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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

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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Access NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.5

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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.
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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 ?

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

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

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.

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.

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

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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 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

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

Option 1)

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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2 Jobs Available
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