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NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 - Probability

NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 - Probability

#CBSE Class 12th
Updated on 04 Dec 2023, 11:30 AM IST

NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.5

NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 Probability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In this article, you will get NCERT solutions for exercise 13.5 Class 12 Maths chapter 13 prepared by subject matter experts who know how to write answers in board exams. Some important concepts like Bernoulli Trials and Binomial Distribution in probability are covered in this exercise 13.5 Class 12 Maths solutions. There are 3 examples related to Bernoulli trials and Binomial distribution are given in the NCERT textbook. First, try to solve these examples, you will get to know about Bernoulli trials. There are 15 questions given in the NCERT textbook exercise 13.5. You should try to solve all the NCERT Maths syllabus problems on your own. If you find difficulties while solving these problems, you can take help from Class 12 Maths chapter 13 exercise 13.5 solutions.

12th class Maths exercise 13.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Probability Class 12 Chapter 13-Exercise: 13.5

Question:1(i) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

5 successes?

Answer:

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.

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

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

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

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

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

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

$= \frac{3}{32}$

Question:1(ii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

at least 5 successes?

Answer:

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.

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

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

$P(X=x)=^6C_6-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})$

$= \frac{7}{64}$

Question:1(iii) A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of

at most 5 successes?

Answer:

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.

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

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

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

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

$=1-P(X> 5)$

$=1-P(X= 5)$

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

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

$= \frac{63}{64}$

Question:2 A pair of dice is thrown $4$ times. If getting a doublet is considered a success, find the probability of two successes

Answer:

A pair of dice is thrown $4$ 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

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

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

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

Question:3 There are $5^{o}/_{o}$ defective items in a large bulk of items. What is the probability that a sample of $10$ items will include not more than one defective item?

Answer:

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}$

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

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

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

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

Question:4(i) Five cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. What is the probability that

all the five cards are spades?

Answer:

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 $52$ cards.

We have 13 spades.

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

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

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}$

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

$= \frac{1}{1024}$

Question:4(ii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

only 3 cards are spades?

Answer:

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 $52$ cards.

We have 13 spades.

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

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

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}$

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

$=\frac{45}{512}$

Question:4(iii) Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

none is a spade?

Answer:

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

We have 13 spades .

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

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

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}$

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

$= \frac{243}{1024}$

Question:5(i) The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05.$. Find the probability that out of $5$ such bulbs

none will fuse after $150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ 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}$

$=(0.95)^{5}$

Question:5(ii) The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05.$ Find the probability that out of $5$ such bulbs

not more than one will fuse after $150$ days of use.

Answer:

Let X represent a number of the bulb that will fuse after $150$ 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}$

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

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

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

Question:5(iii) The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05.$ Find the probability that out of $5$ such bulbs

more than one will fuse after$150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ 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})$

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

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

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

Question:5(iv) The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. Find the probability that out of $5$ such bulbs

at least one will fuse after $150$ days of use.

Answer:

Let X represent number of bulb that will fuse after $150$ 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}$

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

Question:6 A bag consists of $10$ balls each marked with one of the digits $0$ to $9.$If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit $0$?

Answer:

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}$

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

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

Question:7 In an examination,$20$ questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.

Answer:

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

The coin falls heads, he answers 'true'; if it falls tails, he answers 'false'.

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

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

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}$

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

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

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

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

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

Question:8 Suppose X has a binomial distribution $B\left [ 6,\frac{1}{2} \right ].$ Show that $X=3$ is the most likely outcome.

(Hint : $P(X=3)$ is the maximum among all of $P(x_{i})$ ,$x_{i}=0,1,2,3,4,5,6$ )

Answer:

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.

Question:9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

Answer:

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}$

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

$=\frac{11}{243}$

Question:10(a) A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $\frac{1}{100}.$ What is the probability that he will win a prize

at least once

Answer:

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}$

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

$=1-P(X< 1)$

$=1-P(X= 0)$

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

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

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

Question:10(b) A person buys a lottery ticket in $50$ lotteries, in each of which his chance of winning a prize is $\frac{1}{100}$. What is the probability that he will win a prize

exactly once

Answer:

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}$

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

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

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

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

Question:10(c) A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\frac{1}{100}$ . What is the probability that he will win a prize

at least twice?

Answer:

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}$

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

$=1-P(X< 2)$

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

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

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

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

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

Question:11 Find the probability of getting $5$ exactly twice in $7$ throws of a die.

Answer:

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}$

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

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

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

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

Question:12 Find the probability of throwing at most $2$ sixes in $6$ throws of a single die.

