NCERT Solutions for Class 12 Maths Chapter 13 Probability

# NCERT Solutions for Class 12 Maths Chapter 13 Probability

Edited By Ramraj Saini | Updated on Sep 16, 2023 05:06 PM IST | #CBSE Class 12th

## NCERT Probability Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 13 Probability are provided here. You have already studied the basics of probability class 12 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 therefore probability class 12 ncert solutions become very important.

NCERT Class 12 Maths Chapter 13 solutions include 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 of NCERT Class 12 maths probability books. Questions from all these topics are covered in NCERT solutions for class 12 maths chapter 13 probability.

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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 13 NCERT Class 12 maths solutions PDF. class 12 maths ncert solutions pdf will help you to learn the concept of probability distribution which will be required in higher study. NCERT solutions help students to understan the concepts in a much easy way. Here you will get NCERT solutions for class 12 other subjects also.

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## NCERT Class 12 Maths Chapter 13 Question Answer Probability - Important Formulae

>> Conditional Probability: Conditional probability is the likelihood of an event occurring based on the occurrence of a preceding event. For two events A and B with the same sample space, the conditional probability of event A given that B has occurred (P(A|B)) is defined as:

P(A|B) = P(A ∩ B) / P(B) (when P(B) ≠ 0)

Other conditional probability relationships:

P(S|F) = P(F|F) = 1

P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)

P(E'|F) = 1 − P(E|F)

>> Multiplication Rule: The multiplication rule relates the probability of two events E and F in a sample space S:

P(E ∩ F) = P(E) P(F|E) = P(F) P(E|F) (when P(E) ≠ 0 and P(F) ≠ 0)

>> Independent Events: Two experiments are considered independent if the probability of the events E and F occurring simultaneously is the product of their individual probabilities:

P(E ∩ F) = P(E) * P(F)

>> Bayes’ Theorem: Bayes’ theorem deals with events E1, E2, …, En that form a partition of the sample space S. It allows the calculation of the probability of event Ei given event A:

P(Ei|A) = [P(Ei) * P(A|Ei)] / ∑[P(Ej) * P(A|Ej)], for i = 1, 2, …, n

>> Theorem of Total Probability: Given a partition E1, E2, …, En of the sample space and an event A, the theorem of total probability states:

P(A) = P(E1) * P(A|E1) + P(E2) * P(A|E2) + … + P(En) * P(A|En)

>> Random Variables and their Probability Distributions: A random variable is a real-valued function whose domain is a sample space. The probability distribution of a random variable X consists of possible values x1, x2, …, xn and their corresponding probabilities p1, p2, …, pn:

E(X) = μ = ∑(xi * pi)

Var(X) = σ² = ∑((xi - μ)² * pi)

σ = √Var(X)

## NCERT Probability Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.1

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 $\inline P(A\cap B)=0.32$

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

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

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

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

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

$p(A\cap B) = 0.32$

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

$P(A\cap B)=0.32$

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

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

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

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

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

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

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:

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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 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.

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.

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

$n(B)=18$

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

$n(A\cap B)=2$

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

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

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

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

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

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

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

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

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

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

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

A fair die is rolled.

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

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

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

$n(E\cap G)=2$

$n( G)=4$

$n( E)=3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=\frac{3}{4}$

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

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

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

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

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’.

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

$n(B)=7$

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

$A\cap B= \phi$

$n(A\cap B)= 0$

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

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

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

$P(A| B)=0$

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, is not defined .

Thus, correct option is C.

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.

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.

Given,

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

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

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

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

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

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 $P(B)=p.$

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

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

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

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

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

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

Given : A and B be independent events

So, we have

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

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

$\inline P(A)=0.3$ 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, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

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

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

Given : A and B be independent events

So, we have $P(A\cap B)=0.12$

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

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

$\inline P(A)=0.3$ 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.

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

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

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

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

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

$=0.3-0.18=0.12$

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

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

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

$P(A\cap B)=0.18$

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

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

$=0.3+0.6-0.18$

$=0.9-0.18$

$=0.72$

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

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

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

$=1-0.72$

$=0.28$

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

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

$=\frac{7}{8}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 .

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 .

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

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 $=1-P(H\cup E)$

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

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

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

$=\frac{1}{5}$

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 $=P(E|H)$

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

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

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

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 $= P(H |E)$

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

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

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

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.

Question:18 Two events A and B will be independent, if

Two events A and B will be independent, if

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

Or $P(A'\cap B')=P(A'B')=P(A').P(B')=(1-P(A)).(1-P(B))$

Option B is correct.

NCERT solutions 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 :

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

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.

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

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

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

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) $\frac{1}{16}$ $\frac{1}{4}$ $\frac{3}{8}$ $\frac{1}{4}$ $\frac{1}{16}$

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 P(X) $\frac{9}{16}$ $\frac{3}{8}$ $\frac{1}{16}$

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, 0+k+2k+2k+3k+k^{2}+2k^{2}+7k^{2}+k=1$

$10k^{2}+9k-1=0$

$(10k-1)(k+1)=0$

$k=\frac{1}{10}\, \, and\, \, k=-1$

$P(X< 3)=P(X=0)+P(X=1)+P(X=2)$

$=0+K+2K$

$=3K$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

$P(X> 6)=P(X=7)$

$=7K^2+K$

$=7\times (\frac{1}{10})^2+\frac{1}{10}$

$=\frac{7}{100}+\frac{1}{10}$

$=\frac{17}{100}$

$P(0< X< 3)=P(X=1)+P(X=2)$

$=k+2k$

$=3k$

$=3\times \frac{1}{10}$

$= \frac{3}{10}$

Sum of probabilities of probability distribution of random variable is 1.

