NCERT Solutions for Class 11 Maths Chapter 14 Probability

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 16
Chapter Name : Probability

Exercise 16.1

Q1 Describe the sample space for the indicated experiment: A coin is tossed three times.

Answer. A coin has two faces: head (H) and tail (T).
When a coin is tossed three times, the total number of possible outcomes is 2 = 8 3

Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Page : 386 , Block Name : Exercise 16.1

Q2 Describe the sample space for the indicated experiment: A die is thrown two times.

Answer. When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:
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)}

Page : 386 , Block Name : Exercise 16.1

Q3 Describe the sample space for the indicated experiment: A coin is tossed four times.

Answer. When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 2 = 16 4

Thus, when a coin is tossed four times, the sample space is given by:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT,
TTHH, TTHT, TTTH, TTTT}

Page : 386 , Block Name : Exercise 16.1

Q4 Describe the sample space for the indicated experiment: A coin is tossed and a die is thrown.

Answer. A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Page : 386 , Block Name : Exercise 16.1

Q5 Describe the sample space for the indicated experiment: A coin is tossed and then a die is
rolled only in case a head is shown on the coin.

Answer. A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the
sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T}

Page : 386 , Block Name : Exercise 16.1

Q6 2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for
the experiment in which a room is selected and then a person.

Answer. Let us denote 2 boys and 2 girls in room X as B , B and G , G respectively. Let us
1 2 1 2

denote 1 boy and 3 girls in room Y as B and G , G , G respectively.
3, 3 4 5

Accordingly, the required sample space is given by S = {XB , XB , XG , XG , YB , YG ,
1, 2 1, 2 3 3

YG YG }
4, 5

Page : 386 , Block Name : Exercise 16.1
Q7 One die of red colour, one of white colour and one of blue colour are placed in a bag. One die
is selected at random and rolled, its colour and the number on its uppermost face is noted.
Describe the sample space.

Answer. A die has six faces that are numbered from 1 to 6, with one number on each face. Let us
denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}

Page : 386 , Block Name : Exercise 16.1

Q8 An experiment consists of recording boy-girl composition of families with 2 children.
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order
of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?

Answer. (i) When the order of the birth of a girl or a boy is considered, the sample space is given

by S = {GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or
1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}

Page : 386 , Block Name : Exercise 16.1

Q9 A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession
without replacement. Write the sample space for this experiment.

Answer. It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the
red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacement, the sample space is
given by
S = {RW, WR, WW}

Page : 387 , Block Name : Exercise 16.1

Q10 An experiment consists of tossing a coin and then throwing it second time if a head occurs.
If a tail occurs on the rst toss, then a die is rolled once. Find the sample space.

Answer. A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, in the given experiment, the sample space is given by
S = {HH, HT, T1, T2, T3, T4, T5, T6}

Page : 387 , Block Name : Exercise 16.1

Q11 Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classi ed as
defective (D) or non-defective (N). Write the sample space of this experiment?

Answer. 3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and
classi ed as defective (D) or non-defective (N).
The sample space of this experiment is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}

Page : 387 , Block Name : Exercise 16.1

Q12 A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even
number, the die is thrown again. What is the sample space for the experiment?

Answer. When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63,
H64, H65, H66}

Page : 387 , Block Name : Exercise 16.1

Q13 The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a
box and mixed thoroughly. A person draws two slips from the box, one after the other, without
replacement. Describe the sample space for the experiment.

Answer. If 1 appears on the rst drawn slip, then the possibilities that the number appears on the
second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the rst drawn slip, then the
possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true
for the remaining numbers too.
Thus, the sample space of this experiment is given by S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

Q14 An experiment consists of rolling a die and then tossing a coin once if the number on the die
is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this
experiment.

Answer. A die has six faces that are numbered from 1 to 6, with one number on each face. Among
these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.
A coin has two faces: head (H) and tail (T).
Hence, the sample space of this experiment is given by:
S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}

Page : 387 , Block Name : Exercise 16.1

Q15 A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black
balls. If it shows head, we throw a die. Find the sample space for this experiment.

