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

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

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

(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.

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 Question (FAQs)

1. What is the Probability Theory ?

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

2. What is the importance of probability theory ?

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

3. What is the probabilty of a certain event ?

The probability of a certain event is 1.

4. 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.

5. 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.

6. 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

7. Can I get NCERT solutions for Class 10 Maths ?

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

8. Can I get NCERT solutions for Class 10 Maths probability ?

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

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

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.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Read Complete Answer

khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

Read Complete Answer

Parvati Solanki 02 June,2022

View All

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) $0.67\; J$

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 3.8×10⁷J 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) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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) $K/4$

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 .

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

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

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If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

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Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

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Fashion Blogger

Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns.

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Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available

Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

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Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available

Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available

Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available

Videographer

Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

2 Jobs Available

SEO Analyst

An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.

2 Jobs Available

Product Manager

3 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

Resource Links for Online MBA

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QA Manager

Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

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Reliability Engineer

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available

Safety Manager

2 Jobs Available

Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

Product Manager

3 Jobs Available

ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available

.NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available

Corporate Executive

2 Jobs Available

DevOps Architect

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

2 Jobs Available

Cloud Solution Architect

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

2 Jobs Available