$\therefore \, \, \, \, k+2k+3k+0=1$

$6k=1$

$k=\frac{1}{6}$

$P(X< 2)=P(X=0)+P(X=1)$

$=k+2k$

$=3k$

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

$= \frac{1}{2}$

$P(X\leq 2)=P(X=0)+P(X=1)+p(X=2)$

$=k+2k+3k$

$=6k$

$=6\times \frac{1}{6}$

$=1$

$P(X\geq 2)=P(X=2)+P(X> 2)$

$=3k+0$

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

Let X be the success of getting head.

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

$\therefore$ X can be 0,1,2,3

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

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

$P(X=2)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

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

The probability distribution is as

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

$=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

$= \frac{3}{8}+ \frac{6}{8}+ \frac{3}{8}$

$=\frac{12}{8}$

$=\frac{3}{2}=1.5$

$X$ denotes the number of sixes, when two dice are thrown simultaneously.

X can be 0,1,2.

$\therefore$ Not getting six on dice $P(X)=\frac{25}{36}$

Getting six on one time when thrown twice : $P(X=1)=2\times \frac{5}{6}\times \frac{1}{6}=\frac{10}{36}$

Getting six on both dice : $P(X=2)= \frac{1}{36}=\frac{1}{36}$

 X 0 1 2 P(X) $\frac{25}{36}$ $\frac{10}{36}$ $\frac{1}{36}$

Expectation of X = E(X)

$E(X)=0\times \frac{25}{36}+1\times \frac{10}{36}+2\times \frac{1}{36}$

$E(X)= \frac{12}{36}$

$E(X)= \frac{1}{3}$

Two numbers are selected at random (without replacement) from the first six positive integers in $6\times 5=30$ ways.

$\inline X$ denote the larger of the two numbers obtained.

X can be 2,3,4,5,6.

X=2, obsevations : $(1,2),(2,1)$

$P(X=2)=\frac{2}{30}=\frac{1}{15}$

X=3, obsevations : $(1,3),(3,1),(2,3),(3,2)$

$P(X=3)=\frac{4}{30}=\frac{2}{15}$

X=4, obsevations : $(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)$

$P(X=4)=\frac{6}{30}=\frac{3}{15}$

X=5, obsevations : $(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4)$

$P(X=5)=\frac{8}{30}=\frac{4}{15}$

X=6, obsevations : $(1,6),(6,1),(2,6),(6,2),(3,6),(6,3),(4,6),(6,4),(5,6),(6,5)$

$P(X=6)=\frac{10}{30}=\frac{1}{3}$

Probability distribution is as follows:

 X 2 3 4 5 6 P(X) $\frac{1}{15}$ $\frac{2}{15}$ $\frac{3}{15}$ $\frac{4}{15}$ $\frac{1}{3}$

$E(X)=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{1}{3}$

$E(X)= \frac{2}{15}+ \frac{2}{5}+ \frac{4}{5}+ \frac{4}{3}+ \frac{2}{1}$

$E(X)= \frac{70}{15}$

$E(X)= \frac{14}{3}$

$\inline X$ denote the sum of the numbers obtained when two fair dice are rolled.

Total observations = 36

X can be 2,3,4,5,6,7,8,9,10,11,12

$P(X=2)=P(1,1)=\frac{1}{36}$

$P(X=3)=P(1,2)+P(2,1)=\frac{2}{36}=\frac{1}{18}$

$P(X=4)=P(1,3)+P(3,1)+P(2,2)=\frac{3}{36}=\frac{1}{12}$

$P(X=5)=P(1,4)+P(4,1)+P(2,3)+P(3,2)=\frac{4}{36}=\frac{1}{9}$

$P(X=6)=P(1,5)+P(5,1)+P(2,4)+P(4,2)+P(3,3)=\frac{5}{36}$

$P(X=7)=P(1,6)+P(6,1)+P(2,5)+P(5,2)+P(3,4)+P(4,3)=\frac{6}{36}=\frac{1}{6}$

$P(X=8)=P(2,6)+P(6,2)+P(3,5)+P(5,3)+P(4,4)=\frac{5}{36}$ $P(X=9)=P(3,6)+P(6,3)+P(4,5)+P(5,4)=\frac{4}{36}=\frac{1}{9}$

$P(X=10)=P(4,6)+P(6,4)+P(5,5)=\frac{3}{36}=\frac{1}{12}$

$P(X=11)=P(5,6)+P(6,5)=\frac{2}{36}=\frac{1}{18}$

$P(X=12)=P(6,6)=\frac{1}{36}$

Probability distribution is as follows :

 X 2 3 4 5 6 7 8 9 10 11 12 P(X) $\frac{1}{36}$ $\frac{1}{18}$ $\frac{1}{12}$ $\frac{1}{9}$ $\frac{5}{36}$ $\frac{1}{6}$ $\frac{5}{36}$ $\frac{1}{9}$ $\frac{1}{12}$ $\frac{1}{18}$ $\frac{1}{36}$

$E(X)=2\times \frac{1}{36}+3\times \frac{1}{18}+4\times \frac{1}{12}+5\times \frac{1}{9}+6\times \frac{5}{36}+7\times \frac{1}{6}+8\times \frac{5}{36}+9\times \frac{1}{9}+10\times \frac{1}{12}+11\times \frac{1}{18}+12\times \frac{1}{36}$

$E(X)=\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}$

$E(X)=7$

$E(X^2)=4\times \frac{1}{36}+9\times \frac{1}{18}+16\times \frac{1}{12}+25\times \frac{1}{9}+36\times \frac{5}{36}+49\times \frac{1}{6}+64\times \frac{5}{36}+81\times \frac{1}{9}+100\times \frac{1}{12}+121\times \frac{1}{18}+144\times \frac{1}{36}$

$E(X^2)=\frac{987}{18}=\frac{329}{6}=54.833$

$Variance = E(X^2)-(E(X))^2$

$=54.833-7^2$

$=54.833-49$

$=5.833$

Standard deviation = $=\sqrt{5.833}=2.415$

Total students = 15

probability of selecting a student :

