Class 8 SELINA Solutions Maths Chapter 24: Probability
Probability Exercise Ex. 24
Solution 1(i)
Correct option: (a) 8
When a coin is tossed three times, the number of possible outcomes
= 8 (HHH, HTH, HHT, THH, THT, TTH, HTT, TTT)
Solution 1(ii)
Correct option: (a) 1 – P(A)
P(not getting A) = 1 – P(A)
Solution 1(iii)
Correct option: (c) 
When a coin is tossed once, the number of possible outcomes = 2 (H, T)
For a tail,
Total number of favourable outcomes = 1 (T)
Therefore, required probability
Solution 1(iv)
Correct option: (a) 
When a coin is tossed two times, the number of possible outcomes = 4 (HH, HT, TH, TT)
For atleast one tail,
Total number of favourable outcomes = 3 (HT, TH, TT)
Therefore, required probability
Solution 1(v)
Correct option: (c) 
Total number of possible outcomes = Total number of cards = 52
For a face card,
Total number of favourable outcomes = 12 (3 diamond, 3 heart, 3 spade, 3 club)
Therefore, required probability
Solution 2
When a coin is tossed two times, the number of possible outcomes = 4 (HH, HT, TH, TT)
(i) For exactly one head,
Total number of favourable outcomes = 2 (HT, TH)
Therefore, required probability
(ii) For exactly one tail,
Total number of favourable outcomes = 2 (HT, TH)
Therefore, required probability
(iii) For two tails,
Total number of favourable outcomes = 1 (TT)
Therefore, required probability
(iv) For two heads,
Total number of favourable outcomes = 1 (HH)
Therefore, required probability
Solution 3
Total number of possible outcomes = 6 (p, e, n, c, i, l)
For a consonant,
Total number of favourable outcomes = 4 (p, n, c, l)
Therefore, required probability 
Solution 4
Total number of possible outcomes = Total number of balls
= 1 black ball + 1 red ball + 1 green ball
= 3 balls
= 3
(i) For a red ball,
Total number of favourable outcomes = 1
Therefore, required probability
(ii) For not a red ball,
Total number of favourable outcomes = 2
Therefore, required probability
(iii) For a white ball,
Total number of favourable outcomes = 0
Therefore, required probability = 0
Solution 5
In a single throw of a die, total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
(i) For a number greater than 2,
Total number of favourable outcomes = 4 (3, 4, 5, 6)
Therefore, required probability
(ii) For a number less than or equal to 2,
Total number of favourable outcomes = 2 (1, 2)
Therefore, required probability
(iii) For a number not greater than 2,
Total number of favourable outcomes = 2 (1, 2)
Therefore, required probability 
Solution 6
Total number of possible outcomes = Total number of balls
= 3 white balls + 5 black balls + 2 red balls
= 10 balls
= 10
(i) For a black ball,
Total number of favourable outcomes = 5
Therefore, required probability 
(ii) For a red ball,
Total number of favourable outcomes = 2
Therefore, required probability 
(iii) For a white ball,
Total number of favourable outcomes = 3
Therefore, required probability 
(iv) For not a red ball,
Total number of favourable outcomes
= Number of white balls + Number of black balls
= 3 + 5
= 8
Therefore, required probability 
(v) For not a black ball,
Total number of favourable outcomes
= Number of white balls + Number of red balls
= 3 + 2
= 5
Therefore, required probability 
Solution 7
In a single throw of a die, total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
(i) For an even number,
Total number of favourable outcomes = 3 (2, 4, 6)
Therefore, required probability 
(ii) For an odd number,
Total number of favourable outcomes = 3 (1, 3, 5)
Therefore, required probability 
(iii) For not an even number,
Total number of favourable outcomes = 3 (1, 3, 5)
Therefore, required probability 
Solution 8
In a single throw of a die, total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
(i) For a number 8,
Total number of favourable outcomes = 0
Therefore, required probability = 0
(ii) For a number greater than 8,
Total number of favourable outcomes = 0
Therefore, required probability = 0
(iii) For a number less than 8,
Total number of favourable outcomes = 6 (1, 2, 3, 4, 5, 6)
Therefore, required probability 
Solution 9
The probability of an event can never be less than 0 or more than 1.
