NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

# NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:00 PM IST

NCERT Exemplar Class 11 Maths solutions chapter 16 discusses Probability and its applications in real life. Students who find the problems of Probability difficult or want to practice for the exams more can refer to NCERT Exemplar solutions for Class 11 Maths chapter 16 and follow the given steps as mentioned by our experts. The solutions are given in a step-by-step manner which enables the students to understand the solution easily. The students can practice and understand the steps demonstrated in the NCERT Exemplar Class 11 Maths solutions chapter 16 for their CBSE board examinations.

## NCERT Exemplar Class 11 Maths Solutions Chapter 16: Exercise: 1.3

Question:1

If the letters of the word ALGORITHM are arranged at random in a row what is the probability the letters GOR must remain together as a unit?

ALGORITHM ……. (given word)
Total no. of letters = 9
Thus, total no. of words = 9!
Thus, n(s) = 9!
Considering ‘GOR’ as one group –
A L GOR I T H M
↓ ↓ ↓ ↓ ↓ ↓ ↓
1 2 3 4 5 6 7
Thus, no. of letters = 7
Now, if the GOR group remains together, then the order = 7!
Thus, n (E) = 7!
Now, we know that,
Required probability = No. of favorable outcomes/ Total no. of outcomes
= n (E) / n(S) …..[Since n! = n x (n – 1) x (n – 2)…1]
= 7! / 9!
= 7! / 9 x 8 x 7!
= 1 / 72

Question:2

Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will have nonadjacent desks?

Given: total no. of employees = 6
They can be arranged in 6 ways,
Thus, n(S) = 6!
= 6 x 5 x 4 x 3 x 1
= 720
Now, there are 5 different ways to select two adjacent desks for married couples –
(1,2), (2,3), (3,4), (4,5), (5,6)
They can be arranged in 2! ways in the two desks & the other persons can be arranged in 4! Ways
Thus, the no. of ways = 5 x 2! X 4!
= 5 x 2 x 1 x 4 x 3 x 2 x 1
= 240
Thus,
The no. of ways in which married couples occupy non- adjacent desks
= 6! – 240
= 720 – 240
= 480
= n (E)
Required probability = No. of favorable outcomes/ Total no. of outcomes
= n (E) / n(S)
= 480 / 720
= 2 / 3

Question:3

Suppose an integer from 1 through 1000 is chosen at random, find the probability that the integer is a multiple of 2 or a multiple of 9.

Given: we have integers 1, 2, …….. , 1000
No. of outcomes, n(S) = 1000
No. of the integers that are multiples of 2 –
2, 4, 6, 8, …… , 1000.
Let us consider ‘p’ as the no. of integers,
Now, ap = a + (p – 1)d
On substituting the values, we get,
2 + (p – 1)2 = 1000
2 + 2p – 2 = 1000
Thus, p = 1000/ 2
Thus, p = 500
Thus, no. of the integers that are multiples of 2 = 500
No. of the integers that are multiples of 9 –
9, 18, 27, 35, …… , 999.
Let us consider ‘n’ as the no. of integers,
Now, an = a + (n – 1)d
On substituting the values, we get,
9 + (n – 1)9 = 999
9 + 9n – 9 = 999
Thus, n = 999/ 9
Thus, n = 111
Thus, no. of the integers that are multiples of 9 = 111
Now, let m be the no. of multiples common for both 2 & 9, viz. 18, 36, ……., 990.
Thus, the mth term will be 990
Now, am = a + (m – 1)d
We know that, a = 2 & d = 9
Substituting the respective values, we get,
18 + (m – 1)18 = 990
18 + 18m – 18 = 990
Thus, m = 990/18
Thus, m = 55
Now, the no. of multiples of 2 or 9 will be,
No. of multiples of 2 + no. of multiples of 9 – No. of multiples of both 2 & 9
= 500 + 111 – 55
= 556
= n(E)
Required probability = No. of favorable outcomes/ Total no. of outcomes
= n(E) / n(S)
= 556 / 1000
= 0.556

Question:4

An experiment consists of rolling a die until a 2 appears.

(i) How many elements of the sample space correspond to the event that the 2 appears on the kth roll of the die?
(ii) How many elements of the sample space correspond to the event that the 2 appears not later than the kth roll of the die?

We know that, the no. of outcomes when a die is thrown is 6

1. 2 appear on the kth roll of the die ……… (given)

Thus, the first (k – 1)th roll has 5 outcomes each
Thus, no. of outcomes = 5k-1

1. Let us consider that 2 come before the kth roll of the die and not after that.

Thus,
In the first roll, no. of ways in which 2 appears will be = 1 outcome
In the second roll, no. of ways in which 2 appears will be = 5 x 1 outcome
……. (since the first roll doesn’t result in 2)
In the third roll, no. of ways in which 2 appears will be = 5 x 5 x 1 outcome
……. (since the first two rolls doesn’t result in 2)
In the (k – 1)th roll, no. of ways in which 2 appear will be = [5 x 5 x 1 ….. (k – 1)] outcome
= 5k-1
Now, the possibility of 2 appearing before kth roll = 1 + 5 + 52 + 53 + …… + 5k-1
Now,