$=\frac{1}{15}$

The information given can be represented as frequency table :

 X 14 15 16 17 18 19 20 21 f 2 1 2 3 1 2 3 1

$P(X=14)=\frac{2}{15}$ $P(X=15)=\frac{1}{15}$ $P(X=16)=\frac{2}{15}$

$P(X=17)=\frac{3}{15}=\frac{1}{5}$ $P(X=18)=\frac{1}{15}$ $P(X=19)=\frac{2}{15}$

$P(X=20)=\frac{3}{15}=\frac{1}{5}$ $P(X=21)=\frac{1}{15}$

Probability distribution is as :

 X 14 15 16 17 18 19 20 21 P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$

$E(X)=14\times \frac{2}{15}+15\times \frac{1}{15}+16\times \frac{2}{15}+17\times \frac{1}{5}+18\times \frac{1}{15}+19\times \frac{2}{15}+20\times \frac{1}{5}+21\times \frac{1}{15}$

$E(X)=\frac{263}{15}=17.53$

$E(X^2)=14^2\times \frac{2}{15}+15^2\times \frac{1}{15}+16^2\times \frac{2}{15}+17^2\times \frac{1}{5}+18^2\times \frac{1}{15}+19^2\times \frac{2}{15}+20^2\times \frac{1}{5}+21^2\times \frac{1}{15}$

$E(X^2)=\frac{4683}{15}=312.2$

$Variance =E(X^2)-(E(X))^2$

$Variance =312.2-(17.53)^2$

$Variance =312.2-307.42$

$Variance =4.78$

$Standard\, \, deviation =\sqrt{4.78}=2.19$

Given :

$P(X=0)=30\%=\frac{30}{100}=0.3$

$P(X=1)=70\%=\frac{70}{100}=0.7$

Probability distribution is as :

 X 0 1 P(X) 0.3 0.7

$E(X)=0\times 0.3+1\times 0.7$

$E(X)= 0.7$

$E(X^2)=0^2\times 0.3+1^2\times 0.7$

$E(X^2)= 0.7$

$Variance = E(X^2)-(E(X))^2$

$Variance = 0.7-(0.7)^2$

$Variance = 0.7-0.49=0.21$

X is number representing on die.

Total observations = 6

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

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

$P(X=5)=\frac{1}{6}$

 X 1 2 5 P(X) $\frac{1}{2}$ $\frac{1}{3}$ $\frac{1}{6}$

$E(X)=1\times \frac{1}{2}+2\times \frac{1}{3}+5\times \frac{1}{6}$

$E(X)= \frac{1}{2}+ \frac{2}{3}+ \frac{5}{6}$

$E(X)=\frac{12}{6}=2$

Option B is correct.

X be number od aces obtained.

X can be 0,1,2

There 52 cards and 4 aces, 48 are non-ace cards.

$P(X=0)=P(0 \, ace\, and\, 2\, non\, ace\, cards)=\frac{^4C_2.^4^8C_2}{^5^2C_2}=\frac{1128}{1326}$

$P(X=1)=P(1 \, ace\, and\, 1\, non\, ace\, cards)=\frac{^4C_1.^4^8C_1}{^5^2C_2}=\frac{192}{1326}$

$P(X=2)=P(2 \, ace\, and\, 0\, non\, ace\, cards)=\frac{^4C_2.^4^8C_0}{^5^2C_2}=\frac{6}{1326}$

The probability distribution is as :

 X 0 1 2 P(X) $\frac{1128}{1326}$ $\frac{192}{1326}$ $\frac{6}{1326}$

$E(X)=0\times \frac{1128}{1326}+1\times \frac{192}{1326}+2\times \frac{6}{1326}$

$E(X)=\frac{204}{1326}$

$E(X)=\frac{2}{13}$

Option D is correct.

NCERT solutions for class 12 maths chapter 13 probability-Exercise: 13.5

X be the number of success of getting an odd number.

$P(probability\, of\, getting\, odd\, number)=\frac{3}{6}=\frac{1}{2}$

$\therefore \, \, \, \, q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution.

$(\frac{1}{2})^{6-x} . (\frac{1}{2})^{x}$

$(\frac{1}{2})^{6}$

$P(5\, \, \, success) =P(x=5)$

$= ^6C_5 .(\frac{1}{2})^6$

$= 6 .(\frac{1}{64})$

$= \frac{3}{32}$

X be a number of success of getting an odd number.

$P(probability\, of\, getting\, odd\, number)=\frac{3}{6}=\frac{1}{2}$

$\therefore \, \, \, \, q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution.

$(\frac{1}{2})^{6-x} . (\frac{1}{2})^{x}$

$(\frac{1}{2})^{6}$

$P(At \, \, least \, \, 5\, \, \, success) =P(x\geq 5)$

$=P(X=5)+P(X=6)$

$= ^6C_5 .(\frac{1}{2})^6+^6C_6 .(\frac{1}{2})^6$

$= 6 .(\frac{1}{64})$ $+ (\frac{1}{64})$

$= \frac{7}{64}$

X be a number of success of getting an odd number.

$P(probability\, of\, getting\, odd\, number)=\frac{3}{6}=\frac{1}{2}$

$\therefore \, \, \, \, q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution.

$(\frac{1}{2})^{6-x} . (\frac{1}{2})^{x}$

$(\frac{1}{2})^{6}$

$P(atmost\, \, 5\, \, \, success) =P(x\leq 5)$

$=1-P(X> 5)$

$=1-P(X= 5)$

$= 1-^6C_6 .(\frac{1}{2})^6$

$= 1- (\frac{1}{64})$

$= \frac{63}{64}$

A pair of dice is thrown $\inline 4$ times.X be getting a doublet.