(i) 
(ii) 3.8: Since 3.8 > 1, it cannot be the probability of an event.
(iii) 127%: 1.27% = 1.27, which is greater than 1, it cannot be the probability of an event.
(iv) –0.8: Since –0.8 < 0, it cannot be the probability of an event.
Solution 10
Total number of possible outcomes = Total number of balls
= 6 black balls
= 6
(i) For a white,
Total number of favourable outcomes = 0
Therefore, required probability = 0
(ii) For a black ball,
Total number of favourable outcomes = 6
Therefore, required probability
Solution 11
When three identical coins are tossed, the number of possible outcomes
= 8 (HHH, HTH, HHT, THH, THT, TTH, HTT, TTT)
(i) For all heads,
Total number of favourable outcomes = 1 (HHH)
Therefore, required probability 
(ii) For exactly two heads,
Total number of favourable outcomes = 3 (HTH, HHT, THH)
Therefore, required probability 
(iii) For exactly one head,
Total number of favourable outcomes = 3 (THT, TTH, HTT)
Therefore, required probability
(iv) For no head,
Total number of favourable outcomes = 1 (TTT)
Therefore, required probability
Solution 12
Total number of possible outcomes = Total number of pages = 92
For sum of digits in the page number is 9.
Total number of favourable outcomes = 10 (9, 18, 27, 36, 45, 54, 63, 72, 81, 90)
Therefore, required probability 
Solution 13
When two coins are tossed together, the number of possible outcomes
= 4 (HH, HT, TH, TT)
(i) For atleast one head,
Total number of favourable outcomes = 3 (HH, HT, TH)
Therefore, required probability 
(ii) For both heads or both tails,
Total number of favourable outcomes = Both Heads + Both Tails
= 1 (HH) + 1 (TT)
= 2
Therefore, required probability 
Solution 14
Total number of possible outcomes = Total number of cards
= 10 (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
(i) For a multiple of 2,
Total number of favourable outcomes = 5 (2, 4, 6, 8, 10)
Therefore, required probability 
(ii) For a multiple of 3,
Total number of favourable outcomes = 3 (3, 6, 9)
Therefore, required probability 
(iii) For a multiple of 2 and 3,
Total number of favourable outcomes = 1 (6)
Therefore, required probability 
(iv) For a multiple of 2 or 3,
Total number of favourable outcomes = 7 (2, 3, 4, 6, 8, 9, 10)
Therefore, required probability 
Solution 15
When two dice are thrown at the same time, total number of possible outcomes
= 36
{(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)}
(i) For a sum of number on top is 0,
Total number of favourable outcomes = 0
Therefore, required probability = 0
(ii) For a sum of number on top is 12,
Total number of favourable outcomes = 1 {(6, 6)}
Therefore, required probability
(iii) For a sum of number on top is less than 12,
Total number of favourable outcomes
= Total possible outcomes – Number of outcomes with sum of number on top is 12
= 36 – 1
= 35
Therefore, required probability 
(iv) For a sum of number on top is less than or equal to 12,
Total number of favourable outcomes = 36
Therefore, required probability 
Solution 16
In a single throw of a die, total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
(i) For a prime number,
Total number of favourable outcomes = 3 (2, 3, 5)
Therefore, required probability
(ii) For a number greater than 3,
Total number of favourable outcomes = 3 (4, 5, 6)
Therefore, required probability
(iii) For a number other than 3 and 5,
Total number of favourable outcomes = 4 (1, 2, 4, 6)
Therefore, required probability 
(iv) For a number less than 6,
Total number of favourable outcomes = 5 (1, 2, 3, 4, 5)
Therefore, required probability 
(v) For a number greater than 6,
Total number of favourable outcomes = 0
Therefore, required probability = 0
Solution 17
When two coins are tossed together, the number of possible outcomes
= 4 (HH, HT, TH, TT)
(i) For exactly one tail,
Total number of favourable outcomes = 2 (HT, TH)
Therefore, required probability 
(ii) For atleast one head,
Total number of favourable outcomes = 3 (HH, HT, TH)
Therefore, required probability 
(iii) For no head,
Total number of favourable outcomes = 1 (TT)
Therefore, required probability 
(iv) For atmost one head,
Total number of favourable outcomes = 3 (HT, TH, TT)
Therefore, required probability 
Solution 18
