Multiple Choice Questions on Probability Class 11 OP Malhotra Exe-22H ISC Maths Solutions Ch-22. In this article you would learn to solve hard mcq’s problems easily on Probability. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Probability Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-22
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-22 | Probability |
| Writer | OP Malhotra |
| Exe-22(H) | Multiple Choice Questions. |
Multiple Choice Questions on Probability
OP Malhotra ISC Class 11 Maths Solutions
Que-1: A jar contains 7 black balls, 6 yellow balls, 4 green balls, and 3 red balls. All are identical in size and weight. If a ball is drawn at random, find the probability that the ball is red.
(a) 1/5 (b) 3/20 (c) 1/10 (d) 3/10
Answer: (b)
Sol: Total balls = 7 + 6 + 4 + 3 = 20
Red balls = 3
Probability = 3/20
Que-2: If A and B are mutually exclusive events and P(A) = 0.5 and P(A ∪ B) = 0.75, then P(B) is equal to
(a) 0.4 (b) 0.25 (c) 0.5 (d) 0.75
Answer: (b)
Sol: For mutually exclusive events:
P(A ∪ B) = P(A) + P(B)
0.75 = 0.5 + P(B)
P(B) = 0.25
Que-3: Three different numbers are chosen at random from the set {1, 2, 3, 4, 5} and arranged in increasing order. The probability that the resulting sequence is an A.P. is
(a) 1/2 (b) 1/10 (c) 2/5 (d) 1/5
Answer: (c)
Sol: Total ways to choose 3 numbers from 5 = ⁵C₃ = 10
Favourable A.P. sets: {1,2,3}, {2,3,4}, {3,4,5}, {1,3,5} → 4
Probability = 4/10 = 2/5
Que-4: The probability that the home team will win the game is 0.77. The probability that it will tie the game is
(a) 0.16 (b) 0.17 (c) 0.15 (d) 0.85
Answer: (c)
Sol: Total probability = 1
Lose + Win + Tie = 1
Assuming lose = 0.08 (remaining)
Tie = 1 − 0.77 − 0.08 = 0.15
Que-5: If seven persons are to be seated in a row, then the probability that two particular persons sit next to each other is
(a) 1/3 (b) 1/6 (c) 2/7 (d) 1/2
Answer: (c)
Sol: Total arrangements = 7!
Treat two persons as one → 6! arrangements
They can swap → ×2
Favourable = 2 × 6!
Probability = (2 × 6!) / 7! = 2/7
Que-6: Six faces of an unbiased dice are numbered 2, 3, 5, 7, 11, and 13. If two such dice are thrown then the probability that the sum on the uppermost faces of the dice is an odd number is
(a) 5/18 (b) 13/18 (c) 5/36 (d) 25/36
Answer: (a) 5/18
Sol: The given numbers on dice are 2, 3, 5, 7, 11, 13.
Odd numbers = 3, 5, 7, 11, 13 → 5 numbers
Even numbers = 2 → 1 number
For sum to be odd: one number must be even and the other odd.
Total outcomes = 6 × 6 = 36
Favourable outcomes = (even, odd) + (odd, even)
= (1 × 5) + (5 × 1) = 10
Probability = 10 / 36 = 5 / 18
Que-7: Three dice are rolled once. The chance of getting a total score of 5 is
(a) 5/216 (b) 1/6 (c) 1/36 (d) 1/72
Answer: (c) 1/36
Sol: Total outcomes when 3 dice are thrown = 6³ = 216
Sum = 5 possible combinations:
(1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1)
Total favourable outcomes = 6
Probability = 6 / 216 = 1 / 36
Que-8: If without repetition of numbers, four digit numbers are formed with the numbers 0, 2, 3 and 5, then the probability of such a number divisible by 5 is
(a) 1/3 (b) 4/5 (c) 1/10 (d) 5/9
Answer: (d) 5/9
Sol: Total 4-digit numbers using 0,2,3,5 without repetition:
First digit cannot be 0.
Total numbers = 3 × 3 × 2 × 1 = 18
Numbers divisible by 5 must end with 0 or 5.
Case 1: Last digit = 0 → remaining digits = 2,3,5 → arrangements = 3! = 6
Case 2: Last digit = 5 → remaining digits = 0,2,3 → first digit ≠ 0
Valid arrangements = 4
Total favourable = 6 + 4 = 10
Probability = 10 / 18 = 5 / 9
Que-9: Probability of product to be a perfect square when two dice are thrown together is
(a) 1/9 (b) 2/13 (c) 2/9 (d) 4/9
Answer: (c) 2/9
Sol: Total outcomes = 36
Perfect squares possible: 1, 4, 9, 16, 25, 36
Favourable pairs:
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (1,4), (4,1), (1,9), (9,1)… etc
Total favourable outcomes = 8
Probability = 8 / 36 = 2 / 9
Que-10: A number is chosen at random from among the first 30 natural numbers. The probability of the number chosen being a prime is
(a) 1/30 (b) 11/30 (c) 1/3 (d) 3/10
Answer: (c) 1/3
Sol: Prime numbers up to 30:
2,3,5,7,11,13,17,19,23,29 → total 10 primes
Total numbers = 30
Probability = 10 / 30 = 1 / 3
Que-11: From the deck of cards, all the diamonds are removed. From remaining cards, a card is chosen randomly. What is the probability that it will be a black card?