Probability of getting doublet in a throw of pair of dice :

$P=\frac{6}{36}=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=4

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^4C_x.$ $(\frac{5}{6})^{4-x} . (\frac{1}{6})^{x}$

$P(X=x)=^4C_x.$ $\frac{5^{4-x}}{6^4}$

Put x = 2

$P(X=2)=^4C_2.$ $\frac{5^{4-2}}{6^4}$

$P(X=2)=6\times \frac{25}{1296}$

$P(X=2)= \frac{25}{216}$

There are $5^{o}/_{o}$ defective items in a large bulk of items.

X denotes the number of defective items in a sample of 10.

$\Rightarrow \, \, P=\frac{5}{100}=\frac{1}{2}$

$\Rightarrow \, \, q=1-\frac{1}{20}=\frac{19}{20}$

X has a binomial distribution, n=10.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^1^0C_x.$ $(\frac{19}{20})^{10-x} . (\frac{1}{20})^{x}$

$P(not\, more\, than\, one\, defective item)=p(X\leq 1)$

$=P(X=0)+P(X=1)$

$=^{10}C_0(\frac{19}{20})^{10} . (\frac{1}{20})^{0}+^{10}C_1(\frac{19}{20})^{9} . (\frac{1}{20})^{1}$

$=(\frac{19}{20})^{9} . (\frac{29}{20})^{1}$

Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $\inline 52$ cards.

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=5 ,

$P(X=5)=^5C_5.$ $(\frac{3}{4})^{0} . (\frac{1}{4})^{5}$

$=1\times \frac{1}{1024}$

$= \frac{1}{1024}$

Let X represent a number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $\inline 52$ cards.

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=3 ,

$P(X=3)=^5C_3.$ $(\frac{3}{4})^{2} . (\frac{1}{4})^{3}$

$=10\times \frac{9}{16}\times \frac{1}{64}$

$=\frac{45}{512}$

Let X represent number of spade cards among five drawn cards. Five cards are drawn successively with replacement from a well-shuffled deck of $\inline 52$ cards.

$P=\frac{13}{52}=\frac{1}{4}$

$q=1-P=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{3}{4})^{5-x} . (\frac{1}{4})^{x}$

Put X=0 ,

$P(X=0)=^5C_0.$ $(\frac{3}{4})^{5} . (\frac{1}{4})^{0}$

$=1\times \frac{243}{1024}$

$= \frac{243}{1024}$

Let X represent number of bulb that will fuse after $\inline 150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(0.95)^{5-x} . (0.05)^{x}$

Put X=0 ,

$P(X=0)=^5C_0.$ $(0.95)^{5} . (0.05)^{0}$

$=(0.95)^{5}$

Let X represent a number of the bulb that will fuse after $\inline 150$ days of use. Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(0.95)^{5-x} . (0.05)^{x}$

Put $X\leq 1$ ,

$P(X\leq 1)=P(X=0)+P(X=1)$

$=^5C_0.$ $(0.95)^{5} . (0.05)^{0}+^5C_1 (0.95)^{4} . (0.05)^{1}$

$=(0.95)^{5}+ (0.25)(0.95)^4$

$=(0.95)^{4}(0.95+ 0.25)$

$=(0.95)^{4}\times 1.2$

Let X represent number of bulb that will fuse after $\inline 150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(0.95)^{5-x} . (0.05)^{x}$

Put $X> 1$ ,

$P(X> 1)=1-(P(X=0)+P(X=1))$

$=1-(^5C_0.$ $(0.95)^{5} . (0.05)^{0}+^5C_1 (0.95)^{4} . (0.05)^{1})$

$=1-((0.95)^{5}+ (0.25)(0.95)^4)$

$=1-((0.95)^{4}(0.95+ 0.25))$

$=1-(0.95)^{4}\times 1.2$

Let X represent number of bulb that will fuse after $\inline 150$ days of use .Trials =5

$P=0.005$

$q=1-0.005=1-0.005=0.95$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(0.95)^{5-x} . (0.05)^{x}$

Put $X\geq 1$ ,

$P(X\geq 1)=1-P(X< 1)$

$P(X\geq 1)=1-P(X=0)$

$=1-^5C_0.$ $(0.95)^{5} . (0.05)^{0}$

$=1-(0.95)^{5}$

Let X denote a number of balls marked with digit 0 among 4 balls drawn.

Balls are drawn with replacement.

X has a binomial distribution,n=4.

$P=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^4C_x.$ $(\frac{9}{10})^{4-x} . (\frac{1}{10})^{x}$

Put X = 0,

$P(X=0)=^4C_0.$ $(\frac{9}{10})^{4} . (\frac{1}{10})^{0}$

$= 1.(\frac{9}{10})^{4}$

$= (\frac{9}{10})^4$

Let X represent the number of correctly answered questions out of 20 questions.

$P=\frac{1}{2}$

$q=1-P=1-\frac{1}{2}=\frac{1}{2}$

X has a binomial distribution,n=20

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^2^0C_x.$ $(\frac{1}{2})^{20-x} . (\frac{1}{2})^{x}$

$P(X=x)=^2^0C_x.$ $(\frac{1}{2})^{20}$

$P(at\, \, least\, 12\, \,questions \, \, answered\, \, correctly)=P(X\geq 12)$

$=P(X=12)+P(X=13)..................+P(X=20)$

$=^{20}C_1_2 (\frac{1}{2})^{20}+^{20}C_1_3(\frac{1}{2})^{20}+..........^{20}C_2_0 (\frac{1}{2})^{20}$

$=(\frac{1}{2})^{20}(^{20}C_1_2 +^{20}C_1_3+..........^{20}C_2_0 )$

X is a random variable whose binomial distribution is $B\left [ 6,\frac{1}{2} \right ].$

Here , n=6 and $P=\frac{1}{2}$ .