Thus, here, a = 1 & r = 5/1 = 5 >1
Thus,
= 1 x (5k – 1) / 5 – 1
= 5k – 1 / 4

Question:5

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Probability of odd nos. = 2 x (probability of even no.) ……….. (given)
Thus, P (Odd) = 2 x P (Even)
P(Odd) + P(Even) = 1
2P (Even) + P (Even) = 1
3P (Even) = 1
Thus, P (Even) = 1 / 3
Thus, P (Odd) = 1 – 1/3
= 3 – 1/ 3
= 2 / 3
Total no. occurring on a single roll = 6
& 4, 5 & 6 are the nos. greater than 3
Let P(no. greater than 3) = P (G)
= P (no. is 4, 5 or 6)
Here, 4 & 6 – Even & 5 – Odd
Thus, P (G) = 2 x P (Even) x P (Odd)
= 2 x 1/3 x 2/3
= 4/9
Therefore, 4/9 is the required probability.

Question:6

In a large metropolitan area, the probabilities are .87, .36, .30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Let us consider that,
E1 be the element that a family owns a colour TV & E2 be the event that a family owns black & white TV.
Now P(E1) = 0.87 & P(E2) = 0.36 …….. (given)
& P (E1 ∩ E2) = 0.30
To find: probability that the family owns either anyone or both kinds of sets.
Now, by the general rule –
P (A U B) = P(A) + P(B) – P(A ∩ B)
We have,
P (E1 U E2) = P(E1) + P(E2) – P(E1 ∩ E2)
= 0.87 + 0.36 – 0.30
= 1.23 – 0.30
= 0.93
Therefore, 0.93 is the required probability

Question:7

If A and B are mutually exclusive events, P (A) = 0.35 and P (B) = 0.45, find
(a) P (A′)
(b) P (B′)
(c) P (A ∪ B)
(d) P (A ∩ B)
(e) P (A ∩ B′)
(f) P (A′∩ B′)

P(A) = 0.35 & P(B) = 0.45 ….. (given)
P(A ∩ B) = 0 ……. (since A & B are mutually exclusive)

1. P(A’)

Now, we know that,
P (A) + P (A’) = 1
0.35 + P (A’) = 1
P(A’) = 1 – 0.35
P (A’) = 0.65

1. P (B’)

Now, we know that,
P (B) + P (B’) = 1
0.45 + P (B’) = 1
P (B’) = 1 – 0.45
P (B’) = 0.55

1. P (a u b)

Now, we know that,
P (A U B) = P(A) + P(B) – P(A ∩ B)
P (A U B) = 0.35 + 0.45 – 0
P (A U B) = 0.80

1. P (A ∩ B)

Since A & B are mutually exclusive events,
Thus, P (A ∩ B) = 0

1. P (A ∩ B’)

P (A ∩ B’) = P (A) - P (A ∩ B)
= 0.35 – 0
= 0.35

1. P (A’ ∩ B’)

P (A’ ∩ B’) = P (A U B)’
= 1 – P (A U B)
= 1 – 0.8 …… [from (c)]
= 0.2

Question:8

A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26, .08. Find the probabilities that a particular surgery will be rated
(a) complex or very complex;
(b) neither very complex nor very simple;
(c) routine or complex
(d) routine or simple

Given:
P (E1) = 0.15
P (E2) = 0.20
P (E3) = 0.31
P (E4) = 0.26
P (E5) = 0.08
Let us consider that –
E1 → Event that surgeries are rated as very complex
E2 → Event that surgeries are rated as complex
E3 → Event that surgeries are rated as routine
E4 → Event that surgeries are rated as simple
E5 → Event that surgeries are rated as very simple

1. P (complex or very complex) = P (E1 or E2)

= P (E1 U E2)
Now, by the general rule –
P (A U B) = P(A) + P(B) – P(A ∩ B)
We have,
P (E1 U E2) = P(E1) + P(E2) – P(E1 ∩ E2)
= 0.15 + 0.2 – 0
= 0.35

1. P (neither very complex nor very simple) = P (E1’ ∩ E5’)

= 1 - P (E1 ∩ E5) ….. (by complement rule)
= 1 – [P (E1) + P (E5) – P (E1 ∩ E5) … (general addn rule)
= 1 – [0.15 + 0.08 – 0]
= 1 – 0.23
= 0.77

1. P (routine or complex) = P (E3 ∩ E2)

= P (E3) + P (E2) - P (E3 ∩ E2) …… (by general addition rule)
= 0.31 + 0.2 – 0
= 0.51

1. P (routine or simple) = P (E3 ∩ E4)

= P (E3) + P (E4) - P (E3 ∩ E4) …… (by general addition rule)
= 0.31 + 0.26 - 0
= 0.57

Question:9

Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that
(a) C will be selected?
(b) A will not be selected?

