NCERT Solutions Class 12 Mathematics Part II . 674 Pages (668-1342). Free Flip-Book by Study Innovations

Class XII Chapter 13 – Probability Maths
X 012
P (X)

Question 8:
A random variable X has the following probability distribution.

X 01 2 3 4 5 6 7

P (X) 0 k 2k 2k 3k k2 2k2 7k2 + k

Determine
(i) k
(ii) P (X < 3)
(iii) P (X > 6)
(iv) P (0 < X < 3)
Answer
(i) It is known that the sum of probabilities of a probability distribution of random
variables is one.

k = − 1 is not possible as the probability of an event is never negative.
∴
(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

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(iii) P (X > 6) = P (X = 7)

(iv) P (0 < X < 3) = P (X = 1) + P (X = 2)

Question 9:
The random variable X has probability distribution P(X) of the following form, where k is
some number:

(a) Determine the value of k.
(b) Find P(X < 2), P(X ≥ 2), P(X ≥ 2).

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Class XII Chapter 13 – Probability Maths

Answer
(a) It is known that the sum of probabilities of a probability distribution of random
variables is one.

∴ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒k=
(b) P(X < 2) = P(X = 0) + P(X = 1)

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Class XII Chapter 13 – Probability Maths

Question 10:
Find the mean number of heads in three tosses of a fair coin.
Answer
Let X denote the success of getting heads.
Therefore, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that X can take the value of 0, 1, 2, or 3.

∴ P (X = 1) = P (HHT) + P (HTH) + P (THH)

∴P(X = 2) = P (HHT) + P (HTH) + P (THH)
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Class XII Chapter 13 – Probability Maths

Therefore, the required probability distribution is as follows.
X 0123
P(X)

Mean of X E(X), µ =

Question 11:
Two dice are thrown simultaneously. If X denotes the number of sixes, find the
expectation of X.
Answer
Here, X represents the number of sixes obtained when two dice are thrown
simultaneously. Therefore, X can take the value of 0, 1, or 2.
∴ P (X = 0) = P (not getting six on any of the dice) =

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Class XII Chapter 13 – Probability Maths

P (X = 1) = P (six on first die and no six on second die) + P (no six on first die and six
on second die)

P (X = 2) = P (six on both the dice) =
Therefore, the required probability distribution is as follows.

X 012
P(X)

Then, expectation of X = E(X) =

Question 12:
Two numbers are selected at random (without replacement) from the first six positive
integers. Let X denotes the larger of the two numbers obtained. Find E(X).
Answer
The two positive integers can be selected from the first six positive integers without
replacement in 6 × 5 = 30 ways
X represents the larger of the two numbers obtained. Therefore, X can take the value of
2, 3, 4, 5, or 6.
For X = 2, the possible observations are (1, 2) and (2, 1).

For X = 3, the possible observations are (1, 3), (2, 3), (3, 1), and (3, 2).

For X = 4, the possible observations are (1, 4), (2, 4), (3, 4), (4, 3), (4, 2), and (4, 1).
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Class XII Chapter 13 – Probability Maths

For X = 5, the possible observations are (1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5,
2), and (5, 1).

For X = 6, the possible observations are (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 4), (6,
3), (6, 2), and (6, 1).

Therefore, the required probability distribution is as follows.
X 2 3456

P(X)

Question 13:
Let X denotes the sum of the numbers obtained when two fair dice are rolled. Find the
variance and standard deviation of X.
Answer
When two fair dice are rolled, 6 × 6 = 36 observations are obtained.

P(X = 2) = P(1, 1) =

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Class XII Chapter 13 – Probability Maths

P(X = 3) = P (1, 2) + P(2, 1) =

P(X = 4) = P(1, 3) + P(2, 2) + P(3, 1) =

P(X = 5) = P(1, 4) + P(2, 3) + P(3, 2) + P(4, 1) =

P(X = 6) = P(1, 5) + P (2, 4) + P(3, 3) + P(4, 2) + P(5, 1) =

P(X = 7) = P(1, 6) + P(2, 5) + P(3, 4) + P(4, 3) + P(5, 2) + P(6, 1)

P(X = 8) = P(2, 6) + P(3, 5) + P(4, 4) + P(5, 3) + P(6, 2) =

P(X = 9) = P(3, 6) + P(4, 5) + P(5, 4) + P(6, 3) =

P(X = 10) = P(4, 6) + P(5, 5) + P(6, 4) =

P(X = 11) = P(5, 6) + P(6, 5) =

P(X = 12) = P(6, 6) =
Therefore, the required probability distribution is as follows.