$\therefore \, \,q=1-P= 1-\frac{1}{2}= \frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^{6}C_x .(\frac{1}{2})^{6-x}(\frac{1}{2})^x$

$=^{6}C_x (\frac{1}{2})^6$

$P(X=x)$ is maximum if $^{6}C_x$ is maximum.

$^{6}C_0 =^{6}C_6 =\frac{6!}{0!.6!}=1$

$^{6}C_1 =^{6}C_5 =\frac{6!}{1!.5!}=6$

$^{6}C_2 =^{6}C_4 =\frac{6!}{2!.4!}=15$

$^{6}C_3 =\frac{6!}{3!.3!}=20$

$^{6}C_3$ is maximum so for x=3 , $P(X=3)$ is maximum.

Let X represent number of correct answers by guessing in set of 5 multiple choice questions.

Probability of getting a correct answer :

$P=\frac{1}{3}$

$\therefore q=1-P=1-\frac{1}{3}=\frac{2}{3}$

X has a binomial distribution,n=5.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{2}{3})^{5-x} . (\frac{1}{3})^{x}$

$P(guessing \, \, more\, \, than \, 4\, correct\, answer)=P(X\geq 1)$

$=P(X=4)+P(X=5)$

$=^5C_4.$ $(\frac{2}{3})^{1} . (\frac{1}{3})^{4}+^5C_5(\frac{2}{3})^{0} . (\frac{1}{3})^{5}$

$=\frac{10}{243}+\frac{1}{243}$

$=\frac{11}{243}$

Let X represent number of winning prizes in 50 lotteries .

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$ $(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning \:at\: least\: once)=P(X\geq 1)$

$=1-P(X< 1)$

$=1-P(X= 0)$

$=1- ^{50}C_0 (\frac{99}{100})^{50}$

$=1- 1. (\frac{99}{100})^{50}$

$=1- (\frac{99}{100})^{50}$

Let X represent number of winning prizes in 50 lotteries .

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$ $(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning\, exactly\, once)=P(X= 1)$

$=^{50}C_1 (\frac{99}{100})^{49}.\frac{1}{100}$

$= 50.(\frac{99}{100})^{49}\frac{1}{100}$

$=\frac{1}{2} (\frac{99}{100})^{50}$

Let X represent number of winning prizes in 50 lotteries.

$P=\frac{1}{100}$

$q=1-P=1-\frac{1}{100}=\frac{99}{100}$

X has a binomial distribution,n=50.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^50C_x.$ $(\frac{1}{100})^{50-x} . (\frac{99}{100})^{x}$

$P(winning \, \, at \, \, least \, \, twice)=P(X\geq 2)$

$=1-P(X< 2)$

$=1-P(X\leq 1)$

$=1-(P(X=0)+P(X=1))$

$=1- (\frac{99}{100})^{50}-\frac{1}{2}(\frac{99}{100})^{49}$

$=1- (\frac{99}{100})^{49}(\frac{1}{2}+\frac{99}{100})$

$=1- (\frac{99}{100})^{49}\frac{149}{100}$

Let X represent number of times getting 5 in 7 throws of a die.

Probability of getting 5 in single throw of die=P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=7

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^7C_x.$ $(\frac{5}{6})^{7-x} . (\frac{1}{6})^{x}$

$P(getting\, \, exactly\, \, twice)=P(X= 2)$

$=^{7}C_2 (\frac{5}{6})^{5}\frac{1}{6}^2$

$=21 (\frac{5}{6})^{5}\frac{1}{36}$

$= (\frac{5}{6})^{5}\frac{7}{12}$

Let X represent number of times getting 2 six in 6 throws of a die.

Probability of getting 6 in single throw of die=P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=6

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^6C_x.$ $(\frac{5}{6})^{6-x} . (\frac{1}{6})^{x}$

$P(getting\, \, atmost\, \, two\, six)=P(X\leq 2)$

$=P(X=0)+P(X=1)+P(X=2)$

$=^{6}C_0 (\frac{5}{6})^{6}\frac{1}{6}^0+^{6}C_1 (\frac{5}{6})^{5}\frac{1}{6}^1+^{6}C_2 (\frac{5}{6})^{4}\frac{1}{6}^2$

$=1.(\frac{5}{6})^{6}+6. (\frac{5}{6})^{5}\frac{1}{6}+15 (\frac{5}{6})^{4}\frac{1}{36}$

$=(\frac{5}{6})^{6}+ (\frac{5}{6})^{5}+ (\frac{5}{6})^{4}\frac{5}{12}$

$=(\frac{5}{6})^{4}( \frac{5}{6}^{2}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{25}{36}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{70}{36})$

$=(\frac{5}{6})^{4}( \frac{35}{18})$

Let X represent a number of times selecting defective items out of 12 articles.

Probability of getting a defective item =P

$P=10\%=\frac{10}{100}=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

X has a binomial distribution,n=12

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^1^2C_x.$ $(\frac{9}{10})^{12-x} . (\frac{1}{10})^{x}$

$P(selectting\, \, 9\,defective \, items)=$

$=^{12}C_9 (\frac{9}{10})^{3}\frac{1}{10}^9$

$=220 (\frac{9^3}{10^3})\frac{1}{10^9}$

$=22 \times \frac{9^3}{10^1^1}$

Let X represent a number of defective bulbs out of 5 bulbs.

Probability of getting a defective bulb =P

$P=\frac{10}{100}=\frac{1}{10}$

$q=1-P=1-\frac{1}{10}=\frac{9}{10}$

X has a binomial distribution,n=5

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{9}{10})^{5-x} . (\frac{1}{10})^{x}$

$P(non\,of \,bulb\,is\, defective \,)=P(X=0)$

$=^{5}C_0 (\frac{9}{10})^{5}\frac{1}{10}^0$

$=1.\frac{9}{10}^5$

$=(\frac{9}{10})^5$

Let X represent number students out of 5 who are swimmers.