Answer. The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R , R and 1 2

the 3 black balls as B , B , and B .
1 2 3

The sample space of this experiment is given by
S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}

Page : 387 , Block Name : Exercise 16.1

Q16 A die is thrown repeatedly until a six comes up. What is the sample space for this
experiment?

Answer. In this experiment, six may come up on the rst throw, the second throw, the third throw
and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2, 5, 6),
… ,(5, 1, 6), (5, 2, 6), …}

Page : 387 , Block Name : Exercise 16.1

Exercise 16.2

Q1 A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”.
Are E and F mutually exclusive?

Answer. When a die is rolled, the sample space is given by
S = {1, 2, 3, 4, 5, 6}
Accordingly, E = {4} and F = {2, 4, 6}
It is observed that E ∩ F = {4} ≠ Φ
Therefore, E and F are not mutually exclusive events.

Page : 393 , Block Name : Exercise 16.2

Q2 A die is thrown. Describe the following events:
(i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3
(iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3
Also nd A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A − C, D − E, E ∩ F , F ′ ′

Answer. When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.
Accordingly:
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = Φ
(iii) C = {3, 6}
(iv) D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}
(v) E = {6}
(vi) F = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ
B ∪ C = {3, 6}, E ∩ F = {6}
D ∩ E =Φ, A – C = {1, 2, 4, 5}
D – E = {1, 2, 3}, F = {1, 2}, E ∩ F = ϕ
′ ′

Page : 393 , Block Name : Exercise 16.2

Q3 An experiment involves rolling a pair of dice and recording the numbers that come up.
Describe the following events:
A: the sum is greater than 8, B: 2 occurs on either die
C: The sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?

Answer. When a pair of dice is rolled, the sample space is given by
S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6}

Accordingly,

A = {(3, 6) ⋅ (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}

B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (6, 2)}

C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

It is observed that
A ∩ B =Φ
B ∩ C =Φ
C ∩ A = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} ≠ ϕ

Hence, events A and B and events B and C are mutually exclusive.

Page : 393 , Block Name : Exercise 16.2

Q4 Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event
“two heads and one tail show”. C denote the event “three tails show” and D denote the event ‘a
head shows on the rst coin”. Which events are
(i) mutually exclusive? (ii) simple? (iii) compound?

Answer. When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Accordingly,
A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ
B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ
C∩D=Φ
(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A
and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event.

Thus, B and D are compound events.

Page : 393 , Block Name : Exercise 16.2

Q5 Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.

Answer. When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two events that are mutually exclusive can be
A: getting no heads and B: getting no tails
This is because sets A = {TTT} and B = {HHH} are disjoint.
(ii) Three events that are mutually exclusive and exhaustive can be
A: getting no heads
B: getting exactly one head
C: getting at least two heads
I.e., A = {TTT}
B = {HTT, THT, TTH}
C = {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S
(iii) Two events that are not mutually exclusive can be
A: getting three heads
B: getting at least 2 heads
i.e.,A = {HHH}
B = {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ
(iv) Two events which are mutually exclusive but not exhaustive can be
A: getting exactly one head
B: getting exactly one tail
That is
A = {HTT, THT, TTH}
B = {HHT, HTH, THH}
It is because, A ∩ B =Φ, but A ∪ B ≠ S
(v) Three events that are mutually exclusive but not exhaustive can be
A: getting exactly three heads
B: getting one head and two tails
C: getting one tail and two heads
i.e.,
A = {HHH}
B = {HTT, THT, TTH}
C = {HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ S

Page : 393 , Block Name : Exercise 16.2

Q6 Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the rst die.
B: getting an odd number on the rst die.
C: getting the sum of the numbers on the dice ≤ 5
Describe the events
(i) (ii) not B (iii) A or B
(iv) A and B (v) A but not C (vi) B or C
(vii) B and C (viii)A ∩ B ∩ C ′ ′

Answer. When two dice are thrown, the sample space is given by
S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6}

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
A = { }
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
B = { }
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 4), (5, 4), (5, 5), (5, 6)

C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3),
′
A = { } = B
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 4), (5, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (4, 1), (4, 1), (4, 3), (4, 3),
′
Not B = B = { } = A
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 4), (6, 4), (6, 6)

(iii)

( iv ) A and B = A ∩ B = ϕ

(vii) B and C = B ∩ C{(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
⎧ (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2) ⎫
⎪ ⎪