X 2 3 4 5 6 7 8 9 10 11 12
P(X)

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Class XII Chapter 13 – Probability Maths

Question 14:
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17,
16, 19 and 20 years. One student is selected in such a manner that each has the same
chance of being chosen and the age X of the selected student is recorded. What is the
probability distribution of the random variable X? Find mean, variance and standard
deviation of X.
Answer

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Class XII Chapter 13 – Probability Maths

There are 15 students in the class. Each student has the same chance to be chosen.

Therefore, the probability of each student to be selected is
.

The given information can be compiled in the frequency table as follows.

X 14 15 16 17 18 19 20 21

f2 1 2 3 1 2 3 1

P(X = 14) = , P(X = 15) = , P(X = 16) = , P(X = 16) = ,

P(X = 18) = , P(X = 19) = , P(X = 20) = , P(X = 21) =
Therefore, the probability distribution of random variable X is as follows.
X 14 15 16 17 18 19 20 21
f

Then, mean of X = E(X)

E(X2) =

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Class XII Chapter 13 – Probability Maths

Question 15:
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A
member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in
favour. Find E(X) and Var(X).
Answer

It is given that P(X = 0) = 30% =

Therefore, the probability distribution is as follows.
X 01

P(X) 0.3 0.7

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Class XII Chapter 13 – Probability Maths

It is known that, Var (X) =
= 0.7 − (0.7)2
= 0.7 − 0.49
= 0.21

Question 16:
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2
on two faces and 5 on one face is

(A) 1 (B) 2 (C) 5 (D)
Answer
Let X be the random variable representing a number on the die.
The total number of observations is six.

Therefore, the probability distribution is as follows.
X 125

P(X)

Mean = E(X) =

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The correct answer is B.
Question 17:
Suppose that two cards are drawn at random from a deck of cards. Let X be the number
of aces obtained. Then the value of E(X) is
(A) (B) (C) (D)
Answer
Let X denote the number of aces obtained. Therefore, X can take any of the values of 0,
1, or 2.
In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

∴ P (X = 0) = P (0 ace and 2 non-ace cards) =

P (X = 1) = P (1 ace and 1 non-ace cards) =

P (X = 2) = P (2 ace and 0 non- ace cards) =
Thus, the probability distribution is as follows.

X0 1 2

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Class XII Chapter 13 – Probability Maths
P(X)

Then, E(X) =

Therefore, the correct answer is D.

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Class XII Chapter 13 – Probability Maths
Exercise 13.5

Question 1:
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes? (ii) at least 5 successes?
(iii) at most 5 successes?
Answer
The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of
getting odd numbers in an experiment of 6 trials.

Probability of getting an odd number in a single throw of a die is,

X has a binomial distribution.
Therefore, P (X = x) =

(i) P (5 successes) = P (X = 5)

(ii) P(at least 5 successes) = P(X ≥ 5)

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Class XII Chapter 13 – Probability Maths

(iii) P (at most 5 successes) = P(X ≤ 5)

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
The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of
times of getting doublets in an experiment of throwing two dice simultaneously four
times.
Probability of getting doublets in a single throw of the pair of dice is

Clearly, X has the binomial distribution with n = 4,

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Class XII Chapter 13 – Probability Maths

∴ P (2 successes) = P (X = 2)

Question 3:
There are 5% 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
Let X denote the number of defective items in a sample of 10 items drawn successively.
Since the drawing is done with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 10 and
P(X = x) =

P (not more than 1 defective item) = P (X ≤ 1)

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Class XII Chapter 13 – Probability Maths

Question 4:
Five cards are drawn successively with replacement from a well-shuffled deck of 52
cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?
Answer
Let X represent the number of spade cards among the five cards drawn. Since the
drawing of card is with replacement, the trials are Bernoulli trials.
In a well shuffled deck of 52 cards, there are 13 spade cards.