Probability of student who are not swimmers =q

$q=\frac{1}{5}$

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

X has a binomial distribution,n=5

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^5C_x.$ $(\frac{1}{5})^{5-x} . (\frac{4}{5})^{x}$

$P(4\, \,students\, \, are\, swimmers)=P(X= 4)$

$=^5C_4(\frac{1}{5})^{1} . (\frac{4}{5})^{4}$

Option A is correct.

NCERT solutions for class 12 maths chapter 13 probability-Miscellaneous Exercise

A and B are two events such that $\inline P(A)\neq 0.$

$A\subset B$

$\Rightarrow \, \, \, \, A\cap B=A$

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

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

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

$P(B|A)=1$

A and B are two events such that $\inline P(A)\neq 0.$

$P(A\cap B)=P(B\cap A)=0$

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

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

$P(B|A)=0$

Question:2(i) A couple has two children,

A couple has two children,

sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$

Let A be both children are males and B is at least one of the children is male.

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

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

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

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

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

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

Question:2(ii) A couple has two children,

A couple has two children,

sample space $=\left \{ (b,b),(g,g),(b,g),(g,b) \right \}$

Let A be both children are females and B be the elder child is a female.

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

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

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

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

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

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

We have $\inline 5^{o}/_{o}$ of men and $\inline 0.25^{o}/_{o}$ of women have grey hair.

Percentage of people with grey hairs $=(5+0.25)\%=5.25\%$

The probability that the selected haired person is male :

$=\frac{5}{5.25}=\frac{20}{21}$

$\inline 90^{o}/_{o}$ of people are right-handed.

$P(right-handed)=\frac{9}{10}$

$P(left-handed)=q=1-\frac{9}{10}=\frac{1}{10}$

at most $\inline 6$ of a random sample of $\inline 10$ people are right-handed.

the probability that more than $\inline 6$ of a random sample of $\inline 10$ people are right-handed is given by,

$\frac{9}{10}^r .(\frac{1}{10})^{10-r}$

the probability that at most $\inline 6$ of a random sample of $\inline 10$ people are right-handed is given by

. $\frac{9}{10}^r .(\frac{1}{10})^{10-r}$

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

$P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}$

$P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}$

$\inline 6$ balls are drawn with replacement.

Let Z be a random variable that represents a number of balls with Y mark on them in the trial.

Z has a binomial distribution with n=6.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(Z=0)=^6C_0 (\frac{2}{5})^{6} \frac{3}{5}^0$

$P(Z=0)=^6C_0 (\frac{2}{5})^{6}$

Total balls in urn = 25

Balls bearing mark 'X' =10

Balls bearing mark 'Y' =15

$P(ball\, \, bearing\, mark\, 'X')=\frac{10}{25}=\frac{2}{5}$

$P(ball\, \, bearing\, mark\, 'Y')=\frac{15}{25}=\frac{3}{5}$

$\inline 6$ balls are drawn with replacementt.

Let Z be random variable that represents number of balls with Y mark on them in trial.

Z has binomail distribution with n=6.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(not\, more\, than\, 2\, bear\, Y) =P(Z\leq 2)$

$=P(Z=0)+P(Z=1)+P(Z=2)$

$=^6C_0 (\frac{2}{5})^{6} (\frac{3}{5})^0+^6C_1 (\frac{2}{5})^{5} (\frac{3}{5})^1+^6C_2 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{6} +6 (\frac{2}{5})^{5} (\frac{3}{5})^1+15 (\frac{2}{5})^{4} (\frac{3}{5})^2$

$= (\frac{2}{5})^{4} [(\frac{2}{5})^{2}+6 (\frac{2}{5}) (\frac{3}{5})+15(\frac{3}{5})^2]$

$= (\frac{2}{5})^{4} [\frac{4}{25}+\frac{36}{25}+\frac{135}{25}]$

$= (\frac{2}{5})^{4} [\frac{175}{25}]$

$= (\frac{2}{5})^{4} [7]$

$P(atleast\, one\, bear\, \, Y\, mark) =P(Z\geq 1)=1-P(Z=0)$

$=1-(\frac{2}{5})^6$

$P(equal\, to\, the \, number\, of\, balls \, with \, X \, mark \, and\, Y \, mark )=P(Z=3)$

$P(Z=3)=^6C_3.(\frac{2}{5})^2.(\frac{3}{5})^3$

$P(Z=3)=\frac{20\times 8\times 27}{15625}$

$P(Z=3)=\frac{864}{3125}$

Let p and q respectively be probability that the player will clear and knock down the hurdle.

$p=\frac{5}{6}$

$q=1-p=1-\frac{5}{6}=\frac{1}{6}$

Let X represent random variable that represent number of times the player will knock down the hurdle.

$P(Z=z)=^nC_Z P^{n-Z}q^Z$

$P(Z< 2)=P(Z=0)+P(Z=1)$

$=^{10}C_0 .( \frac{5}{6})^{10}.( \frac{1}{6})^{0}+^{10}C_1 .( \frac{5}{6})^{9}.( \frac{1}{6})^{1}$

$=( \frac{5}{6})^{10}+10.( \frac{5}{6})^{9}.( \frac{1}{6})$

$=( \frac{5}{6})^{9}( \frac{5}{6}+10\times . \frac{1}{6})$

$=( \frac{5}{6})^{9} \times \frac{5}{2}$

$=\frac{5^1^0}{2\times 6^9}$

Probability of 6 in a throw of die =P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

Probability that 2 sixes come in first five throw of die :

$=^5C_2 (\frac{5}{6})^3 (\frac{1}{6})^2$

$=\frac{10\times 5^3}{6^5}$

Probability that third six comes in sixth throw :

$=\frac{10\times 5^3}{6^5}\times \frac{1}{6}$

$=\frac{10\times 125}{6^6}$

$=\frac{1250}{23328}$

In a leap year, there are 366 days.