Given:
A is twice likely to be selected as B, P(A) = 2 P(B)
& C is twice likely to be selected as D, P(C) = 2 P(D)
It is given that B & C have about the same chance
Thus, P(B) = P(C)
Now, sum of all probabilities is 1,
Thus,
P(A) + P(B) + P(C) + P(D) = 1
P(A) + P(B) + P(B) + P(D) = 1
Thus,
P(A) + P(A)/2 + P(A)/2 + P(C)/2 = 1
[2 P(A) + P(A) + P(A) + P(B)] /2 = 1
4 P(A) + P(A) / 2 = 2
[8 P(A) + P(A)] / 2 = 2
9 P(A) = 4
P(A) = 4/9
Now, (a) P (C will be selected) = P (C)
= P (B)
= 4/9 x ½
= 2/9
(b) P (A will not be selected) = P (A’)
= 1 – P (A) ……. (by complement rule)
= 1 – 4/9
= 9-4/ 9
= 5 / 9

Question:10

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted} You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(a) Determine P (John promoted)
P (Rita promoted)
P (Aslam promoted)
P (Gurpreet promoted)
(b) If A = {John promoted or Gurpreet promoted}, find P (A).

Given:
Sample Space (S) = John promoted, Rita promoted, Aslam promoted, Gurpreet promoted
Chances of John’s promotion is same as Gurpreet’s, P (E1) = P (E4)
Rita’s chances of promotion as twice as john’s, P (E2) = 2 P(E1)
& chances of Aslam’s promotion are four times that of John’s, P (E3) = 4 P(E­1)
Now, let us consider that,
E1 → events that John promoted
E2 → events that Rita promoted
E3 → events that Aslam promoted
E4 → events that Gurpreet promoted
Now, we know that,
Sum of all probabilities = 1
Thus, P(E1) + P(E2) + P(E3) + P(E4) = 1
P(E1) + 2 P(E1) + 4 P(E1) + P(E1) = 1
8 P(E1) = 1
P (E1) = 1/8
Now,

1. P (John promoted) = P (E1) = 1/8

P (Rita promoted) = P (E2) = 2P (E1)
= 2 x 1/8
= 1/4
P (Aslam promoted) = P (E3) = 4P (E1)
= 4 x 1/8
= 1/2
P (Gurpreet promoted) = P (E4) = P (E1)
= 1/8

2. A = John or Gurpreet promoted …….. (Given)

Thus, A = E1 U E2
P (A) = P (E1 U E2)
= P (E1) + P (E4) – P (E1 ∩ E2) ……. (general addition rule)
= P (E1) + P (E1) – 0
= 1/8 + 1/8
= 2/8
= 1/4

Question:11

The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections (for instance, P (A ∩ B) = .07. Determine

(a) P (A)
(b)
(c) P (A ∪ B)
(d) P (A ∩ B)
(e) P (B ∩ C)
(f) Probability of exactly one of the three occurs.

P (A ∩ B) = 0.07 ……. (given)

1. P (A) = 0.13 + 0.7 …… (by given Venn diagram)

= 0.20

2. P (B ∩ ) = P (B) – P (B ∩ C)

= 0.07 + 0.10 + 0.15 – 0.15
= 0.07 + 0.10
= 0.17

3. P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)

P (A U B) = 0.20 + (0.07 +0.10 +0.15) – 0.07
P (A U B) = 0.20 + 0.25
P (A U B) = 0.45

4. P (A ∩) = P (A) – P (A ∩ B)

= 0.20 – 0.07
= 0.13

5. P (B ∩ C) = 0.15 …… (from venn diagram)

6. P (exactly one of the 3 occurs) = 0.13 + 0.10 + 0.28 = 0.51

Question:12

One urn contains two black balls (labelled B1 and B2) and one white ball. A second urn contains one black ball and two white balls (labelled W1 and W2). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.
(a) Write the sample space showing all possible outcomes
(b) What is the probability that two black balls are chosen?
(c) What is the probability that two balls of opposite colour are chosen?

Given:
One urn contains 2 black balls & 1 white ball
The other urn contains 1 black ball & 2 white balls
If one of the 2 urns is chosen then a ball is randomly chosen from urn & without replacing the first ball, the second ball is also chosen from the same urn.

1. Now, we know that,

Sample Space, S = {B1B2, B1W, B2W, B2B1, B2W, WB1, WB2, W1W2, W1B, W2B, W2B, W2W1, BW1, BW2}
Thus, total no. of sample space = 12

1. Now, if two black balls are chosen,

Favorable outcomes = B1B2 & B2B1
& total no. of favorable outcomes = 2
Now,
Probability = no. of favorable outcomes / total no. of outcomes
= 2 /12 = 1 /6

1. Now, if two balls of opposite colours are chosen,

Favorable outcomes = B1W, B2W, WB1, WB2, W1B, W2B, BW1, BW2
& total no. of favorable outcomes = 8
Now,
Probability = no. of favorable outcomes / total no. of outcomes
= 8 /12 = 2 /3