(viii) C ′
= ⎨ (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)
′ ′ ′ ′
∴ A ∩ B ∩ C = A ∩ A ∩ C = A ∩ C

(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5)
= { }
(4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Page : 393 , Block Name : Exercise 16.2

Q7 Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the rst die.
B: getting an odd number on the rst die.
C: getting the sum of the numbers on the dice ≤ 5
State true or false: (give reason for your answer)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
′
(iii) A = B

(iv) A and C are mutually exclusive

′
(v) A and B are mutually exclusive

′ ′
(vi) A , B , C are mutually exclusive and exhaustive.

Answer.
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3)
A = { }
(4, 4) ⋅ (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3)
B = { }
(3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(i) It is observed that A ∩ B = Φ
∴ A and B are mutually exclusive.
Thus, the given statement is true.
(ii) It is observed that A ∩ B = Φ and A ∪ B = S
∴ A and B are mutually exclusive and exhaustive.

Thus, the given statement is true.
(iii) It is observed that
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3),
′
B = { } = A
(4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Thus, the given statement is true.
(iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠Φ
∴A and C are not mutually exclusive.
Thus, the given statement is false.
′
(v) A ∩ B = A ∩ A = A

′
∴ A ∩ B ≠ ϕ

′
∴ A and B are not mutually exclusive.

Thus, the given statement is false.

′ ′
(vi) It is observed that A ∪ B ∪ C = S

′
However, B ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ ϕ

Therefore, events A , B and C ′ ′
are not not mutually exclusive and exhaustive.
Thus, the given statement is false.

Page : 393 , Block Name : Exercise 16.2

Exercise 16.3

Q1 Which of the following can not be valid assignment of probabilities for outcomes of sample
space S = {ω , ω , ω , ω , ω , ω , ω }
1 2 3 4 5 6 7

Answer.

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities
= p (ω1 ) + p (ω2 ) + p (ω3 ) + p (ω4 ) + p (ω5 ) + p (ω6 ) + p (ω7 )

= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

= 1

Thus, the assignment is valid.

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities
= p (ω1 ) + p (ω2 ) + p (ω3 ) + p (ω4 ) + p (ω5 ) + p (ω6 ) + p (ω7 )

1 1 1 1 1 1 1
= + + + + + = 7 × = 1
7 7 7 7 7 7 7

Thus, the assignment is valid.

Here, each of the numbers p(ωi) is positive and less than 1.
Sum of probabilities
= p (ω1 ) + p (ω2 ) + p (ω3 ) + p (ω4 ) + p (ω5 ) + p (ω6 ) + p (ω7 )

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7

= 2.8 ≠ 1

Thus, the assignment is not valid.

Here, p(ω1) and p(ω5) are negative.
Hence, the assignment is not valid.

Here, p (ω ) =
7
15

14
> 1

Hence, the assignment is not valid.

Page : 403 , Block Name : Exercise 16.3

Q2 A coin is tossed twice, what is the probability that at least one tail occurs?

Answer. When a coin is tossed twice, the sample space is given by
S = {HH, HT, TH, TT}
Let A be the event of the occurrence of at least one tail.
Accordingly, A = {HT, TH, TT}
Number of outcomes favourable to A
∴ P(A) =
Total number of possible outcomes

n(A)
=
n(S)

3
=
4

Page : 404 , Block Name : Exercise 16.3

Q3 A die is thrown, nd the probability of following events:
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.

Answer. The sample space of the given experiment is given by
S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number.
Accordingly, A = {2, 3, 5}
n(A)
Number of outcomes favourable to A 3 1
∴ P(A) = = = =
Total number of possible outcomes n(S) 6 2

(ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B =
{3, 4, 5, 6}
n(B)
Number of outcomes favourable to B 4 2
∴ P(B) = = = =
Total number of possible outcomes n(S) 6 3

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C =
{1}

Number of outcomes favourable to C n(C) 1
∴ P(C) = = =
Total number of possible outcomes n(S) 6

(iv) Let D be the event of the occurrence of a number greater than 6.
Accordingly, D = Φ
Number of outcomes favourable to D n(D) 0
∴ P(D) = = = = 0
Total number of possible outcomes n(S) 6

(v) Let E be the event of the occurrence of a number less than 6.
Accordingly, E = {1, 2, 3, 4, 5}
Number of outcomes favourable to E n(E) 5
∴ P (E) = = =
Total number of possible outcomes n(S) 6

Page : 404 , Block Name : Exercise 16.3

Q4 A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace (ii) black card.

