Probability Class 10 Concise Exe-25A ICSE Maths Selina Solutions. In this article you will learn how to solve problems on Some Basic Terms and Measurement of Probability.. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Probability Class 10 Concise Exe-25A ICSE Maths Selina Solutions Ch-25

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-25	Probability
Writer	R.K. Bansal
Exe-25A	Some Basic Terms and Measurement of Probability.
Edition	2025-2026

 Some Basic Terms and Measurement of Probability

Class 10 Concise Exe-25A ICSE Maths Selina Solutions Ch-25 Probability

Que-1: A coin is tossed once. Find the probability of: (i) getting a tail    (ii) not getting a tail

Sol: In a random experiment of tossing coin once, total number of possible outcomes are 2 which are Head (H) and Tail (T)
(i) Favourable outcome is ‘getting a tail’.
∴ Number of favourable outcome = 1.
P(getting a tail) = No. of favourable outcomes / No. of possible outcomes = 1/2
Hence, the probability of getting a tail = 1/2​.

(ii) Favourable outcome is ‘not getting a tail’ or ‘we can say getting a head’.
∴ Number of favourable outcome = 1.
P(not getting a tail) = No. of favourable outcomes / No. of possible outcomes  = 1/2
Hence, the probability of not getting a tail = 1/2​.

Que-2: A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.

Sol: Total number of ball  3+5+2 = 10
Total number of events = p(n) = 10
(i) There are 5 black balls
favourable number of events = p(A) = 5
Hence, p(getting a black ball)=𝑃⁡(𝐴)/𝑃⁡(𝑛) = 5/10 = 1/2

(ii) a red ball = 2
favourable number of events = p(A) = 2
Hence, p(getting a black ball)=𝑃⁡(𝐴)/𝑃⁡(𝑛) = 2/10 = 1/5

(iii) a white ball = 3
favourable number of events = p(A) = 3
Hence, p(getting a black ball)=𝑃⁡(𝐴)/𝑃⁡(𝑛) = 3/10.

(iv) not a red ball = 3+5 = 8
favourable number of events = p(A) = 8
Hence, p(getting a black ball)=𝑃⁡(𝐴)/𝑃⁡(𝑛) = 8/10 = 4/5

(v) not a black ball = 3+2 = 5
favourable number of events = p(A) = 5
Hence, p(getting a black ball)=𝑃⁡(𝐴)/𝑃⁡(𝑛) = 5/10 = 1/2

Que-3: In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.

Sol: (i) Sample space = {1, 2, 3, 4, 5, 6}
n(s) = 6
E = event of getting a number more greater than 4 = {5,6}
n(E) = 4
Probability of a number more greater than 4 = 𝑛⁡(𝐸)/𝑛⁡(𝑠) = 2/6 = 1/3

(ii) E = event of getting a number less than or equal to 4 = {1, 2, 3, 4}
n(E) = 4
Probability of a number not greater than 4 = 𝑛⁡(𝐸)/𝑛⁡(𝑠) = 4/6 = 2/3

(iii) E = event of getting a number not greater than 4 = {1, 2, 3, 4}
n(E) = 4
Probability of a number not greater than 4 = 𝑛⁡(𝐸)/𝑛⁡(𝑠) = 4/6 = 2/3

Que-4: In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.

Sol: A die has six numbers: 1, 2, 3, 4, 5, 6
∴ Number of possible outcome = 6
(i) Number of favourable outcomes = an even number i.e. 2, 4, 6 which are 3 in numbers
∴ P(E) = Number of favourable outcome / Number of all possible outcome
= 3/6 = 1/2

(ii) & (iii) Number of favourable outcome = not an even number i.e. odd numbers : 1,3,5 which are 3 in numbers
∴ P(E) = Number of favourable outcome / Number of all possible outcome
= 3/6 = 1/2

Que-5: From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.

Sol: Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26
Probability of drawing a black card = P(E)/P(s) = 26/52 = 1/2

(ii) Number of red cards in a deck = 26
Therefore, number of non-red cards = 52 – 26 = 26
P(E) = favourable outcomes for the event of not drawing a red card = 26
Probability of not drawing a red card = 𝑃⁡(𝐸)/𝑃⁡(𝑠) = 26/52 = 1/2

(iii) Number of red cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26
Probability of drawing a black card = P(E)/P(s) = 26/52 = 1/2

(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12
Probability of drawing a black card = P(E)/P(s) = 12/52 = 3/13

(v) There are 26 cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens, and 2 jacks).
P(E) = 6
Probability of drawing a black card = P(E)/P(s) = 6/52 = 3/26

Que-6: (i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Sol: (i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Que-7: In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta     (ii) not winning of Ritu

Sol: (i) Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27

Que-8: In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh       (ii) winning of John

Sol: (i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46

Que-9: (i) Write the probability of a sure event
(ii) Write the probability of an event when impossible
(iii) For an event E, write a relation representing the range of values of P(E)

Sol: (i) The probability of a sure event is 1
i.e. P(S) = 1 where ‘S’ is the sure event.
Proof: In a sure event n(E) = n(S)
[Since number of elements in Event ‘E’ will be equal to the number of element in sample space.]
By definition of probability:
𝑃⁡(𝑆) = 𝑛⁡(𝐸)/𝑛⁡(𝑆) = 1
P(S) = 1

(ii) The probability of an impossible event is ‘0’
i.e. P(S) = 0
Proof: Since E has no element, n(E) = 0
From definition of probability:
𝑃⁡(𝑆) = 𝑛⁡(𝐸)/𝑛⁡(𝑆) = 0/𝑛⁡(𝑆)
P(S) = 0

(iii) The probability of an event lies between ‘0’ and ‘1’.
i.e 0 ≤ P(E) ≤ 1.
Proof: Let ‘S’ be the sample space and ‘E’ be the event.
Then
0 ≤ n(E) ≤ n(S)
0/𝑛⁡(𝑆) ≤ 𝑛⁡(𝐸)/𝑛⁡(𝑆) ≤ 𝑛⁡(𝑆)/𝑛⁡(𝑆)
or 0 ≤ P(E) ≤ 1
The number of elements in ‘E’ can’t be less than ‘0’ i.e. negative and greater than the number of elements in S.