In 52 weeks, there are 52 Tuesdays.

The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday.

The remaining 2 days can be :

1. Monday and Tuesday

2. Tuesday and Wednesday

3. Wednesday and Thursday

4. Thursday and Friday

5.friday and Saturday

6.saturday and Sunday

7.sunday and Monday

Total cases = 7.

Favorable cases = 2

Probability of having 53 Tuesday in a leap year = P.

$P=\frac{2}{7}$

Probability of success is twice the probability of failure.

Let probability of failure be X

then Probability of success = 2X

Sum of probabilities is 1.

$\therefore \, \, \, X+2X=1$

$\Rightarrow \, \, \, 3X=1$

$\Rightarrow \, \, \, X=\frac{1}{3}$

Let $P=\frac{1}{3}$ and $q=\frac{2}{3}$

Let X be random variable that represent the number of success in six trials.

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X\geq 4)=P(X=4)+P(X=5)+P(X=6)$

$=^6C_4 \left [ \frac{2}{3} \right ]^4\left [ \frac{1}{3} \right ]^2+^6C_5 \left [ \frac{2}{3} \right ]^5\left [ \frac{1}{3} \right ]^1+^6C_6 \left [ \frac{2}{3} \right ]^6\left [ \frac{1}{3} \right ]^0$

$=\frac{15\times 2^4}{3^6}+\frac{6\times 2^5}{3^6}+\frac{2^6}{3^6}$

$=\frac{ 2^6}{3^6}(15+12+4)$

$=\frac{ 2^6}{3^6}(31)$

$=\frac{31}{9}\left [ \frac{2}{3} \right ]^4$

Let the man toss coin n times.

Probability of getting head in first toss = P

$P=\frac{1}{2}$

$q=\frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^nC_x.\left ( \frac{1}{2} \right )^{n-x}.\left ( \frac{1}{2} \right )^x$

$P(getting\, atleast\, one\, head)> \frac{90}{100}$

$P(X\geq 1)> 0.9$

$1-P(X=0)> 0.9$

$1-^nC_0 \frac{1}{2^n}> 0.9$

$^nC_0 \frac{1}{2^n}< 0.1$

$\frac{1}{2^n}< 0.1$

$\frac{1}{0.1}< 2^n$

$10< 2^n$

The minimum value to satisfy the equation is 4.

The man should toss a coin 4 or more times.

In a throw of die,

probability of getting six = P

$P=\frac{1}{6}$

probability of not getting six = q

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

There are three cases :

1. Gets six in the first throw, required probability is $\frac{1}{6}$

The amount he will receive is Re. 1

2.. Does not gets six in the first throw and gets six in the second throw, then the probability

$=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$

The amount he will receive is - Re.1+ Re.1=0

3. Does not gets six in first 2 throws and gets six in the third throw, then the probability

$=\frac{5}{6}\times\frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$

Amount he will receive is -Re.1 - Re.1+ Re.1= -1

Expected value he can win :

$=1\times \frac{1}{6}+0\times \frac{5}{36}+(-1)\times \frac{25}{216}$

$= \frac{1}{6}-\frac{25}{216}$

$= \frac{11}{216}$

'

Let R be the event of drawing red marble.

Let $E_A,E_B,E_C$ respectively denote the event of selecting box A, B, C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box A is $P(E_A|R)$

$P(E_A|R)=\frac{P(E_A\cap R)}{P(R)}$

$=\frac{\frac{1}{40}}{\frac{3}{8}}$

$=\frac{1}{15}$

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box B is $P(E_B|R)$

$P(E_B|R)=\frac{P(E_B\cap R)}{P(R)}$

$=\frac{\frac{6}{40}}{\frac{3}{8}}$

$=\frac{2}{5}$

Let R be event of drawing red marble.

Let $E_A,E_B,E_C$ respectivly denote event of selecting box A,B,C.

Total marbles = 40

Red marbles =15

$P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing red marble from box C is $P(E_C|R)$

$P(E_C|R)=\frac{P(E_C\cap R)}{P(R)}$

$=\frac{\frac{8}{40}}{\frac{3}{8}}$

$=\frac{8}{15}$

Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted

the drug prescription.

$\therefore \, \, \, \, P(A)=0.40$

$\therefore \, \, \, \, P(E1)=P(E2)=\frac{1}{2}$

$P(A|E1)=0.40\times 0.70=0.28$

$P(A|E2)=0.40\times 0.75=0.30$

the probability that the patient followed a course of meditation and yoga is $P(E1,A)$

$P(E1,A)=\frac{P(E1).P(E1|A)}{P(E1).P(E1|A)+P(E2).P(E2|A)}$

$P(E1,A)=\frac{0.5\times 0.28}{0.5\times 0.28 + 0.5\times 0.30}$

$=\frac{14}{29}$

Total number of determinant of second order with each element being 0 or 1 is $2^4=16$

The values of determinant is positive in the following cases $\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 0\\ 1 & 1\end{bmatrix}$

Probability is

$=\frac{3}{16}$

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_A\, and \, E_B)=0.15$

$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$

$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$

$\Rightarrow \, \, \, P(E_B)=0.3$

$P(E_A|E_B)=\frac{P(E_A\cap E_B)}{P(E_B)}$

$=\frac{0.15}{0.3}=0.5$

Let event in which A fails and B fails be $E_A,E_B$

$P(E_A)=0.2$

$P(E_A\, and \, E_B)=0.15$

$P(B\, fails\, alone)=P(E_B)-P(E_A\, and\, E_B)$

$\Rightarrow \, \, \, 0.15=P(E_B)-0.15$

$\Rightarrow \, \, \, P(E_B)=0.3$

$P(A\, fails\, \, alone)=P(E_A)-P(E_A\, and\, E_B)$

$=0.2-0.15=0.05$

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.