Question:13

A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the Probability that:
(a) All the three balls are white
(b) All the three balls are red
(c) One ball is red and two balls are white

Given:
No. of reds balls = 8
No. of white balls = 5
Thus, total no. of balls, n = 13
Now, 3 balls are drawn at random, thus,
r = 3
Thus,
n (S) = nCr
= 13C3

1. If all balls are white,

P (A) = n (A)/ n(S)
= no. of favorable outcomes/ sample space
Now, total white balls are = 5
Thus,
P (all the three balls are white) = 5C3 / 13C3

= 5 /143

1. All three balls are red,

P (A) = n (A)/ n(S)
= no. of favorable outcomes/ sample space
Now, total red balls are = 8
Thus,
P (all the three balls are white) = 8C3 / 13C3

= 28 /143

1. One ball is red and two balls are white

P (A) = n (A)/ n(S)
= no. of favorable outcomes/ sample space
Now, total white balls are = 5
Thus,
P (One ball is red and two balls are white) = 8C1 x 5C2 / 13C3

= 40 /143

Question:14

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) Four S’s come consecutively in the word
(b) Two I’s and two N’s come together
(c) All A’s are not coming together
(d) No two A’s are coming together.

Given word: ASSASINATION
Total no. of letters in the word = 13
Viz., 3 A’s, 4 S’s, 2 I’s, 1 T & 1 O
No. of ways in which these letters can be arranged –
n(S) = 13! / 3! 4! 2! 2!

1. 4 S’s come consecutively in ASSASINATION

Then the word becomes-

 S S S S A A I N A T I O N

No. of letters = 1 + 9 = 10Now,
Thus,
n(E) = 10! / 3! 2! 2!
Now,
Required probability =
= 10! / 3! 2! 2! x 3! 4! 2! 2! / 13!
= 10! x 4!/ 13 x 12 x 11 x 10!
= 2 /143

1. 2 I’s & 2 N’s come together

Then the word becomes-

 I I N N A S S A S S A T O

No. of letters = 1 + 9 = 10Now,
Thus,
n(E) = 4! / 2! 2! x 10!/ 3! 4!
Now,
Required probability =
= 10! 4! / 3! 4! 2! 2! x 3! 4! 2! 2! / 13!
= 10! x 4!/ 13 x 12 x 11 x 10!
= 2 /143

1. All A’s are not coming together

Then the word becomes-

 A A A S S S S I N T I O N

Now,
No. of letters = 1 + 10 = 11
Thus,
When all A’s come together no. of words = 11! / 4! 2! 2!
Now,
probability =
= 11! / 4! 2! 2! x 3! 4! 2! 2! / 13!
= 11! x 3!/ 13 x 12 x 11!
= 1 / 26
Now,
P (All A’s don’t come together) = 1 – P (all A’s comes together)
= 1 – 1/26
= 25 / 26

1. No two A’s are coming together,

Then the word becomes-

 S S S S I N T I O N

No. of ways of arranging except A = 10! / 4! 2! 2!Now,
There are 11 vacant places
Total no. of A’s in ASSASINATION = 3
Thus, the 3 A’s can be placed in 11C3 ways
= 11! / 3! (11 – 3)!
= 11! / 3! 8!
No. of ways when 2 A’s are not together
= 11! / 3! 8! x 10! / 4! 2! 2! x 3! 4! 2! 2!/ 13!
= 11! X 10 x 9 x 8! / 8! x 13 x 12 x 11!
= 10 x 9 / 13 x 12
= 15 / 26

Question:15

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

We know that, in a deck,
Total no. of cards = 52
No. of kings = 4
No. of heart cards = 13
& total no. of red cards = 13 + 13 = 26
Thus, favorable outcomes = 4 + 13 + 26 – 13 – 2
= 28
Now,
Probability = no. of favorable outcomes / total no. of outcomes
= 28 / 52
= 7 / 13

Question:16

A sample space consists of 9 elementary outcomes e1, e2, ..., e9 whose probabilities are
P(e1) = P(e2) = .08, P(e3) = P(e4) = P(e5) = .1
P(e6) = P(e7) = .2, P(e8) = P(e9) = .07
Suppose A = {e1, e5, e8}, B = {e2, e5, e8, e9}
(a) Calculate P (A), P (B), and P (A ∩ B)
(b) Using the addition law of probability, calculate P (A ∪ B)
(c) List the composition of the event A ∪ B, and calculate P (A ∪ B) by adding the probabilities of the elementary outcomes.
(d) Calculate P () from P (B), also calculate P () directly from the elementary outcomes of

Given data:
S = {e1, e2, e3, e4, e5, e6, e7, e8, e9}
A = {e1, e5, e8}
B = { e2, e5 e8, e9}
P (e1) = P (e2) = 0.08
P (e3) = P (e4) = P (e5) = 0.1
P (e6) = P (e7) = 0.2
P (e8) = P (e9) = 0.07

(a) P (A), P (B) & P (A ∩ B)