Que-10: In a single throw of die, find the probability of getting:
(i) 5    (ii) 8     (iii) a number less than 8      (iv) a prime number

Sol: On a die, the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6

(i) Number of a favourable outcomes = 1
P(E) = Number of favourable outcomes / Number of all possible outcome = 1/6.

(ii) Number of a favourable outcomes = 0 (∵ 8 is not possible)
∴ P(E) = Number of favourable outcomes / Number of all possible outcome = 0/6 = 0

(iii) Number less than 8 will be 1,2,3,4,5,6
∴ P(E) = Number of favourable outcomes / Number of all possible outcome = 6/6 = 1

(iv) Number of prime number outcomes = {2, 3, 5}
∴ P(E) = Number of favourable outcomes / Number of all possible outcome = 3/6 = 1/2

Que-11: A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3

Sol: (i) Total number of outcomes = 6.
Out of the given numbers, even numbers are 2, 4 and 6.
Numbers of favourable outcomes = 3.
∴ P(getting an even number) = Number of favourable outcomes / Number of all possible outcomes
= 3/6 = 1/2
Thus, the probability of getting an even number is 1/2.
Hence, the correct answer is option 1/2.

(ii) Sample space = {1, 2, 3, 4, 5, 6}
n(S) = 6
E = the possible even numbers between 3 and 8 = {4, 5, 6}
n(E) = 3
Probability of getting an even number between 3 and 8 = P(S)
= n⁡(E)n⁡(S) = 3/6 = 1/2

(iii) The number of possible outcomes is six: 1, 2, 3, 4, 5 and 6.
Among these numbers; 2, 4, 6 are the even numbers and 3, 6 are the multiples of 3.
Favourable outcomes are {2, 3, 4, 6}.
So, P (getting an even number or a multiple of 3)  =Number of favourable outcomes / Total number of all possible outcomes = 4/6 = 2/3

Que-12: Which of the following cannot be the probability of an event?
(i) 3/5     (ii) 2.7     (iii) 43%      (iv) -0.6     (v) -3.2      (vi) 0.35

Sol: The probability of an event lies between ‘0’ and ‘1’ i.e. 0 ≤ P(E) ≤ 1.
(i) 35 =0.6
∵ 0 ≤ 0.6 ≤ 1
Hence, it can be the probability of an event.

(ii) 2.7
∵ 0 ≤ 1 ≤ 2.7
Hence, it cannot be the probability of an event.

(iii) 43% =43100 =0.43
0 ≤ 0.43 ≤ 1
Hence, it can be the probability of an event.

(iv) – 0.6
– 0.6 ≤ 0 ≤ 1
Hence, it cannot be the probability of an event.

(v) – 3.2
– 3.2 ≤ 0 ≤ 1
Hence, it cannot be the probability of an event.

(vi) 0.35
0 ≤ 0.35 ≤ 1
Hence, it can be the probability of an event.

Que-13: A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball         (ii) a black ball

Sol: There are six identical black balls.
∴ No. of possible outcomes = 6.
(i) A white ball
As there is no white ball in the bag.
∴ No. of favourable outcomes (of getting a white ball) = 0.
P(getting a white ball) = No. of favourable outcomes / No. of possible outcomes = 0/6 = 0
Hence, the probability of getting a white ball = 0.

(ii) There are 6 identical black balls.
∴ No. of favourable outcomes (of getting a black ball) = 6.
P(getting a black ball) = No. of favourable outcomes / No. of possible outcomes = 6/6 = 1
Hence, the probability of getting a black ball = 1.

Que-14: A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.

Sol: Possible outcomes = S = {‘P’, ‘r’, ‘o’, ‘b’, ‘a’, ‘b’, ‘i’, ‘l’, ‘i’, ‘t, ‘y’}
n(S) = 11
Event of selection of vowels = E = {‘o’, ‘a’, ‘i’, ‘i’}
n(E) = 4
Probability of selection of a vowel = P(S)
= n(E) / n(S)
= 4/11

Que-15: Ramesh chooses a date at random in January for a party (see the following figure).

Find the probability that he chooses :
(i) a Wednesday     (ii) a Friday       (iii) a Tuesday or a Saturday.

Sol: Number of possible outcomes = number of day in the month = 31
n(S) = 31
(i) E = event of selection of a Wednesday = {1, 8, 15, 22, 29}
n(E) = 5
Probability of selection of a Wednesday = P(S)
= n(E) / n(S)
= 5/31

(ii) E = event of selection of a Friday = {3, 10, 17, 24, 31}
n(E) = 5
Probability of selection of a Friday = P(S)
= n(E) / n(S)
= 5/31

(iii) E = event of selection of a Tuesday or a Saturday = {4, 7, 11, 14, 18, 21, 25, 28}
n(E) = 8
Probability of selection of a Tuesday or a Saturday = P(S)
= n(E) / n(S)
= 8/31

— : End of Probability Class 10 Concise Exe-25A ICSE Maths Selina Solutions. :–

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