# ICSE Class 10 Maths Probability Previous Years Questions

Are you gearing up for your ICSE Class 10 exams and seeking comprehensive preparation material for the Probability section? Look no further! In this article, we have compiled a collection of ICSE Class 10 Maths Probability previous years questions from the year 2010 to 2023 to help you sharpen your understanding of this crucial topic. Whether you’re familiar with concepts like sample space, events, or conditional probability or need a refresher, these questions will reinforce your knowledge and boost your confidence.

By utilizing the previous years’ questions on Probability for ICSE Class 10 provided in this article, you can enhance your understanding and preparation for the upcoming exams. Strengthen your grasp of essential concepts, practice diverse problem-solving techniques, and boost your confidence in tackling Probability-related questions.

Remember, a solid foundation in Probability will not only help you excel in exams but also equip you with valuable skills for real-life decision-making scenarios. So, dive into the world of Probability and embark on a journey towards mastering this fascinating subject.

## ICSE Class 10 Maths Previous Years’ Questions on Probability

Q1. A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:
(a) multiple of 5
(b) a perfect square
(c) a prime number? [2023]

Solution: (a) 1/5 (b) 1/5 (c) 9/25

Step-by-step Explantion:

Let S be the sample space i.e bag contains cards numbered from 1 to 25.

n(S) = 25 .

(a) A event of getting the number on this card multiple of 5 is

n(A)= { 5, 10, 15, 20, 25} = 5

P(A) = 5/25 = 1/5

(b) Number of perfect squares between 1 to 25 = 5. The perfect squares are 1, 4, 9, 16, 25.

Hence P (a perfect square) = 5/25 = 1/5

(c) number of prime number between 1 to 25 = 9 (2, 3, 5, 7, 11, 13, 17, 19, 23)

Hence P (a prime number) = 9/25

Q2. A letter is chosen at random from all the letters of the English alphabets. The probability that the letter chosen is a vowel, is:
(a) 4/26
(b) 5/26
(c) 21/26
(d) 5/24 [2023]

Solution: (b) 5/26

Step-by-step Explantion: Total number of English Alphabets are 26, namely, A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z.

Let S be the sample space i.e. n (S) = 26

Vowels in English alphabets are 5- A, E, I, O, U.

Hence, P (a vowel) = 5/26 (Option b)

Q3. A letter of the word ‘SECONDARY’ is selected at random. What is the probability that the letter selected is not a vowel? [2022 Semester-2]

Solution: 2/3

Step-by-step Explantion: Number of letters in the word SECONDARY = 9.

Let S be the sample space. Hence, n (S) = 9.

letters which are not vowel are S, C, N, D, R, Y = 6

Hence, P (not a vowel) = 6/9 = 2/3

Watch the video solution of the PYQs HERE.

Q4. The bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball? [2022 Semester-2]

Solution: 1/5

Step-by-step Explantion: Total number of balls in the bag = 5 + 2 + 3 =10

Let S be the sample space. Hence, n(S) = 10

Number of red balls = 2.

Hence, P (a red ball) = 2/10 = 1/5

Q5. The Probability of getting a number divisible by 3 in throwing a dice is:
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3 [2022 Semester-2]

Solution: (b) 1/3

Step-by-step Explantion: The numbers in a dice are 1, 2, 3, 4, 5, 6 = 6.

Let S be the sample space. Hence, n(S)= 6.

Numbers divisible by 3 in a dice are = 3 and 6 = 2

Hence P(getting a number divisible by 3) = 2/6 = 1/3

Hence option (b) is correct.

Q6. Each of the letters of the word ‘AUTHORIZES’ is written on identical circular discs and put in a bag. They are well shuffled. If a disc is drawn at random from the bag, what is the probability that the letter is: [2020]
(i) a vowel
(ii) one of the first 9 letters of the English alphabet which appears in the given word.
(iii) one of the last 9 letters of the English alphabet which appears in the given word.