Answer. (a) When a card is selected from a pack of 52 cards, the number of possible outcomes is
52 i.e., the sample space contains 52 elements.
Therefore, there are 52 points in the sample space.
(b) Let A be the event in which the card drawn is an ace of spades.
Accordingly, n(A) = 1
n(A)
Number of outcomes favourable to A 1
∴ P(A) = = =
Total number of possible outcomes n(S) 52

(c) (i)Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards, n(E) = 4
n(E)
Number of outcomes favourable to E 4 1
∴ P(E) = = = =
Total number of possible outcomes n(S) 52 13

(ii)Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26
n(F)
Number of outcomes favourable to F 26 1
∴ P(F) = = = =
Total number of possible outcomes n(S) 52 2

Page : 404 , Block Name : Exercise 16.3

Q5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find
the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Answer. Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces
that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12

(i) Let A be the event in which the sum of numbers that turn up is 3.
Accordingly, A = {(1, 2)}
n(A)
Number of outcomes favourable to A 1
∴ P(A) = = =
Total number of possible outcomes n(S) 12

(ii) Let B be the event in which the sum of numbers that turn up is 12.
Accordingly, B = {(6, 6)}
Number of outcomes favourable to B n(B) 1
∴ P(B) = = =
Total number of possible outcomes n(S) 12

Page : 404 , Block Name : Exercise 16.3

Q6 There are four men and six women on the city council. If one council member is selected for a
committee at random, how likely is it that it is a woman?

Answer. There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10
(4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
Number of outcomes favourable to A n(A) 6 3
∴ P (A) = = = =
Total number of possible outcomes n(S) 10 5

Page : 404 , Block Name : Exercise 16.3

Q7 A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each
tail that turns up. From the sample space calculate how many different amounts of money you
can have after four tosses and the probability of having each of these amounts.

Answer. Since the coin is tossed four times, there can be a maximum of 4 heads or tails.
When 4 heads turns up, is the gain.
When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.
When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.
When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the
loss.
When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.
There are 2 = 16 elements in the sample space S, which is given by:
4

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT,
THTT, TTHT, TTTH, TTTT}
∴n(S) = 16
The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.
∴Probability (of winning Rs 4.00) = 16
1

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH,
HTHH, THHH} occurs.
∴Probability (of winning Rs 1.50) = =
16
4 1

4

The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH,
TTHH, HTHT, THTH, THHT} occurs.
∴Probability (of losing Re 1.00) = =
6

16
3

8

The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT,
TTHT, TTTH} occurs.

Probability (of losing Rs 3.50) = 4

16
=
1

4

The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.
Probability (of losing Rs 6.00) = 1

16

Page : 404 , Block Name : Exercise 16.3

Q8 Three coins are tossed once. Find the probability of getting
(i) 3 heads (ii) 2 heads (iii) at least 2 heads
(iv) at most 2 heads (v) no head (vi) 3 tails
(vii) exactly two tails (viii) no tail (ix) at most two tails.

Answer. When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴Accordingly, n(S) = 8
It is known that the probability of an event A is given by
n(A)
Number of outcomes favourable to A
P(A) = =
Total number of possible outcomes n(S)

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}
n(B)
1
∴ P(B) = =
n(S) 8

(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}
n(C) 3
∴ P (C) = =
n(S) 8

(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}
n(D) 4 1
∴ P (D) = = =
n(S) 8 2

(iv) Let E be the event of the occurrence of at most 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
n(E) 7
∴ P(E) = =
n(S) 8

(v) Let F be the event of the occurrence of no head.
Accordingly, F = {TTT}
n(F) 1
∴ P(F) = =
n(S) 8

(vi) Let G be the event of the occurrence of 3 tails.
Accordingly, G = {TTT}
n(G) 1
∴ P(G) = =
n(S) 8

(vii) Let H be the event of the occurrence of exactly 2 tails.
Accordingly, H = {HTT, THT, TTH}
n(H) 3
∴ P(H) = =
n(S) 8

(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}
n(I) 1
∴ P(l) = =
n(S) 8

(ix) Let J be the event of the occurrence of at most 2 tails.

Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}
n(J) 7
∴ P(J) = =
n(S) 8

Page : 404 , Block Name : Exercise 16.3

Q9 If 2

11
is the probability of an event, what is the probability of the event ‘not A’.

Answer. It is given that P(A) = 2

11

Accordingly, P(not A) = 1 – P(A) = 1 − 2

11
=
9

11

Page : 404 , Block Name : Exercise 16.3

Q10 A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that
letter is (i) a vowel (ii) an consonant

Answer. There are 13 letters in the word ASSASSINATION.
∴Hence, n(S) = 13
(i) There are 6 vowels in the given word.
∴Probability (vowel) = 6

13

(ii) There are 7 consonants in the given word.
∴Probability (consonant) = 7

13

Page : 404 , Block Name : Exercise 16.3

Q11 In a lottery, person choses six different natural numbers at random from 1 to 20, and if these
six numbers match with the six numbers already xed by the lottery committee, he wins the
prize. What is the probability of winning the prize in the game? [Hint: order of the numbers is
not important.]

Answer. Total number of ways in which one can choose six different numbers from 1 to 20
⌊20 ⌊20 20×19×18×17×16×15
20
= C6 = = = = 38760
⌊6|20−6 ⌊6⌊14 1⋅2⋅3⋅4⋅5⋅6

Hence, there are 38760 combinations of 6 numbers.
Out of these combinations, one combination is already xed by the lottery committee.
∴ Required probability of winning the prize in the game = 1

38760

Page : 405 , Block Name : Exercise 16.3

Q12 Check whether the following probabilities P(A) and P(B) are consistently de ned
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

Answer. (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
However, here, P(A ∩ B) > P(A).
Hence, P(A) and P(B) are not consistently de ned.
(ii)P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B).
Hence, P(A) and P(B) are consistently de ned.

Page : 405 , Block Name : Exercise 16.3

Q13 Fill in the blanks in following table:

Answer. (i) Here, P (A) = 1

3
, P (B) =
1

5
, P (A ∩ B) =
1

15

We know that P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
1 1 1 5+3−1 7
∴ P(A ∪ B) = + − = =
3 5 15 15 15

(ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴0.6 = 0.35 + P(B) – 0.25
⇒ P(B) = 0.6 – 0.35 + 0.25
⇒ P(B) = 0.5
(iii)Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴0.7 = 0.5 + 0.35 – P(A ∩ B)
⇒ P(A ∩ B) = 0.5 + 0.35 – 0.7
⇒ P(A ∩ B) = 0.15

Page : 405 , Block Name : Exercise 16.3

Q14 Given P(A) =and P(B) =. Find P(A or B), if A and B are mutually exclusive events.

Answer. Here, P(A) = 3

5
, P (B) =
1

5

For mutually exclusive events A and B,
P(A or B) = P(A) + P(B)
∴ P(A or B) = + =
3

5
1

5
4

5

Page : 405 , Block Name : Exercise 16.3

Q15 If E and F are events such that P (E) = 1

4
, P (F ) =
1

2
and P (E and F ) =
1

8
, nd:(i) P(E or
F), (ii) P(not E and not F).

Answer. Here, P (E) = 1

4
, P (F ) =
1

2
, and P (E and F ) =
1

8

(i) We know that P(E or F) = P(E) + P(F) – P(E and F)
∴P(E or F) = = + − =
1 1 1 2+4−1 5
=
4 2 8 8 8

(ii) From (i), P(E or F) = P (E ∪ F) = 5

8

We have (E ∪ F) = (E ∩ F )
′ ′ ′
[ By De Morgan's law ]
′ ′ ′
∴ P (E ∩ F ) = P(E ∪ F)

′ 5 3
Now, P(E ∪ F) = 1 − P(E ∪ F) = 1 − =
8 8

′ ′ 3
∴ P (E ∩ F ) =
8

3
Thus, P( not E and not F) =
8

Page : 405 , Block Name : Exercise 16.3

Q16 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually
exclusive.

