Latest Maharashtra State Board (SSC & HSC) 2026-27 Syllabus Digest & Solutions Updated!
Class 10 (SSC Board)Mathematics & Statistics2026-27 Syllabus

Chapter 5 Probability Practice Set 5.4 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 5 Probability Practice Set 5.4. Step-by-step solved exercises, numerical problems, and digest answers.

5 Solved Questions811 words

Practice Set 5.4 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1 Maharashtra Board Solution
If two coins are tossed, find the probability of the following events. i. Getting at least one head. ii. Getting no head.
Solution & Step-by-Step Answer:
Sample space, S = {HH, HT, TH, TT} ∴ n(S) = 4

i. Let A be the event of getting at least one head.
∴ A = {HT, TH, HH}
∴ n(A) = 3
∴ P(A) =
∴ P(A) =

ii. Let B be the event of getting no head.
∴ B = {TT}
∴ n(B) = 1
∴ P(B) =
∴ P(B) =
∴ P(A) = ; P(B) =

Question 2 Maharashtra Board Solution
If two dice are rolled simultaneously, find the probability of the following events. i. The sum of the digits on the upper faces is at least 10. ii. The sum of the digits on the upper faces is 33. iii. The digit on the first die is greater than the digit on second die.
Solution & Step-by-Step Answer:
Sample space, 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)} ∴ n(S) = 36

i. Let A be the event that the sum of the digits on the upper faces is at least 10.
∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = =
∴ P(A) =

ii. Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B) = =
∴ P(B) = 0

iii. Let C be the event that the digit on the first die is greater than the digit on the second die.
C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5),
∴ n(C) = 15
∴ P(C) = =
∴ P(C) =
∴ P(A) = ; P(B) = 0; P(C) =

Question 3 Maharashtra Board Solution
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn: i. shows an even number. ii. shows a number which is a multiple of 5.
Solution & Step-by-Step Answer:
Sample space, S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15} ∴ n(S) = 15

i. Let A be the event that the ticket drawn shows an even number.
∴ A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
∴ P(A) =
∴ P(A) =

ii. Let B be the event that the ticket drawn shows a number which is a multiple of 5.
∴ B = {5, 10, 15}
∴ n(B) = 3
∴ P(B) = =
∴ P(B) =
∴ P(A) = ; P(B) =

Question 4 Maharashtra Board Solution
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is i. an odd number? ii. a multiple of 5?
Solution & Step-by-Step Answer:
Sample space (S) = {23, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97} ∴ n(S) = 20 i. Let A be the event that the number formed is an odd number. ∴ A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75,79,93,95,97} ∴ n(A) = 16 ∴ P(A) = = ∴ P(A) =

ii. Let B be the event that the number formed is a multiple of 5.
∴ B = {25,35,75,95}
∴ n(B) = 4
∴ P(B) = =
∴ P(B) =
∴ P(A) = ; P(B) =

Question 5 Maharashtra Board Solution
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is i. an ace. ii. a spade.
Solution & Step-by-Step Answer:
There are 52 playing cards. ∴ n(S) = 52 i. Let A be the event that the card drawn is an ace. ∴ n(A) = 4 ∴ P(A) = = ∴ P(A) =

ii. Let B be the event that the card drawn is a spade.
∴ n(B) = 13
∴ P(B) = =
∴ P(B) =
∴ P(A) = ; P(B) =