A = {e1, e5, e8} …….. (given)
Thus,
P (A) = P (e1) + P (e5) + P (e8)
= 0.08 + 0.1 + 0.07
= 0.25
Now, B = { e2, e5 e8, e9} ……. (given)
P (B) = P (e2) + P (e5) + P (e8) + P (e9)
Now, P (A ∩ B)
A ∩ B = {e5, e8}
Thus, P (A ∩ B) = P (e5) + P (e8)
= 0.1 + 0.07
= 0.17

(b) P (A U B)

P (A) = 0.25
P (B) = 0.32
P (A ∩ B) = 0.17
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
= 0.25 + 0.32 – 0.17
= 0.40
(c) A = {e1, e5, e8}
B = { e2, e5 e8, e9} ……. (given)
Thus, A U B = {e1, e2, e5, e8, e9}
Thus, P (A U B) = P (e1) + P (e2) + P (e5) + P (e8) + P (e9)
= 0.08 + 0.08 + 0.1 + 0.07 + 0.07
= 0.40
(d) P ()
P () = 1 – 0.32
= 0.68
Now, we have,
B = { e2, e5 e8, e9}
Thus, = { e1, e3, e4, e6, e7}
P () = P (e1) + P (e3) + P (e4) + P (e6) + P (e7)
= 0.08 + 0.1 + 0.1 + 0.2 +0.2
= 0.68

Question:17

(a) We know that, the possible outcomes of a fair die are-
S = {1, 2, 3, 4, 5, 6}
Thus, total no. of outcomes = 6
Here, 1, 3 & 5 are odd nos.
Thus, favorable outcomes = 3
Probability = no. of favorable outcomes/ total no. of outcomes
= 3 / 6
= 1/2
(b) When a fair coin is tossed twice,
S = {HH, HT, TH, TT}
Thus, total n, of outcomes = 4
For at least one head to appear, the possible cases will be – HH, HT, TH
Thus, favorable outcomes = 3
Now,
Probability = no. of favorable outcomes/ total no. of outcomes
= 3/ 4

When we roll a pair of dice,

S = {(1, 1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2) ,(2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4),(3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Thus, n(S) = 36
For sum to be 6, the possible outcomes will be –
(1,5), (2,4), (3,3), (4,2), (5,1)
Thus, favorable outcomes = 5
Probability = no. of favorable outcomes/ total no. of outcomes
= 5/ 36

Question:18

In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is
A. 1/7
B. 2/7
C. 3/7
D. none of these

B
(i) A non-leap year cofltains 365 days. So, on dividing it by 7, we get 52 weeks and 1 more day.
So, since 52 weeks are there, it means 52 Tuesdays will also be there necessarily with probability I and 1 more day may be either Sun. or Mon. or Tue. or Wed. or Thur. or Fri. or Sat.
So, to get 53 Tuesdays, we have to select one more Tuesday from these 7 possibilities with Probability 1/7.
Therefore, probability of having 53 Tuesdays or 53 wednesdat in a non-leap year $=\frac{1}{7}+ \frac{1}{7}=\frac{2}{7}$

Question:19

The set has 3 consecutive nos. from 1 to 20,
Thus, it is – (1,2,3), (2,3,4), (3,4,5), …….. , (18,19,20)
Now, if we consider 3 nos. as a single digit, there will be 18 nos.
Now, choosing 3 nos. out of 20 can be done in 20C3 ways
This, n(S) = 20C3
Required event – the 3 nos. chosen must be consecutive, thus,
P (nos. are consecutive) = 18/ 30C3

Now,
P (nos. that are not consecutive) = 1 – 3/190
= 190 – 3/ 190
= 187 / 190
Thus, option B is the correct answer.

Question:20

There are 26 red cards & 26 black cards in a pack of 52 cards.
Now,
2 cards are accidently dropped …….. (given)
Thus,
Probability of dropping a red card first = 26 /52
& probability of dropping a red card second = 26 / 51
…….. (since one card is already dropped, we are left with 51 cards)
Now,
Probability of dropping a black card first = 26 /52
Probability of dropping a black card second= 26 /51
Thus,
P (both cards are of diff colour) = 26/52 x 26/51 + 26/52 + 26/51
= 2 x 26/52 x 26/51
= 26 /51
Thus, option C is the correct answer.

Question:21

7 persons are to be seated in a row ….. (given)
Let us consider 2 persons as 1 group since 2 person sit next to each other
Thus, we have to arrange 6 persons,
Thus, the no. of arrangement = 2! X 6!
Thus, total no. of arrangement of 7 persons = 7!
Now,
Probability = no. of favorable outcomes/ total no. of outcomes
= 2! X 6! / 7!
= 2 x 1 x 6! / 7 x 6!
= 2/ 7
Thus, option C is the correct answer.

