NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

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?

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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 = 5^k-1

2. 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
= 5^k-1
Now, the possibility of 2 appearing before kth roll = 1 + 5 + 5² + 5³ + …… + 5^k-1
Now,

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

Answer:

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?

Answer:

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′)

Answer:

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

2. 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

3. 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

4. P (A ∩ B)

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

5. P (A ∩ B’)

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

6. 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

Answer:

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

2. 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

3. 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

4. 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?

Answer:

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).

Answer:

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.

Answer:

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?

Answer:

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

2. 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

3. 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

Answer:

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) = ⁿC_r
= 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

2. 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

3. 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.

Answer:

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-

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

2. 2 I’s & 2 N’s come together

Then the word becomes-

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

3. All A’s are not coming together

Then the word becomes-

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

4. No two A’s are coming together,

Then the word becomes-

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.

Answer:

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

Answer:

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

Determine the probability p, for each of the following events.
(a) An odd number appears in a single toss of a fair die.
(b) At least one head appears in two tosses of a fair coin.
(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
(d) The sum of 6 appears in a single toss of a pair of fair dice.
Answer:

(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

Answer:

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

Question:19

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive

A.
B.
C.
D.

Answer:

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

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours
A. 29/52
B. 1/2
C. 26/51
D. 27/51

Answer:

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

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is
A. 1/3
B. 1/6
C. 2/7
D. 1/2

Answer:

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

Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is
A. 1/5
B. 4/5
C. 1/30
D. 5/9

Answer:

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

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,

Here, we can fill the first two places in 2 ways & the 2^nd and 3^rd 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

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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

If the probabilities for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A or B fails is
A. >0 . 5
B. 0.5
C. ≤ 0.5
D. 0

Answer:

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

The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P () + P () is
A. 0.4
B. 0.8
C. 1.2
D. 1.6

Answer:

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)

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

A & B are two events
A ∩ B 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.

Answer:

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.

Answer:

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 _____.

Answer:

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
Answer = 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 ______.

Answer:

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
Answer: 0.3

Question:39

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

Answer:

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}
Answer: {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 is _______.

Answer:

Given data:
P (A) = 0.3
P (B) = 0.2
P (A ∩ B) = 0.1
Now, we know that,
= 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 ________.

Answer:

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 :

Answer:

(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

Answer:

Answers –
(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 E₂
Then, (E₁ ∩ E₂) = E₁
Thus, (d) → (i).