Answer. It is given that P (not E or not F) = 0.25
I.e., P (E ∪ F ) = 0.25
′ ′

′ ′ ′ ′
⇒ P(E ∩ F) = 0.25 [E ∪ F = (E ∩ F) ]
′
Now, P(E ∩ F) = 1 − P(E ∩ F)

⇒ P(E ∩ F) = 1 − 0.25

⇒ P(E ∩ F) = 0.75 ≠ 0

⇒ E ∩ F ≠ ϕ

Thus, E and F are not mutually exclusive.

Page : 405 , Block Name : Exercise 16.3

Q17 A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i)
P(not A), (ii) P (not B) and (iii) P(A or B).

Answer. It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16
(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74

Page : 405 , Block Name : Exercise 16.3

Q18 In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10%
of the class study both Mathematics and Biology. If a student is selected at random from the
class, nd the probability that he will be studying Mathematics or Biology.

Answer. Let A be the event in which the selected student studies Mathematics and B be the event
in which the selected student studies Biology.

40 2
Accordingly, P(A) = 40% = =
100 5

30 3
P(B) = 30%c6 = =
100 10

P(A and B) = 10% = 100
10
=
10
1

We know that P(A or B) = P(A) + P(B) - P(A and B)
2 3 1 6
∴ P(A or B) = + − = = 0.6
5 10 10 10

Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6.

Page : 405 , Block Name : Exercise 16.3

Q19 In an entrance test that is graded on the basis of two examinations, the probability of a
randomly chosen student passing the rst examination is 0.8 and the probability of passing the
second examination is 0.7. The probability of passing at least one of them is 0.95. What is the
probability of passing both?

Answer. Let A and B be the events of passing rst and second examinations respectively.
Accordingly, P(A) = 0.8, P(B) = 0.7 and P(A or B) = 0.95
We know that P(A or B) = P(A) + P(B) – P(A and B)
∴0.95 = 0.8 + 0.7 – P(A and B)
⇒ P(A and B) = 0.8 + 0.7 – 0.95 = 0.55
Thus, the probability of passing both the examinations is 0.55.

Page : 405 , Block Name : Exercise 16.3

Q20 The probability that a student will pass the nal examination in both English and Hindi is
0.5 and the probability of passing neither is 0.1. If the probability of passing the English
examination is 0.75, what is the probability of passing the Hindi examination?

Answer. Let A and B be the events of passing English and Hindi examinations respectively.
Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e.,P (A ∩ B ) = 0.1
′ ′

P(A) = 0.75
Now, (A ∪ B) = (A ∩ B )
′ ′ ′
[ De Morgan's law ]
′ ′ ′
∴ P(A ∪ B) = P (A ∩ B ) = 0.1

′
P(A ∪ B) = 1 − P(A ∪ B) = 1 − 0.1 = 0.9

We know that P(A or B) = P(A) + P(B) – P(A and B)
∴0.9 = 0.75 + P(B) – 0.5
⇒ P(B) = 0.9 – 0.75 + 0.5
⇒ P(B) = 0.65
Thus, the probability of passing the Hindi examination is 0.65.

Page : 405 , Block Name : Exercise 16.3

Q21 In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and
NSS. If one of these students is selected at random, nd the probability that

(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.

Answer. Let A be the event in which the selected student has opted for NCC and B be the event in
which the selected student has opted for NSS.
Total number of students = 60
Number of students who have opted for NCC = 30
30 1
∴ P(A) = =
60 2

Number of students who have opted for NSS = 32
32 8
∴ P(B) = =
60 15

Number of students who have opted for both NCC and NSS = 24
24 2
∴ P(A and B) = =
60 5

(i) We know that P(A or B) = P(A) + P(B) – P(A and B)
1 8 2 15+16−12 19
∴ P(A or B) = + − = =
2 15 5 30 30

Thus, the probability that the selected student has opted for NCC or NSS is 19

30
.
P( not A and not B)

(ii) = P (A and B )
′ ′

′ ′
= P (A ∩ B )

′ ′ ′ ′
= P(A ∪ B) [(A ∩ B ) = (A ∪ B) (by De Morgan's law ) ]