When two dice are thrown simultaneously, total number of possible outcomes
= 36
{(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)}
(i) For same number on both the dice,
Total number of favourable outcomes = 6 {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Therefore, required probability
(ii) For a sum of 7 on the uppermost faces of both the dice,
Total number of favourable outcomes = 6 {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Therefore, required probability
Probability Exercise TEST YOURSELF
Solution 1(i)
Correct option: (c) 
When two dice are rolled simultaneously, the total number of possible outcomes = 36
{(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)}
For a sum equal to 5,
Total number of favourable outcomes = 4 {(1, 4), (2, 3), (3, 2), (4, 1)}
Therefore, required probability 
Solution 1(ii)
Correct option: (b) 0.5
In a single throw of a die, the total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
For an odd number,
Total number of favourable outcomes = 3 (1, 3, 5)
Therefore, required probability 
Solution 1(iii)
Correct option: (c) 
Total number of possible outcomes = Total number of cards = 52
For a club card,
Number of favourable outcomes = Total number of club cards = 13
Therefore, required probability 
Solution 1(iv)
Correct option: (d) 
In a single throw of a dice, the total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
For a number not more than 5,
Total number of favourable outcomes = 5 (1, 2, 3, 4, 5)
Therefore, required probability 
Solution 1(v)
Correct option: (a) 
In a single throw of a dice, the total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
For a prime number,
Total number of favourable outcomes = 3 (2, 3, 5)
Therefore, required probability 
Solution 2
When two dice are rolled together, the total number of possible outcomes = 36
{(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)}
For a sum equal to an odd number,
Total number of favourable outcomes = 18
{(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
Therefore, required probability 
Solution 3
When two dice are rolled together, the total number of possible outcomes = 36
{(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)}
For a total of atleast 11,
Total number of favourable outcomes = 3 {(5, 6), (6, 5), (6, 6)}
Therefore, required probability 
Solution 4
Total number of possible outcomes = Total number of cards = 52
For a black queen,
Number of favourable outcomes = 2 (1 spade queen, 1 club queen)
Therefore, required probability 
Solution 5
For a leap year, number of days = 366 = 52 weeks + 2 days
52 weeks will have 52 Tuesdays.
The remaining two days can be as follows:
(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thur), (Thur, Fri), (Fri, Sat), (Sat, Sun)
For 53 Tuesdays, one of these days has to be Tuesday.
So, we have
Total number of possible outcomes = 7
Total number of favourable outcomes = 2 {(Mon, Tue), (Tue, Wed)}
Therefore, required probability 
Solution 6
Total number of possible outcomes = Total number of slips with numbers 1 to 10 = 10
(i) For getting a number less than 6,
Total number of favourable outcomes = 5 (1, 2, 3, 4, 5)
Therefore, required probability 
(ii) For getting a single digit number,
Total number of favourable outcomes = 9 (1, 2, 3, 4, 5, 6, 7, 8, 9)
Therefore, required probability 
Solution 7
Total number of possible outcomes = Total number of cards from 1 to 100 = 100
For drawing a square number,
Total number of favourable outcomes = 10 (1, 4, 9, 16, 25, 36, 49, 64, 81, 100)
Therefore, required probability 
Solution 8
When a dice is tossed once, total number of possible outcomes
= 6 (1, 2, 3, 4, 5, 6)
For a number 7, total number of favourable outcomes = 0
Therefore, required probability = 0
Solution 9
Total number of possible outcomes = Total number of sectors
= 5 (3 green, 1 blue, 1 red)
For getting a non-blue sector,
Total number of favourable outcomes = 3 green + 1 red = 4
Therefore, required probability 
Solution 10
Total number of possible outcomes = Total number of cards numbered 1, 2, 3, ….. 21 = 21
For a number on the card divisible by 2 or 3,
Total number of favourable outcomes = 14 (2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21)
Therefore, required probability 