Question:22

Given digits: 0, 2, 3, 5
Now, we now that if a no. is divisible by 5, then the digit in the units place should be either 0 or 5
Now, if units place is 0, then

 3 2 1 1

Now, the first three places can be filled in 3! Ways
= 3 x 2 x 1 x 1
= 6
Now, if units place is 5, then,

 2 1 1 1

Here, we can fill the first two places in 2 ways & the 2nd and 3rd place in 2! Ways
= 2 x 2 x 1 x 1
= 4
Thus, the total no. of ways = 6 + 4
= 10
= n (E)
Now, total no. of ways to arrange the given digits to form 4 – digits without repetition is 3 x 3 x 2 x 1
= 18
Probability = no. of favorable outcomes/ total no. of outcomes
= 10 /18
= 5 /9
Thus, option D is the correct answer.

Question:23

If A and B are mutually exclusive events, then
A. P (A) ≤ P ()
B. P (A) ≥ P ()
C. P (A) < P ()
D. none of these

A & B are mutually exclusive events …… (given)
Thus, P (A ∩ B) = 0
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
Thus, P (A U B) = P (A) + P (B) - P (A ∩ B)
P (A U B) = P (A) + P (B) – 0
P (A U B) = P (A) + P (B)
Now, for all events A, B
0 ≤P(A)≤1
Thus, P (A) + P (B) ≤ 1
P (A) ≤ 1 – P (B)
Thus, by complement rule,
P (A) ≤ P ()
Thus, option A is the correct answer.

Question:24

If P (A ∪ B) = P (A ∩ B) for any two events A and B, then
A. P (A) = P
B. (B) P (A) > P (B)
C. P (A) < P (B)
D. none of these

P (A ∩B) = P (A U B)
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
P (A ∩ B) = P (A) + P (B) – P (A ∩ B)
Thus, [P (A) – P (A ∩ B)] + [P (B) – P (A ∩ B)] = 0
Now,
P (A) – P (A ∩ B) ≥ 0
& P (B) – P (A ∩ B) ≥ 0
P (A) – P (A ∩ B) = 0
P (B) – P (A ∩ B) = 0
Thus, P (A) = P (A ∩ B)
& P (B) = P (A ∩ B)
Therefore, we get
P(A) = P (B)
Thus, option A is the correct answer.

Question:25

6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is
A. 1/432
B. 12/431
C. 1/132
D. none of these

Let us consider group 1 where all girls sit together,
Thus,

 G G G G G G 6 BOYS

Total persons = 1 + 6
= 7
Now, total no. of arrangements in row of 7 = 7!
Also, the ways in which girls interchange their seats is 6!
Probability = no. of favorable outcomes/ total no. of outcomes
= 6! 7! / 12!
= 6 x 5 x 4 x 3 x 2 x 1 x 7! / 12 x 11 x 10 x 9 x 8 x 7!
= 6 x 5 x 4 x 3 x 2 / 12 x 11 x 10 x 9 x 8
= 1 / 2 x 11 x 2 x 3 x 2
= 1 / 132
Thus, option C is the correct answer.

Question:26

A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is
A. 1/3
B. 4/11
C. 2/11
D. 3/11

Total no. of alphabet in PROBABILITY = 11
& no. of vowels in it = 4, viz., (O, A, I, I)
Probability = no. of favorable outcomes/ total no. of outcomes
Now, P (letter is vowel) = 4 / 11
Thus, option B is the correct answer.

Question:27

Let us consider that-
E1 → event that A fails in the examination
& E2 → event that B fails in the examination
Thus,
P (E1) = 0.2
P (E2) = 0.3
To find: P (either E1 or E2 fails)
Thus, P (either E1 or E2 fails) = P (E1) + P (E2) – P (E1 ∩ E2)
≤ P(E1) + P (E2)
≤ 0.2 + 0.3
≤ 0.5
Thus, option C is the correct answer.

Question:28

Given:
P ( at least one of A or B occurs) = 0.6, thus, P (A U B) = 0.6
& P (A & B occurs simultaneously) = 0.2, thus, P (A ∩ B) = 0.2
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
Thus, P (A) + P (B) – 0.2 = 0.6
P (A) + P (B) = 0.6 + 0.2
P (A) + P (B) = 0.8
Now,
P (A) = 1 – P (A’) ……… by complement rule
Similarly, P (B) = 1 – P (B’)
Therefore,
1 – P (A’) + 1 – P (B’) = 0.8
2 – [P (A’) + P (B’)] = 0.8
2 – 0.8 = P (A’) + P (B’)
P (A’) + P (B’) = 1.2
Thus, option C is the correct answer.

Question:29

If M and N are any two events, the probability that at least one of them occurs is
A. P (M) + P (N) – 2 P (M ∩ N)
B. P (M) + P (N) – P (M ∩ N)
C. P (M) + P (N) + P (M ∩ N)
D. P (M) + P (N) + 2P (M ∩ N)

M and N are two events ……. (given)
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
Thus, P (M U N) = P (M) + P (N) – P (M ∩ N)
Thus, option B is the correct answer.

Question:30

State whether the statements are True or False
The probability that a person visiting a zoo will see the giraffe is 0.72, the probability that he will see the bears is 0.84 and the probability that he will see both is 0.52.