Solution: (i) 1/2 (ii) 2/5 (iii) 1/2

Step-by-step Explantion:

Number of letters in the word ‘AUTHORIZES’ is 10.

Let S be the sample space. Hence n(S) = 10.

(i) Vowels in the word AUTHORIZES = A, U, O, I, E = 5

Hence P (a vowel) = 5/10 = 1/2

(ii) One of the first 9 letters of the English alphabet which appears in the word AUTHORIZES are = A, H, I, E = 4

Hence P (one of the first 9 letters of the English alphabet which appears in the word) = 4/10 = 2/5

(iii) One of the last 9 letters of the English alphabet which appears in the word AUTHORIZES are = U, T, R, Z, S = 5

Hence, P(One of the last 9 letters of the English alphabet which appears in the word) = 5/10 = 1/2

ICSE Class 10 Mathematics Board Chapter-wise PYQs with Solution

Q7. There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is: [3]
(i) an odd number
(ii) divisible by 2 and 3 both
(iii) a number less than 16. [2019]

Solution: (i) 13/25 (ii) 4/25 (iii) 3/5

Step-by-step Explantion: There are 25 discs numbered 1 to 25. Let S be the sample space.

Hence, n(S) = 25.

(i) odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 = 13

Hence, P (odd number) = 13/25.

(ii) Numbers divisible by 2 and 3 both are = 6, 12, 18, 24 = 4

Hence, P(Number divisible by 2 and 3 both) = 4/25.

(iii) Numbers less than 16 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 = 15

Hence, P(Number less than 16) = 15/25 = 3/5.

Q8. Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is :
(i) a prime number.
(ii) a number divisible by 4.
(iii) a number that is a multiple of 6.
(iv) an odd number [4] [2018]

Solution: (i) 1/10 (ii) 1/2 (iii) 3/10 (iv) 0

Step-by-step Explantion: Total number of cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 =10.

Let S be the sample space. Hence n(S) = 10.

(i) Prime numbers in the cards is only number 2. Hence total number of cards having prime number = 1

Hence, P (a prime number) = 1/10

(ii) numbers divisible by 4 in cards are 4, 8, 12, 16, 20 = 5.

Hence, P(number divisible by 4) = 5/10 = 1/2.

(iii) numbers multiple of 6 in the cards are 6, 12, 18 = 3

Hence, P(a number that is a multiple of 6) = 3/10.

(iv) odd numbers in the cards are 0.

Hence, P(an odd number) = 0/10 = 0.

Q9. Sixteen cards are labelled as a, b, c … m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is : [3]
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median [3] [2017]

Solution: (i) 1/4 (ii) 3/4 (iii) 5/8

Step-by-step Explantion: Sixteen cards are labelled as a, b, c … m, n, o, p.

Let S be the sample space. Hence n(S) = 16.

(i) Vowels in the cards are a, e, i, o = 4

Hence, P(a vowel) =4/16 = 1/4.

(ii) consonants in the cards are b, c, d, f, g, h, j, k, l, m, n, p = 12

Hence, P(a consonant) = 12/16 = 3/4.

(iii) none of the letters of the word ‘median’ in the cards are b, c, f, g, h, j, k, l, o, p = 10.

Hence, P(none of the letters of the word median) = 10/16 = 5/8.

Q10. A game of number has cards marked with 11, 12, 13, 40. A card is drawn at random. Find the Probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7. [3] [2016]

Solution: (i) 0 (ii) 0

Step-by-step Explantion: A game of number has cards marked with 11, 12, 13, 40.

Let S be the sample space. Hence, n (S) = 4

(i) Perfect squares in the numbers 11, 12, 13 40 are = 0.

Hence P(a perfect square) = 0/4 = 0.

(ii) Numbers divisible by 7 in the numbers 11, 12, 13 40 are = 0.

Hence P(divisible by 7) = 0/4 = 0.

Q11. A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball.
(ii) a white or a red ball
(iii) is neither a green ball nor a white ball. [2015]

Solution: (i) 9/20 (ii) 11/20 (iii) 3/10

Step-by-step Explantion: Let S be the sample space. Total number of balls in the bag = 5 + 6 + 9 = 20.