= 1 − P(A ∪ B)

= 1 − P(A or B)

19
= 1 −
30

11
=
30

Thus, the probability that the selected students has neither opted for NCC nor NSS is 11

30
.
(iii) The given information can be represented by a Venn diagram as

It is clear that
Number of students who have opted for NSS but not NCC
= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8
Thus, the probability that the selected student has opted for NSS but not for NCC
= 60
8
=
2

15

Page : 406 , Block Name : Exercise 16.3

Miscellaneous Exercise

Q1 A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles
are drawn from the box, what is the probability that
(i) all will be blue? (ii) at least one will be green?

Answer. Total number of marbles = 10 + 20 + 30 = 60
Number of ways of drawing 5 marbles from 60 marbles = C 60
5

(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.
5 blue marbles can be drawn from 20 blue marbles in C ways. 60
5
20

∴Probability that all marbles will be blue =
Cs

60
C5

(ii) Number of ways in which the drawn marble is not green = (20+10)
C5 =
30
C5
30

∴Probability that no marble is green =
Cs

60
C5
20

∴Probability that at least one marble is green = 1 −
Cs

60
C5

Page : 408 , Block Name : Miscellaneous Exercise

Q2 4 cards are drawn from a well – shuf ed deck of 52 cards. What is the probability
of obtaining 3 diamonds and one spade?

Answer.Number of ways of drawing 4 cards from 52 cards = s2
C4

In a deck of 52 cards, there are 13 diamonds and 13 spades.
∴Number of ways of drawing 3 diamonds and one spade = 13
C3 ×
13
C1
13 13

Thus, the probability of obtaining 3 diamonds and one spade =
C3 × C1

52C
4

Page : 408 , Block Name : Miscellaneous Exercise

Q3 A die has two faces each with number ‘1’, three faces each with number ‘2’ and
one face with number ‘3’. If die is rolled once, determine
(i) P(2) (ii) P(1 or 3) (iii) P(not 3)

Answer. Total number of faces = 6
(i) Number faces with number ‘2’ = 3
3 1
∴ P (2) = =
6 2

(ii) P (1 or 3) = P (not 2) = 1 − P (2) = 1 − 1

2
=
1

2

(iii) Number of faces with number ‘3’ = 1
1
∴ P(3) =
6

1 5
Thus, P( not 3) = 1 − P(3) = 1 − =
6 6

Page : 409 , Block Name : Miscellaneous Exercise

Q4 In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded.

What is the probability of not getting a prize if you buy
(a) one ticket (b) two tickets (c) 10 tickets.

Answer. Total number of tickets sold = 10,000
Number prizes awarded = 10
(a) If we buy one ticket, then
P (getting a prize) = =
10000
10

1000
1

∴P (not getting a prize) = 1 − 1

1000
=
999

1000

(b) If we buy two tickets, then
Number of tickets not awarded = 10,000 − 10 = 9990
9990

P (not getting a prize) =
C2

10000
C2

(c) If we buy 10 tickets, then
9990

P (not getting a prize) =
C10

10000
C10

Page : 409 , Block Name : Miscellaneous Exercise

Q5 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend
are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

Answer. My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students = C 100
2

(a) The two of us will enter the same section if both of us are among 40 students or among 60
students.
∴ Number of ways in which both of us enter the same section = C + C 40
2
60
2

∴ Probability that both of us enter the same section
140 160
40 60 +
C2 + C2 ⌊2⌊38 ⌊2⌊58

= =
100 |100
C2
⌊2⌊98

(b) P(we enter different sections)
= 1 − P(we enter the same section)
17 16
= 1 − =
33 33

Page : 409 , Block Name : Miscellaneous Exercise

Q6 Three letters are dictated to three persons and an envelope is addressed to each
of them, the letters are inserted into the envelopes at random so that each envelope
contains exactly one letter. Find the probability that at least one letter is in its
proper envelope.