Answer:

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}$

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

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

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

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

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

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

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

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

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

Question:13 It is known that $10^{o}/_{o}$ of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles,$9$ are defective?

Answer:

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}$

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

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

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

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

Question:14 In a box containing $100$ bulbs, $10$ are defective. The probability that out of a sample of $5$ bulbs, none is defective is

(A) $10^{-1}$

(B) $\left ( \frac{1}{5} \right )^{5}$

(C) $\left ( \frac{9}{10} \right )^{5}$

(D) $\frac{9}{10}$

Answer:

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}$

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

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

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

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

The correct answer is C.

Question:15 The probability that a student is not a swimmer is $\frac{1}{5}.$ Then the probability that out of five students, four are swimmers is

In the following, choose the correct answer:

(A) $^{5}C_{4}\left ( \frac{4}{5} \right )^{4}\frac{1}{5}$

(B) $\left ( \frac{4}{5} \right )^{4}\frac{1}{5}$

(C) $^{5}C_{1}\frac{1}{5}\left ( \frac{4}{5} \right )^{4}$

(D) None of these

Answer:

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.

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More About NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.5:-

Bernoulli trials is a very important concept for the finding probability of independents trials where outcomes are only success or failure. In exercise 13.4 Class 12 Maths you will get questions related to Bernoulli Trials and Binomial Distribution. There are 3 examples and 15 questions in this exercise. You must be thorough with Class 12 Maths ch 13 ex 13.5 in order to perform well in board exams. Basic knowledge of permutations and combinations is also required for this exercise.

Also Read| Probability Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.5:-

  • NCERT solutions for Class 12 Maths chapter 13 Exercise 13.5 are very helpful for the students to get conceptual clarity as all the questions are solved in a step-by-step manner.
  • Class 12th Maths chapter 13 exercise 13.5 solutions are also helpful for the students to revise important concepts.
  • NCERT solutions for Class 12 Maths chapter 13 exercise 13.5 are useful for the students who are stuck with these problems.
  • You don't need to buy any other reference book as CBSE mostly asks questions from the NCERT textbook in the board exams.
  • You must be thorough with the NCERT textbook, you can take NCERT solutions for Class 12 Maths chapter 13 exercise 13.5 as reference.
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Key Features Of NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 13.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 13.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 13.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 13.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 13.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 13.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

Q: What is the Probability Theory ?
A:

Probability theory is a branch of mathematics that deals with the numerical measurement of the degree of uncertainty.

Q: What is the importance of probability theory ?
A:

Probability theory is useful in Meteorologists, weather prediction, health, sports, gambling, etc.  

Q: What is the probabilty of a certain event ?
A:

The probability of a certain event is 1.

Q: what is probability of getting head when a fair coin is tossed ?
A:

The probability of getting head is 0.5 when a fair coin is tossed.

Q: what is probability of getting two consecutive heads when a fair coin is tossed two times ?
A:

The probability of getting two consecutive heads when a fair coin is tossed two times is 0.25.

Q: If the probability of getting tail is 0.6 when a biased coin is tossed then what is probability of getting head ?
A:

Probability of getting Head = 1 - Probability of  getting tail

  = 1 - 0.6

= 0.4

Q: Can I get NCERT solutions for Class 10 Maths ?
A:

Yes, click here to get NCERT Solutions for Class 10 Maths.

Q: Can I get NCERT solutions for Class 10 Maths probability ?
A:
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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?
how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer
G
gopi
4 Sep'25
Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

  • B.Tech in Computer Science Engineering (CSE) – most popular choice.

  • BCA (Bachelor of Computer Applications) – good for software and IT jobs.

  • B.Sc. Computer Science / IT – good for higher studies and research.

  • B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer
P
Prachi Kumari
4 Sep'25
I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

Read Complete Answer
G
gopi
3 Sep'25
I gave my 12th exam this year in which I got less marks. Then I gave improvement exam in which also I could not score full marks. So can I become a private candidate and appear for the exam again next year or not?

Hello

Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

Read Complete Answer
S
SEJAL JAIN
20 Aug'25
i gave improvement 2025 as the roll no is different from my previous year marksheet so will i get new migration certificate

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Read Complete Answer
G
gopi
12 Aug'25
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Popular CBSE Class 12th Questions

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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read Complete Answer

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

 
Read Complete Answer

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)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read Complete Answer

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

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read Complete Answer

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read Complete Answer

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
Read Complete Answer

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.
Read Complete Answer

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.
Read Complete Answer

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
Read Complete Answer

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
Read Complete Answer

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