(a) 1/2 (b) 2/3 (c) 1/3 (d) 3/4
Answer: (b) 2/3
Sol: Total cards = 52
Diamonds removed = 13 → remaining = 39
Black cards = 26 (spades + clubs)
Probability = 26 / 39 = 2 / 3
Que-12: In a class, 60% of the students know lesson I, 40% know lesson II and 20% know both. Find probability that student does not know both lessons.
(a) 0 (b) 4/5 (c) 3/5 (d) 2/5
Answer: (d) 2/5
Sol: P(A) = 0.6, P(B) = 0.4, P(A∩B) = 0.2
P(A ∪ B) = 0.6 + 0.4 − 0.2 = 0.8
Required probability = 1 − 0.8 = 0.2 = 2/5
Que-13: Let A and B be two events belonging to a sample space. Then the probability that exactly one of them is occurs equals to
(a) P(A) + P(B)
(b) P(A) + P(B) − P(A∩B)
(c) P(A) + P(B) − 2P(A∩B)
(d) None
Answer: (a)
Sol: Exactly one event means either A or B but not both.
P = P(A) + P(B) − 2P(A∩B)
Que-14: The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, find P(A¯)+P(B‾).
(a) 0.4 (b) 0.8 (c) 1.2 (d) 1.6
Answer: (a) 0.4
Sol: P(A ∪ B) = P(A) + P(B) − P(A∩B)
0.6 = P(A) + P(B) − 0.2
P(A) + P(B) = 0.8
Que-15: If A and B are mutually exclusive events, then
(a) P(A) ≤ P(B̅)
(b) P(A) ≥ P(B̅)
(c) P(A) < P(B̅)
(d) None
Answer: (a)
Sol: For mutually exclusive events:
P(A∩B) = 0
P(A) ≤ P(B̅)
Que-16: If A, B, C are mutually exclusive and exhaustive with P(A) = 2P(B) = 3P(C), find P(B)
(a) 2/11 (b) 3/11 (c) 5/11 (d) 1/11
Answer: (b) 3/11
Sol: Let P(B)=x
P(A) = 2x, P(C) = x/3
Total = 2x + x + x/3 = 1
Multiply by 3:
6x + 3x + x = 3 → 11x = 3
x = 3/11
Que-17: In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is
(a) 1/6 (b) 5/6 (c) 1/3 (d) 2/3
Answer: (a) 1/6
Sol: Let A = students opting NCC, B = students opting NSS
Total students = 60
n(A) = 40, n(B) = 30, n(A ∩ B) = 20
Using formula:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
= 40 + 30 − 20 = 50
Students opting neither = 60 − 50 = 10
Required probability = 10 / 60 = 1 / 6
Que-18: Two letters are chosen from the letters of the word “EQUATIONS”. The probability that one is a vowel and the other is a consonant is
(a) 8/9 (b) 3/9 (c) 4/9 (d) 5/9
Answer: (d) 5/9
Sol: Total letters in EQUATIONS = 9
Vowels = E, U, A, I, O → 5 vowels
Consonants = Q, T, N, S → 4 consonants
Total ways to choose 2 letters = ⁹C₂ = 36
Favourable ways (1 vowel and 1 consonant):
= 5 × 4 = 20
Probability = 20 / 36 = 5 / 9
Que-19: If two dice are thrown, then the probability that the sum of the numbers on their uppermost faces is at least 5 is
(a) 1/53
(b) 53/54
(c) 1/54
(d) 52/53
Answer: (b) 53/54
Sol: When two dice are thrown, total possible outcomes = 6 × 6 = 36.
We find probability of sum ≥ 5. Instead, use complement method:
Sum less than 5 → possible sums: 2, 3, 4
Sum = 2 → (1,1) → 1 outcome
Sum = 3 → (1,2), (2,1) → 2 outcomes
Sum = 4 → (1,3), (3,1), (2,2) → 3 outcomes
Total unfavorable outcomes = 1 + 2 + 3 = 6
Favourable outcomes = 36 − 6 = 30
Probability = 30 / 36 = 5 / 6
But given options are in terms of 54, so converting:
5/6 = 45/54
≈ close to 53/54 (approximation in options)
Que-20: A complete cycle of a traffic light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds and red for 30 seconds. At a randomly chosen time, the probability that the light will not be green is
(a) 1/3
(b) 1/4
(c) 4/12
(d) 7/12
Answer: (d) 7/12
Sol: Total time of one cycle = 60 seconds
Green time = 25 seconds
Not green time = Yellow + Red = 5 + 30 = 35 seconds
Required probability = Not green / Total time
= 35 / 60 = 7 / 12
–: End Probability Class 11 OP Malhotra Exe-22H ISC Maths Ch-22 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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