X has a binomial distribution with n = 5 and

(i) P (all five cards are spades) = P(X = 5)
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Class XII Chapter 13 – Probability Maths

(ii) P (only 3 cards are spades) = P(X = 3)
(iii) P (none is a spade) = P(X = 0)

Question 5:
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05.
What is the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one
will fuse after 150 days of use.
Answer
Let X represent the number of bulbs that will fuse after 150 days of use in an experiment
of 5 trials. The trials are Bernoulli trials.
It is given that, p = 0.05

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Class XII Chapter 13 – Probability Maths

X has a binomial distribution with n = 5 and p = 0.05

(i) P (none) = P(X = 0)

(ii) P (not more than one) = P(X ≤ 1)

(iii) P (more than 1) = P(X > 1)
(iv) P (at least one) = P(X ≥ 1)

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Class XII Chapter 13 – Probability Maths

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 the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 4 and

P (none marked with 0) = P (X = 0)

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.

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Class XII Chapter 13 – Probability Maths

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the
true answer and “tail” represents the false answer, the correctly answered questions are
Bernoulli trials.

∴p=

X has a binomial distribution with n = 20 and p =

P (at least 12 questions answered correctly) = P(X ≥ 12)

Question 8:

Suppose X has a binomial distribution . Show that X = 3 is the most likely

outcome.

(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)
Answer

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Class XII Chapter 13 – Probability Maths

X is the random variable whose binomial distribution is .
Therefore, n = 6 and

It can be seen that P(X = x) will be maximum, if will be maximum.

The value of is maximum. Therefore, for x = 3, P(X = x) is maximum.
Thus, X = 3 is the most likely outcome.

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
The repeated guessing of correct answers from multiple choice questions are Bernoulli
trials. Let X represent the number of correct answers by guessing in the set of 5 multiple
choice questions.

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Class XII Chapter 13 – Probability Maths

Probability of getting a correct answer is, p

Clearly, X has a binomial distribution with n = 5 and p

P (guessing more than 4 correct answers) = P(X ≥ 4)

Question 10:
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a

prize is . What is the probability that he will in a prize (a) at least once (b) exactly
once (c) at least twice?
Answer
Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli
trials.

Clearly, X has a binomial distribution with n = 50 and

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Class XII Chapter 13 – Probability Maths

(a) P (winning at least once) = P (X ≥ 1)

(b) P (winning exactly once) = P(X = 1)
(c) P (at least twice) = P(X ≥ 2)

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Class XII Chapter 13 – Probability Maths

Question 11:
Find the probability of getting 5 exactly twice in 7 throws of a die.
Answer
The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of
getting 5 in 7 throws of the die.
Probability of getting 5 in a single throw of the die, p

Clearly, X has the probability distribution with n = 7 and p

P (getting 5 exactly twice) = P(X = 2)

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Class XII Chapter 13 – Probability Maths

Question 12:
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Answer
The repeated tossing of the die are Bernoulli trials. Let X represent the number of times
of getting sixes in 6 throws of the die.

Probability of getting six in a single throw of die, p

Clearly, X has a binomial distribution with n = 6

P (at most 2 sixes) = P(X ≤ 2)

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Class XII Chapter 13 – Probability Maths

Question 13:
It is known that 10% of certain articles manufactured are defective. What is the
probability that in a random sample of 12 such articles, 9 are defective?
Answer
The repeated selections of articles in a random sample space are Bernoulli trails. Let X
denote the number of times of selecting defective articles in a random sample space of
12 articles.