Let us consider that,
E1 → event that person see the giraffe
E2 → event that person see the bear
We have,
P (E1) = 0.72
& P (E2) = 0.84
P (person will see giraffe as well as bear) = 0.52
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
P (E1 U E2) = P (E1) + P (E2) – P (E1 ∩ E2)
= 0.72 + 0.84 – 0.52
= 1.04 viz. not possible
Thus, the given statement is false.

Question:31

State whether the statements are True or False
The probability that a student will pass his examination is 0.73, the probability of the student getting a compartment is 0.13, and the probability that the student will either pass or get compartment is 0.96.

Let us consider that,
A → event that student pass examination
B → event that student get compartment
We have,
P (A) = 0.73
P (B) = 0.13
& P (A U B) = 0.96
To find: P (A ∩ B)
Now, P (A U B) = P (A) + P (B) – P (A ∩ B) ……. (by general addition rule)
P (A U B) = 0.73 + 0.13 – 0
= 0.86
but given P (A U B) = 0.96
Thus, the given statement is False.

Question:32

State whether the statements are True or False.
The probabilities that a typist will make 0, 1, 2, 3, 4, 5 or more mistakes in typing a report are, respectively, 0.12, 0.25, 0.36, 0.14, 0.08, 0.11.

Let us consider that –

A → event that typist will make 0 mistake

B → event that typist will make 1 mistake

C → event that typist will make 2 mistake

D → event that typist will make 3 mistake

E → event that typist will make 4 mistake

F → event that typist will make 5 mistake

We have,

P (A) = 0.12

P (B) = 0.25

P (C) = 0.36

P (D) = 0.14

P (E) = 0.08

P (F) = 0.11

Now,

Sum of all probabilities = 1

Thus,

P (A) + P (B) + P (C) + P (D) + P (E) + P (F)

= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11

= 1.06 viz. > 1

Thus, the given statement is False.

Question:33

State whether the statements are True or False
If A and B are two candidates seeking admission in an engineering College. The probability that A is selected is .5 and the probability that both A and B are selected is at most .3. Is it possible that the probability of B getting selected is 0.7?

Let us consider that,
E1 → event that A is selected in engineering college
E2 → event that B is selected in engineering college
We have,
P (E1) = 0.5
& P (E2) = 0.7
P (E1 ∩ E2) ≤ 0.3
Thus,
0.5 x P (E2) ≤ 0.3
P (E2) ≤ 0.6
But, we know that,
P (E2) = 0.7
Thus, the given statement is False.

Question:34

State whether the statements are True or False
The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

A & B are two events
A ∩ B $\subset$ A
P (A ∩ B) ≤ P (A)
Thus, the given statement is true.

Question:35

State whether the statements are True or False.
The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probability of occurrence of both is .4.

Given data:
P (occurrence of event A) = 0.7
P (occurrence of event B) = 0.3
P (occurrence of both) = 0.4
Now,
P (A ∩ B) = P (A) x P (B)
P (A ∩ B) = 0.7 x 0.3
= 0.21
But, given
P (A ∩ B) = 0.4
Thus, the given statement is False.

Question:36

State whether the statements are True or False
The sum of probabilities of two students getting distinction in their final examinations is 1.2.

We know that,
Probability of each student getting distinction in final examination is ≤ 1
Now, the given events are not related to the sample space
Thus, 1.2 may be the sum of their probabilities.
Thus, the given statement is true

Question:37

Fill in the blanks
The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is _____.

Let us consider that,
A → event that home team will win football game
B → event that home team will tie football game
C → event that home team will lose football game
It is given that,
P (A) = 0.77
& P (B) = 0.08
To find: P (C)
Now, we know that,
Sum of all probabilities = 1
Thus,
P (A) + P (B) + P (C) = 1
0.77 + 0.08 + P (C) = 1
P (C) = 1 – (0.77 + 0.08)
= 0.15

Question:38

Fill in the blanks
If e1, e2, e3, e4 are the four elementary outcomes in a sample space and P(e1) =.1, P(e2) = .5, P (e3) = .1, then the probability of e4 is ______.

Given data:
P (e1) = 0.1
P (e2) = 0.5
P (e3) = 0.1
Now, we know that,
Sum of all probabilities = 1
Thus,
P (e1) + P (e2) + P (e3) + P (e4) = 1
0.1+ 0.5 + 0.1 + P (e4) = 1
P (e4) = 1 – 0.7
Thus, P (e4) = 0.3

Question:39

Fill in the blanks
Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then is _________.

Given data:
S = {1, 2, 3, 4, 5, 6}
& E = {1, 3, 5}
To find:
Now,
= S – E {w:w ? S & w ?E}
= {2, 4, 6}

Question:40

Fill in the blanks
If A and B are two events associated with a random experiment such that P (A) = 0.3, P (B) = 0.2 and P (A ∩ B) = 0.1, then the value of $P(A \cap \bar{B})$ is _______.

Given data:
P (A) = 0.3
P (B) = 0.2
P (A ∩ B) = 0.1
Now, we know that,
$P(A \cap \bar{B})$ = P (A) - P (A ∩ B)
= 0.3 – 0.1
= 0.2

Question:41

Fill in the blanks
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is ________.