Answer. L , L , L be three letters and E , E , and E be their corresponding envelops
1 2 3 1 2 3

respectively.
There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

L1 E 1 , L2 E 3 , L3 E 2
⎤
L2 E 2 , L1 E 3 , L3 E 1
⎥
⎥
⎥
L3 E 3 , L1 E 2 , L2 E 1 ⎥

⎦
L1 E 1 , L2 E 2 , L3 E 3

L1 E 2 , L2 E 3 , L3 E 1

L1 E 3 , L2 E 1 , L3 E 2

There are 4 ways in which at least one letter is inserted in a proper envelope.
Thus, the required probability is = . 4

6
2

3

Page : 409 , Block Name : Miscellaneous Exercise

Q7 A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.
Find (i) P(A ∪ B) (ii) P(A´ ∩ B´) (iii) P(A ∩ B´) (iv) P(B ∩ A´)

Answer. It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35
(i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B)
∴P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88
(ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law]
∴P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12
(iii) P(A ∩ B′) = P(A) − P(A ∩ B)
= 0.54 − 0.35
= 0.19
′
n(B∩A )=n(B)−n(A∩B)
(iv) We know that
′
n(B∩A ) n(B) n(A∩B)
⇒ = −
n(S) n(S) n(S)

′
∴ P (B ∩ A ) = P(B) − P(A ∩ B)

′
∴ P (B ∩ A ) = 0.69 − 0.35 = 0.34

Page : 409 , Block Name : Miscellaneous Exercise

Q8 From the employees of a company, 5 persons are selected to represent them in
the managing committee of the company. Particulars of ve persons are as follows:

A person is selected at random from this group to act as a spokesperson. What is
the probability that the spokesperson will be either male or over 35 years?

Answer. Let E be the event in which the spokesperson will be a male and F be the event in which
the spokesperson will be over 35 years of age.
Accordingly, P(E) = and P(F) =
3

5
2

5

Since there is only one male who is over 35 years of age,
We know that P(E ∪ F) = P(E) + P(F) − P(E ∩ F)

3 2 1 4 4
∴ P(E ∪ F) = + − = =
5 5 5 5 5

Thus, the probability that the spokesperson will either be a male or over 35 years of age is 4

5

Page : 409 , Block Name : Miscellaneous Exercise

Q9 If 4-digit numbers greater than 5,000 are randomly formed from the digits
0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,
(i) the digits are repeated? (ii) the repetition of digits is not allowed?

Answer.(i) When the digits are repeated
Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.
The remaining 3 places can be lled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is
allowed.
∴Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 − 1
= 250 − 1 = 249
[In this case, 5000 can not be counted; so 1 is subtracted]
A number is divisible by 5 if the digit at its units place is either 0 or 5.
∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 − 1 =
100 − 1 = 99
Thus, the probability of forming a number divisible by 5 when the digits are repeated is
=
99

249
=
33

83
.
(ii) When repetition of digits is not allowed

The thousands place can be lled with either of the two digits 5 or 7.
The remaining 3 places can be lled with any of the remaining 4 digits.
∴Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2
= 48
When the digit at the thousands place is 5, the units place can be lled only with 0 and the tens
and hundreds places can be lled with any two of the remaining 3 digits.
∴Here, number of 4-digit numbers starting with 5 and divisible by 5
=3×2=6
When the digit at the thousands place is 7, the units place can be lled in two ways (0 or 5) and
the tens and hundreds places can be lled with any two of the remaining 3 digits.
∴ Here, number of 4-digit numbers starting with 7 and divisible by 5
= 1 × 2 × 3 × 2 = 12
∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18
Thus, the probability of forming a number divisible by 5 when the repetition of digits is not
allowed is 18

48
=
3

8
.

Page : 409 , Block Name : Miscellaneous Exercise

Q10 The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e.,
from 0 to 9. The lock opens with a sequence of four digits with no repeats. What
is the probability of a person getting the right sequence to open the suitcase?

Answer. The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.
Number of ways of selecting 4 different digits out of the 10 digits = C 10
4

Now, each combination of 4 different digits can be arranged in ⌊4 ways.
⌊10
∴Number of four digits with no repetitions = 10
C4 × ⌊4=
⌊4⌊6
× ⌊4 = 7 × 8 × 9 × 10 = 5040

There is only one number that can open the suitcase.
Thus, the required probability is 1
.
5040

Page : 409 , Block Name : Miscellaneous Exercise