IPMPrepA bag has 6 red, 2 blue and 5 green balls. Two are drawn at random. What is the probability of one red and one blue?
Solution
Favourable: 6 × 2 = 12 out of C(13, 2) = 78.
Probability = 213.
Answer: (C) 213
Quantitative Ability · Counting and Data
Probability: IPMAT practice questions
24 questions. Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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A bag has 4 red, 4 blue and 6 green balls. Three are drawn at random. What is the probability that at least one is green?
Solution
P(no green) = C(8, 3)/C(14, 3) = 56364.
P(at least one green) = 1 − 213 = 1113.
Answer: (D) 1113
A hits a target with probability 23 and B with probability 12, independently. Each fires once and the target is hit. What is the probability that A hit it?
Solution
P(target hit) = 1 − 13 × 12 = 56.
P(A hit | target hit) = 23 ÷ 56 = 45.
Answer: (A) 45
A hits a target with probability 34 and B with probability 45, independently. Each shoots once. What is the probability that the target is hit?
Solution
P(both miss) = 14 × 15 = 120.
P(hit) = 1 − 120 = 1920.
Answer: (B) 1920
4 fair coins are tossed. What is the probability of exactly 3 heads?
Solution
C(4, 3) = 4 favourable outcomes out of 24 = 16.
Probability = 14.
Answer: (D) 14
4 fair coins are tossed. What is the probability of at least one head?
Solution
P(no head) = (12)4 = 116.
P(at least one head) = 1 − 116 = 1516.
Answer: (D) 1516
Bag I has 5 white and 5 black balls; bag II has 3 white and 5 black. A bag is picked at random and a ball drawn from it is white. What is the probability it came from bag I?
Solution
P(white) = (12)(510) + (12)(38) = 716.
P(bag I | white) = (12)(510) ÷ 716 = 47.
Answer: (B) 47
3 fair coins are tossed. What is the probability of exactly 1 heads?
Solution
C(3, 1) = 3 favourable outcomes out of 23 = 8.
Probability = 38.
Answer: (C) 38
Two fair dice are thrown. What is the probability that the sum is a multiple of 4?
Solution
There are 36 equally likely outcomes.
Favourable outcomes: 9.
Probability = 936 = 14.
Answer: (D) 14
A solves a problem with probability 34 and B with probability 13, independently. What is the probability that exactly one of them solves it?
Solution
A only: 34 × 23; B only: 13 × 14.
Sum = 712.
Answer: (D) 712
A hits a target with probability 34 and B with probability 15, independently. Each shoots once. What is the probability that the target is hit?
Solution
P(both miss) = 14 × 45 = 15.
P(hit) = 1 − 15 = 45.
Answer: (C) 45
Bag I has 6 white and 6 black balls; bag II has 3 white and 6 black. A bag is picked at random and a ball drawn from it is white. What is the probability it came from bag I?
Solution
P(white) = (12)(612) + (12)(39) = 512.
P(bag I | white) = (12)(612) ÷ 512 = 35.
Answer: (C) 35
A bag has 3 red, 6 blue and 1 green balls. Three are drawn at random. What is the probability that at least one is green?
Solution
P(no green) = C(9, 3)/C(10, 3) = 84120.
P(at least one green) = 1 − 710 = 310.
Answer: (D) 310
Two fair dice are thrown. What is the probability that the sum is 9?
Solution
There are 36 equally likely outcomes.
Favourable outcomes: 4.
Probability = 436 = 19.
Answer: (C) 19
3 fair coins are tossed. What is the probability of at least 2 heads?
Solution
Add C(3, j) for j = 2 to 3: 4 outcomes out of 8.
Probability = 12.
Answer: (D) 12
A number is chosen at random from 1 to 120. What is the probability that it is divisible by 4 or 5?
Solution
Divisible by 4: 30; by 5: 24; by both (by 20): 6.
Favourable = 30 + 24 − 6 = 48.
Probability = 48120 = 25.
Answer: (B) 25
A card is drawn from a standard pack of 52. What is the probability that it is red or a king?
Solution
26 red cards + 4 kings − 2 red kings = 28.
Probability = 2852 = 713.
Answer: (A) 713
A number is chosen at random from 1 to 150. What is the probability that it is divisible by 2 or 5?
Solution
Divisible by 2: 75; by 5: 30; by both (by 10): 15.
Favourable = 75 + 30 − 15 = 90.
Probability = 90150 = 35.
Answer: (A) 35
Two fair dice are thrown. What is the probability that the numbers differ by 2?
Solution
There are 36 equally likely outcomes.
Favourable outcomes: 8.
Probability = 836 = 29.
Answer: (B) 29
Bag I has 4 white and 4 black balls; bag II has 2 white and 5 black. A bag is picked at random and a ball drawn from it is white. What is the probability it came from bag I?
Solution
P(white) = (12)(48) + (12)(27) = 1128.
P(bag I | white) = (12)(48) ÷ 1128 = 711.
Answer: (D) 711
Bag I has 6 white and 3 black balls; bag II has 2 white and 4 black. A bag is picked at random and a ball drawn from it is white. What is the probability it came from bag I?
Solution
P(white) = (12)(69) + (12)(26) = 12.
P(bag I | white) = (12)(69) ÷ 12 = 23.
Answer: (C) 23
A number is chosen at random from 1 to 60. What is the probability that it is divisible by 2 or 5?
Solution
Divisible by 2: 30; by 5: 12; by both (by 10): 6.
Favourable = 30 + 12 − 6 = 36.
Probability = 3660 = 35.
Answer: (C) 35
A bag has 6 red, 5 blue and 4 green balls. Two are drawn at random. What is the probability of one red and one blue?
Solution
Favourable: 6 × 5 = 30 out of C(15, 2) = 105.
Probability = 27.
Answer: (B) 27
Two fair dice are thrown. What is the probability that the product is even?
Solution
There are 36 equally likely outcomes.
Favourable outcomes: 27.
Probability = 2736 = 34.
Answer: (A) 34
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.