Given:
P (happening of an event A) = 0.5
P (happening of an event B) = 0.3
A & B are mutually exclusive events
Thus, P (A ∩ B) = 0
To find: P (neither A nor B)

= 1 – P (A U B)
= 1 – [P (A) + P (B)]
= 1 – (0.5 + 0.3)
= 1 – 0.8
= 0.2

Question:42

Match the proposed probability under Column C1 with the appropriate written description under column C2 :

 C1 Probability C2 Probability a 0.95 i An incorrect assignment b 0.02 ii no chance of happening c -0.3 iii as much chance of happening as not d 0.5 iv very likely to happen e 0 v very little chance of happening

(a) Since 0.95 is very close to 1, it is very likely to happen.
Thus, (a) → (iv)
(b) The probability of 0.02 is very low, hence there is very little chance of happening, thus, (b) → (v)
(c) Since the probability of -0.3 is negative, and we know that probability can never be negative, thus it is an incorrect assignment, thus, (c) → (i)
(d) For 0.5 the sum of chances of happening and not happening is 1, hence it has as much chance of happening as not, thus, (d) → (iii)
(e) For 0, there is no chance of happening, thus, (e) →(ii)

Question:43

Match the following

 a if E1 and E2 are the two mutually exclusive events i E1 ∩ E2 = E1 b if E1 and E2 are the mutually exclusive and exhaustive events ii (E1 – E2) U (E1 ∩ E2) = E1 c If E1 & E2 have common outcomes then, iii E1 ∩ E2 = E1 U E2 = S d If E1 and E2 are two events such that E1 ? E2 iv E1 ∩ E2 = φ

(a) E1 & E2 are mutually exclusive events,
Thus, P (E1 ∩ E2) = φ
Thus, (a) → (iv)
(b) E1 & E2 are mutually exclusive and exhaustive events,
Thus, (b) → (iii)
(c) If E1 & E2 have common outcomes then,
(E1 – E2) U (E1 ∩ E2) = E1
Thus, (c) → (ii)
(d) If E1 & E2 are such that,
E1$\subset$ E2
Then, (E1 ∩ E2) = E1
Thus, (d) → (i).

## More About NCERT Exemplar Class 11 Maths Chapter 16

NCERT Exemplar Class 11 Maths chapter 16 solutions can be downloaded by the students. They can make use of NCERT Exemplar Class 11 Maths solutions chapter 16 PDF Download function for better convenience. The solutions given in the PDF are explained broadly and every sub-topic is covered.

## List of Topics and Sub-topics in NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability:

· Random Experiments

· Outcomes

· Sample space

· Event

· The occurrence of an event

· Types of events

· Algebra of Events

· Mutually Exclusive Events

· Exhaustive Events

· Axiomatic Approach to Probability

· Probability of an Event

· Probability of equally likely outcomes

· Probability of the event ‘A or B’

· Probability of the event ‘not A’

## NCERT Solutions for Class 11 Mathematics Chapters

 Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 11 Conic Sections Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics

What will students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability?

The students can comprehensively understand the different fundamental concepts of Probability like the different types of events and the axiomatic approach used in probability. Pupils who find the formulas of Probability difficult can refer to the NCERT Exemplar Class 11 Maths solutions chapter 16 as all of the formulas along with their application are explained in a stepwise manner.

Illustrations are used which makes the problem even easier to comprehend. MCQ type questions, miscellaneous questions, HOTS, short answer type questions and long answer type questions are included in.

## Important Topics to Cover in NCERT Exemplar Class 11 Maths Solutions Chapter 16

Types of Events in Probability:

To understand the classification of events, the students must first understand what events are. An event is the subset of the sample space. Sure events, complementary events, impossible events are the different types of events that occur in probability. NCERT Exemplar Class 11 Maths solutions Chapter 16 can be used to understand the basic concepts like sample space, subsets and an empty set of probability.

Axiomatic Approach to Probability:

The problems related to mutually exclusive events and exhaustive events are important and should not be skipped by the students. Probability is an important topic in the CBSE mathematics syllabus for Class 11 and has an average weightage. It is one of the most scoring chapters in mathematics and the students can easily mark it just by applying a formula and writing the required steps to solve the problems.

Check Chapter-Wise NCERT Solutions of Book

 Chapter-1 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

### NCERT Exemplar Class 11 Solutions

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Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

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1. Can the solutions of NCERT Exemplar Class 11 Maths Chapter 16 Solutions be downloaded?

2. Can NCERT Exemplar Class 11 Maths Chapter 16 Solutions be refereed for CBSE final exam?

Yes, Class 11 Maths NCERT exemplar solutions chapter 16  can be referred to for CBSE final exam.

3. Which topics should be covered from the chapter Probability?

The following are the important topics of Probability:

·       Axiomatic Approach to Probability

·       Exhaustive Events

·       Mutually Exclusive Events

·       Types of Events

·       Random Experiments

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