Clearly, X has a binomial distribution with n = 12 and p = 10% =

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Class XII Chapter 13 – Probability Maths

P (selecting 9 defective articles) =

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)

(C)
(D)
Answer
The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote
the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb, p

Clearly, X has a binomial distribution with n = 5 and

P (none of the bulbs is defective) = P(X = 0)

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Class XII Chapter 13 – Probability Maths

The correct answer is C.
Question 15:
The probability that a student is not a swimmer is . Then the probability that out of five
students, four are swimmers is

(A) (B)

(C) (D) None of these
Answer
The repeated selection of students who are swimmers are Bernoulli trials. Let X denote
the number of students, out of 5 students, who are swimmers.
Probability of students who are not swimmers, q

Clearly, X has a binomial distribution with n = 5 and

P (four students are swimmers) = P(X = 4)
Therefore, the correct answer is A.

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Class XII Chapter 13 – Probability Maths

Miscellaneous Solutions

Question 1:
A and B are two events such that P (A) ≠ 0. Find P (B|A), if
(i) A is a subset of B (ii) A ∩ B = Φ
Answer
It is given that, P (A) ≠ 0
(i) A is a subset of B.

(ii)

Question 2:
A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the
children is male.
(ii) Find the probability that both children are females, if it is known that the elder child
is a female.
Answer
If a couple has two children, then the sample space is
S = {(b, b), (b, g), (g, b), (g, g)}
(i) Let E and F respectively denote the events that both children are males and at least
one of the children is a male.

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Class XII Chapter 13 – Probability Maths

(ii) Let A and B respectively denote the events that both children are females and the
elder child is a female.

Question 3:
Suppose that 5% of men and 0.25% of women have grey hair. A haired person is
selected at random. What is the probability of this person being male?
Assume that there are equal number of males and females.
Answer
It is given that 5% of men and 0.25% of women have grey hair.
Therefore, percentage of people with grey hair = (5 + 0.25) % = 5.25%

∴ Probability that the selected haired person is a male

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Class XII Chapter 13 – Probability Maths

Question 4:
Suppose that 90% of people are right-handed. What is the probability that at most 6 of a
random sample of 10 people are right-handed?
Answer
A person can be either right-handed or left-handed.
It is given that 90% of the people are right-handed.

Using binomial distribution, the probability that more than 6 people are right-handed is
given by,

Therefore, the probability that at most 6 people are right-handed
= 1 − P (more than 6 are right-handed)

Question 5:
An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a
mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is
replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear ‘X’ mark.
(ii) not more than 2 will bear ‘Y’ mark.
(iii) at least one ball will bear ‘Y’ mark
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.
Answer
Total number of balls in the urn = 25
Balls bearing mark ‘X’ = 10
Balls bearing mark ‘Y’ = 15

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Class XII Chapter 13 – Probability Maths

p = P (ball bearing mark ‘X’) =

q = P (ball bearing mark ‘Y’) =
Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.
Let Z be the random variable that represents the number of balls with ‘Y’ mark on them
in the trials.

Clearly, Z has a binomial distribution with n = 6 and p = .
∴ P (Z = z) =

(i) P (all will bear ‘X’ mark) = P (Z = 0) =

(ii) P (not more than 2 bear ‘Y’ mark) = P (Z ≤ 2)
= P (Z = 0) + P (Z = 1) + P (Z = 2)

(iii) P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z = 0)
(iv) P (equal number of balls with ‘X’ mark and ‘Y’ mark) = P (Z = 3)

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Class XII Chapter 13 – Probability Maths
=

Question 6:
In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each

hurdle is . What is the probability that he will knock down fewer than 2 hurdles?
Answer
Let p and q respectively be the probabilities that the player will clear and knock down
the hurdle.

Let X be the random variable that represents the number of times the player will knock
down the hurdle.
Therefore, by binomial distribution, we obtain
P (X = x) =
P (player knocking down less than 2 hurdles) = P (X < 2)
= P (X = 0) + P (X = 1)
=

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Class XII Chapter 13 – Probability Maths

Question 7:
A die is thrown again and again until three sixes are obtained. Find the probability of
obtaining the third six in the sixth throw of the die.
Answer
The probability of getting a six in a throw of die is and not getting a six is .
Let
The probability that the 2 sixes come in the first five throws of the die is

∴ Probability that third six comes in the sixth throw =

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Class XII Chapter 13 – Probability Maths

