=1$

$x=\frac{1}{4}$

$P(T)=\frac{1}{4}$               and           $P(H)=\frac{3}{4}$

Let  X : number of tails

No tail : $P(X=0)=P(H).P(H)=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

1 tail : $P(X=1)=P(HT)+P(TH)=\frac{3}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{3}{4}=\frac{3}{8}$

2 tail : $P(X=2)=P(TT)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

the probability distribution of number of tails are

 X 0 1 2