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

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 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 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

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

$F=\{HHH,HHT\}$

$E\cap F=HHH$

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

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

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

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

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

Question 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

$E=\{HT,TH\}=2$

$F=\{HT,TH\}=2$

$E\cap F=\{HT,TH\}=2$

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

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

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

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

$P(E|F)=1$

Question 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 $=6^3=216$

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

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

$n(F)=6$

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

$n(E\cap F)=1$

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

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

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

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

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

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,

$E=\{SMF,SFM,FMS,MFS\}$

$n(E)=4$

$F=\{SFM,MFS\}$

$n(F)=2$

$E\cap F=\{SFM,MFS\}$

$n(E\cap F)=2$

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

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

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

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

$P(E|F)=1$

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 $5.$

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

$n(A)=6$

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

$n(B)=6$

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

$n(A\cap B)=2$

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

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

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

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

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

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

Answer:

A fair die is rolled.

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

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

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

$n(E\cap F)=1$

$n(F)=2$

$n(E)=3$

$P(E)=\frac{3}{6}$ $P(F)=\frac{2}{6}$ 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 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 $G1andG2$ respectively also first and second boy are denoted by $B1andB2$

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

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

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

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

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

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

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

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

Therefore, the required probability is 1/2

Question 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.

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

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

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

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

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

$n(A\cap B)=2$

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

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

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

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

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

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

Total number of outcomes =20

Let A be a event when coin shows a tail.

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

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

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

$n(B)=7$

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

$A\cap B=\varphi$

$n(A\cap B)=0$

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

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

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

$P(A|B)=0$

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

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

(A) $0$

(B) $\frac{1}{2}$

(C) $notdefined$

(D) $1$

Answer:

It is given that

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

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

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

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

Thus, correct option is C.

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

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

(A) $A\subset B$ but $A\ne B$

(B) $A=B$

(C) $A\cap B=\psi$

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

Answer:

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

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

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

Hence, option D is correct.