Question 8:
If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Answer
In a leap year, there are 366 days i.e., 52 weeks and 2 days.
In 52 weeks, there are 52 Tuesdays.
Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the
probability that the remaining 2 days will be Tuesdays.
The remaining 2 days can be
Monday and Tuesday
Tuesday and Wednesday
Wednesday and Thursday
Thursday and Friday
Friday and Saturday
Saturday and Sunday
Sunday and Monday
Total number of cases = 7
Favourable cases = 2

∴Probability that a leap year will have 53 Tuesdays =

Question 9:
An experiment succeeds twice as often as it fails. Find the probability that in the next six
trials, there will be at least 4 successes.
Answer
The probability of success is twice the probability of failure.
Let the probability of failure be x.
∴ Probability of success = 2x

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Class XII Chapter 13 – Probability Maths

Let p = and q =
Let X be the random variable that represents the number of successes in six trials.
By binomial distribution, we obtain

P (X = x) =
Probability of at least 4 successes = P (X ≥ 4)
= P (X = 4) + P (X = 5) + P (X = 6)

Question 10:
How many times must a man toss a fair coin so that the probability of having at least
one head is more than 90%?
Answer
Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability (p) of getting a head at the toss of a coin is .

∴p= ∴q=

It is given that,

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Class XII Chapter 13 – Probability Maths

P (getting at least one head) >
P (x ≥ 1) > 0.9
∴ 1 − P (x = 0) > 0.9

The minimum value of n that satisfies the given inequality is 4.
Thus, the man should toss the coin 4 or more than 4 times.

Question 11:
In a game, a man wins a rupee for a six and loses a rupee for any other number when a
fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets
a six. Find the expected value of the amount he wins/loses.
Answer

In a throw of a die, the probability of getting a six is and the probability of not getting

a 6 is
.

Three cases can occur.

i. If he gets a six in the first throw, then the required probability is .
Amount he will receive = Re 1

ii. If he does not get a six in the first throw and gets a six in the second throw, then

probability =

Amount he will receive = −Re 1 + Re 1 = 0

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Class XII Chapter 13 – Probability Maths

iii. If he does not get a six in the first two throws and gets a six in the third throw,

then probability =

Amount he will receive = −Re 1 − Re 1 + Re 1 = −1

Expected value he can win

+

Question 12:
Suppose we have four boxes. A, B, C and D containing coloured marbles as given below:

Box Marble colour

Red White Black

A1 6 3

B6 2 2

C8 1 1

D0 6 4

One of the boxes has been selected at random and a single marble is drawn from it. If
the marble is red, what is the probability that it was drawn from box A?, box B?, box C?
Answer
Let R be the event of drawing the red marble.
Let EA, EB, and EC respectively denote the events of selecting the box A, B, and C.
Total number of marbles = 40
Number of red marbles = 15

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Class XII Chapter 13 – Probability Maths

Probability of drawing the red marble from box A is given by P (EA|R).

Probability that the red marble is from box B is P (EB|R).

Probability that the red marble is from box C is P (EC|R).

Question 13:
Assume that the chances of the patient having a heart attack are 40%. It is also
assumed that a meditation and yoga course reduce the risk of heart attack by 30% and
prescription of certain drug reduces its chances by 25%. At a time a patient can choose
any one of the two options with equal probabilities. It is given that after going through
one of the two options the patient selected at random suffers a heart attack. Find the
probability that the patient followed a course of meditation and yoga?
Answer
Let A, E1, and E2 respectively denote the events that a person has a heart attack, the
selected person followed the course of yoga and meditation, and the person adopted the
drug prescription.

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Class XII Chapter 13 – Probability Maths

Probability that the patient suffering a heart attack followed a course of meditation and
yoga is given by P (E1|A).

Question 14:
If each element of a second order determinant is either zero or one, what is the
probability that the value of the determinant is positive? (Assume that the individual
entries of the determinant are chosen independently, each value being assumed with

probability ).
Answer
The total number of determinants of second order with each element being 0 or 1 is (2)4
= 16

The value of determinant is positive in the following cases.