Hence, n(S) = 20

(i) Number of green balls = 9.

Hence, P(a green ball) = 9/20.

(ii) Number of white and red balls = 5 + 6 = 11.

Hence, P(a white or a red ball) = 11/20.

(iii) Neither green ball or white balls mean red balls. Number of red balls = 6.

Hence, P(neither a green ball nor a white ball) = 6/20 = 3/10.

Q12. A die has 6 faces marked by the given numbers shown below:

The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer. [2014]

Solution: (i) 1/2 (ii) 5/6 (iii) 1/6

Step-by-step Explantion: Numbers in a die are 6 (1, 2, 3, -1, -2, -3).

Let S be the sample space. Therefore, n(S) = 6.

(i) Number of positive integers in the die is 3 (1, 2, 3).

Hence, P(a positive integer) = 3/6 = 1/2.

(ii) Number of integers in a die which are greater than -3 is 5 (1, 2, 3, -1, -2).

Hence, P(an integer greater than -3) = 5/6.

(iii) the smallest integer is -3.

Hence P(smallest integer) = 1/6.

Q13. A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box. [3] [2013]

Solution: 12

Step-by-step Explantion: Let the number of Black balls be x. Therefore, total number of balls = x + 30.

Let S be the sample space. Hence, n(S) = x + 30.

Number of white balls = 30.

Hence, P(a white ball) = 30/ (x+30)

Number of black balls = x.

Hence, P(a black ball) = x/(x+30)

Given,

$\frac x{x+30}=\frac25\times\frac{30}{x+30}\\\Rightarrow x=\frac{2\times30}5\\\Rightarrow x=12$

Hence, number of black balls is 12.

Q14. Two coins are tossed once. Find the probability of getting:
(ii) At least 1 tail. [2012]

Solution: (i) 1/4 (ii) 3/4

Step-by-step Explantion:

Two coins are tossed. The sample set S will be {HH, HT, TH, TT}. Hence, n(S) = 4

(i) Let A be the event of getting two heads. Then event A will contain the outcome {HH} = 1

Hence, P(A) = 1/4.

(ii) Let B be the event of getting at least 1 tail. Then, B will contain the outcomes {HT, TH, TT} = 3

P(B) = 3/4.

Q15. From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is:
(i) a face card (King, Jack or Queen)
(ii) an even-numbered red card? [3] [2011]

Solution: (i) 3/10 (ii) 1/4

Step-by-step Explantion: From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed.

So, 3, 6, 9 are removed from each set. Hence total cards removed = 3×4 = 12.

Total number of cards left = 52 – 12 = 40.

Let S be the sample space. Hence, n(S) = 40.

(i) Total face cards = 4 kings, 4 queens and 4 Jacks = 12

Hence, P(a face card) = 12/40 =3/10.

(ii) even-numbered red cards (i.e from Hearts and diamonds) = 2(2, 4, 6, 8, 10 )= 2(5) = 2×5 =10.

Hence, P(an even-numbered red card) = 10/40 = 1/4.

Q16. Cards marked with numbers 1, 2, 3, 4 … 20 are well shuffled and a card is drawn at random. What is the probability that the number of the cards is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square? [3] [2010]

Solution: (i) 2/5 (ii) 3/10 (iii) 1/5

Step-by-step Explantion: Cards are marked with numbers 1, 2, 3, 4 … 20. Therefore, total number of cards = 20.

Let S be the sample space. Hence, n(S) = 20.

(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 = 8.

Hence, P(a prime number) = 8/20 = 2/5.

(ii) numbers divisible by 3 are 3, 6, 9, 12, 15, 18 = 6.

Hence, P(divisible by 3) = 6/20 = 3/10.

(iii) Perfect squares are 1, 4, 9, 16 = 4.

Hence, P(a perfect square) = 4/20 = 1/5.

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