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

Q: What is the Probability Theory ?

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 ?

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

Q: What is the probabilty of a certain event ?

The probability of a certain event is 1.

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

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 ?

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 ?

Probability of getting Head = 1 - Probability of getting tail = 1 - 0.6 = 0.4

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

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

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

Click on the link to get NCERT Solutions for Class 10 Maths Probability .

Edited By Ravindra Pindel | Updated on Jul 5, 2022 - 2:59 p.m. IST

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. You can also check for NCERT Solutions.

Also, see

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

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

(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.

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.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 - Probability

Question: What is the Probability Theory ?

Answer:

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

Question: What is the importance of probability theory ?

Answer:

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

Question: What is the probabilty of a certain event ?

Answer:

The probability of a certain event is 1.

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

Answer:

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

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

Answer:

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

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

Answer:

Probability of getting Head = 1 - Probability of getting tail

= 1 - 0.6

= 0.4

Question: Can I get NCERT solutions for Class 10 Maths ?

Answer:

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

Question: Can I get NCERT solutions for Class 10 Maths probability ?

Answer:

Click on the link to get NCERT Solutions for Class 10 Maths Probability.

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

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NCERT Solutions

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

Probability Class 12 Chapter 13-Exercise: 13.5

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

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

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Exercise 13.5 Class 12 Maths Chapter 13 - Probability

Question: What is the Probability Theory ?

Question: What is the importance of probability theory ?

Question: What is the probabilty of a certain event ?

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

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

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

Question: Can I get NCERT solutions for Class 10 Maths ?

Question: Can I get NCERT solutions for Class 10 Maths probability ?

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