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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

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:30 AM IST | #CBSE Class 12th
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CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

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 after150 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.

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)

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 ?

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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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 More

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 More

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 More

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 More

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 More

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 More

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 More

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 More

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 More

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 More

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