$P(E1)=\frac{3}{7}$ and $P(E2)=\frac{4}{7}$

Let A be the event that ball drawn is red.

When a red ball is transfered from bag 1 to bag 2.

$P(A|E1)=\frac{5}{10}=\frac{1}{2}$

When a black ball is transfered from bag 1 to bag 2.

$P(A|E2)=\frac{4}{10}=\frac{2}{5}$

$P(E2|A)=\frac{P(E2).P(A|E2)}{P(E2).P(A|E2)+P(E1).P(A|E1)}$

$=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{4}{7}\times \frac{2}{5}+\frac{3}{7}\times \frac{1}{2}}$

$=\frac{16}{31}$

A and B are two events such that $\inline P(A\neq 0)$ and $\inline P(B\mid A)=1,$

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

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

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

$\Rightarrow \, \, \, A\subset B$

Option A is correct.

$\inline P(A\mid B)> P(A)$

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

$\Rightarrow \, \, P(A\cap B)> P(A).P(B)$

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

$\Rightarrow \, \, P(B|A)> P(B)$

Option C is correct.

$\inline P(A)+P(B)-P(A \; and\; B)=P(A),$

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

$\Rightarrow \, \, \, P(B)-P(A\cap B)=0$

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

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

Option B is correct.

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

If you are looking for probability class 12 ncert solutions of exercise then they are listed below.

### Class 12 Maths Chapter 13 NCERT solutions: Insight

• Generally, two questions( 8 marks) are asked from this chapter in 12th board final examination. You can score these 8 marks very easily with the help of probability Class 12 ncert solutions chapter 13.

• In the NCERT textbook there are 37 solved examples are given, so you can understand the concept easily. In this chapter 13 12 th class, there is a total of 81 questions in 5 exercises. You should try to solve every question given in chapter 13 class 12 maths on your own.

• If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 13 probability. These probability class 12 NCERT questions are solved and explained in a step-by-step manner, so it can be understood very easily.

• This chapter 13 12 th class requires lots of practice to understand it better. So, you are advised to solve miscellaneous exercise of probability class 12 ncert solutions.

## NCERT class 12 maths chapter 13 Solutions Probability - Topics

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 Wise

 chapter 1 NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions chapter 2 NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions chapter 3 NCERT solutions for class 12 maths chapter 3 Matrices chapter 4 NCERT solutions for class 12 maths chapter 4 Determinants chapter 5 NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability chapter 6 NCERT solutions for class 12 maths chapter 6 Application of Derivatives chapter 7 NCERT solutions for class 12 maths chapter 7 Integrals chapter 8 NCERT solutions for class 12 maths chapter 8 Application of Integrals chapter 9 NCERT solutions for class 12 maths chapter 9 Differential Equations chapter 10 NCERT solutions for class 12 maths chapter 10 Vector Algebra chapter 11 NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry chapter 12 NCERT solutions for class 12 maths chapter 12 Linear Programming chapter 13 NCERT solutions for class 12 maths chapter 13 Probability
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## Key Features of NCERT Solutions for Class 12 Maths Chapter 13 Probability

Comprehensive explanations: The probability ncert class 12 solutions provided in NCERT Solutions for Class 12 Maths Chapter 13 Probability are explained in a comprehensive and step-by-step manner. This helps students to understand the concepts better and makes it easy for them to solve similar problems.

Easy to understand: The probability solutions class 12 are written in simple language, making it easy for students to understand and learn the concepts. The solutions are designed to cater to the needs of students of all learning levels.

Covers all the topics: The class 12 probability solutions cover all the topics in Chapter 13 Probability of Class 12 Maths. This helps students to have a comprehensive understanding of the chapter.

Exercise-wise solutions: The class 12 chapter 13 maths solutions are provided exercise-wise, which helps students to focus on specific problems and concepts that they find difficult.

Examples and illustrations: The ch 13 maths class 12 solutions include examples and illustrations to explain the concepts and solutions better. These examples help students to understand the applications of probability in real-life situations.

## NCERT Books and NCERT Syllabus

Benefits of NCERT solutions

• As this chapter has 10% weightage in 12th board final exam. NCERT solutions for class 12 maths chapter 13 probability will help you to score good marks in the final exam.
• NCERT solutions for Class 12 Maths Chapter 13 are very easy to understand as these are prepared and explained in a detailed manner.

• At the end of every chapter, there is an additional exercise called Miscellaneous exercise which is very important for you if you wish to develop a grip on the concepts. In NCERT solutions for class 12 maths chapter 13 probability, you will get solutions for miscellaneous exercise too.

• These NCERT solutions for Class 12 Maths Chapter 13 PDF download are prepared with different approaches so it will give you new ways of solving the problems.

• NCERT solutions for class 12 maths chapter 13 probability are prepared and explained by the experts who know how best to answer the questions in the board exam. So, it will help you to score good marks in the exam.

## NCERT Exemplar Class 12 Solutions

1. What are the important topics in chapter probability?

Basic probability, conditional probability, properties of conditional probability, multiplication theorem on probability, independent events, Bayes' theorem, random variables, and its probability distributions, Bernoulli trials, and Binomial distribution are important topics of this chapter.

2. Does CBSE provides the solutions of NCERT for class 12 maths Chapter 13?

No, CBSE doesn’t provide NCERT solutions for any class or subject. but there are so many coaching institutions which provide solutions freely. if you are interested then you can download these from careers360 official website.

3. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Students consider integration and application of integration as the most difficult chapters in CBSE class 12 maths but with rigorous practice, you will get conceptual clarity and will be able to have a strong grip on them also.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

• Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

• Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

• Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

• Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

• Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

• JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
• NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

• Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
• NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
• Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

• High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

• Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

• Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

• Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

• Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9