∴ Required probability =

Question 15:
An electronic assembly consists of two subsystems, say, A and B. From previous testing
procedures, the following probabilities are assumed to be known:
P(A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities
(i) P(A fails| B has failed) (ii) P(A fails alone)

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Class XII Chapter 13 – Probability Maths

Answer
Let the event in which A fails and B fails be denoted by EA and EB.
P (EA) = 0.2
P (EA ∴ EB) = 0.15
P (B fails alone) = P (EB) − P (EA ∴ EB)
∴ 0.15 = P (EB) − 0.15
∴ P (EB) = 0.3

(i)

(ii) P (A fails alone) = P (EA) − P (EA ∴ EB)
= 0.2 − 0.15
= 0.05

Question 16:
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One
ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so
drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer
Let E1 and E2 respectively denote the events that a red ball is transferred from bag I to II
and a black ball is transferred from bag I to II.

Let A be the event that the ball drawn is red.
When a red ball is transferred from bag I to II,

When a black ball is transferred from bag I to II,

Page 101 of 103

Class XII Chapter 13 – Probability Maths

Question 17:
If A and B are two events such that P (A) ≠ 0 and P(B|A) = 1, then.
(A) A ∴ B
(B) B ∴ A
(C) B = Φ
(D) A = Φ
Answer
P (A) ≠ 0 and

Thus, the correct answer is A.
Question 18:
If P (A|B) > P (A), then which of the following is correct:
(A) P (B|A) < P (B) (B) P (A ∴ B) < P (A).P (B)
(C) P (B|A) > P (B) (D) P (B|A) = P (B)
Answer

Page 102 of 103

Class XII Chapter 13 – Probability Maths

Thus, the correct answer is C.
Question 19:
If A and B are any two events such that P (A) + P (B) − P (A and B) = P (A), then
(A) P (B|A) = 1 (B) P (A|B) = 1
(C) P (B|A) = 0 (D) P (A|B) = 0
Answer

Thus, the correct answer is B.

Page 103 of 103

Class XII www.ncrtsolutions.blogspot.com Maths
Question 1: Chapter 2 – Inverse Trigonometric Functions

Exercise 2.1

Find the principal value of
Answer

Let sin-1 Then sin y =

We know that the range of the principal value branch of sin−1 is

and sin

Therefore, the principal value of
Question 2:

Find the principal value of
Answer

We know that the range of the principal value branch of cos−1 is
.

Therefore, the principal value of .

Question 3:
Find the principal value of cosec−1 (2)
Answer

Page 1 of 38

Class XII Chapter 2 – Inverse Trigonometric Functions Maths

Let cosec−1 (2) = y. Then,
We know that the range of the principal value branch of cosec−1 is
Therefore, the principal value of
Question 4:
Find the principal value of
Answer

We know that the range of the principal value branch of tan−1 is

Therefore, the principal value of
Question 5:
Find the principal value of
Answer

We know that the range of the principal value branch of cos−1 is

Therefore, the principal value of
Page 2 of 38

Class XII Chapter 2 – Inverse Trigonometric Functions Maths

Question 6:
Find the principal value of tan−1 (−1)
Answer

Let tan−1 (−1) = y. Then,
We know that the range of the principal value branch of tan−1 is

Therefore, the principal value of
Question 7:
Find the principal value of
Answer

We know that the range of the principal value branch of sec−1 is

Therefore, the principal value of
Question 8:
Find the principal value of
Answer

Page 3 of 38

Class XII Chapter 2 – Inverse Trigonometric Functions Maths

We know that the range of the principal value branch of cot−1 is (0,π) and

Therefore, the principal value of
Question 9:
Find the principal value of
Answer

We know that the range of the principal value branch of cos−1 is [0,π] and
.

Therefore, the principal value of
Question 10:
Find the principal value of
Answer

We know that the range of the principal value branch of cosec−1 is

Therefore, the principal value of
Page 4 of 38

Class XII Chapter 2 – Inverse Trigonometric Functions Maths
Question 11:

Find the value of
Answer

Question 12:
Find the value of
Answer

Page 5 of 38