Quantitative Aptitude

Probability

1,860 Questions
All Sample Space Setup 558 Permutation & Combination Counting 288 Multiplication Theorem (AND) 234 Addition Theorem (OR) 151 At Least One Probability 134 Complement Rule (P(Not A)) 130 Drawing Without Replacement 126 Conditional Probability 75 Bayes' Theorem 69 Drawing With Replacement 49 Success and Failure Rates 46

Probability Questions

Multiple choice

There is great excitement at the local cricket oval. There is one ball remaining to be bowled and the home team needs four runs to win the match. The new batsman is old Basher Jones. Over his long career, Basher is known to hit a four once in every 25 balls that he faces. The team scorer estimates that he hits a six once in every 100 balls. Unfortunately, Basher is not as young as he was and the chances of him running four is nil. Also, Basher has a 50-50 chance to go out on any ball he faces. Discounting the probability of no-balls, wides and so on, what is the probability that the home team will win?

  1. 0.2

  2. 0.09

  3. 0.05

  4. 0.60

  5. 0.65
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

Direction: The following question has two statements, (1) and (2). Answer the question using the following options: When one piece of fruit is taken at random from a fruit bowl, what is the chance that it is an apple? (1) There are half as many apples as oranges in the fruit bowl. (2) One third of the fruits in the fruit bowl are oranges.

  1. if statement (1) alone is sufficient but statement (2) alone is not

  2. if statement (2) alone is sufficient but statement (1) alone is not

  3. if both (1) and (2) together are sufficient but none of them alone is sufficient

  4. if both independently are sufficient

  5. if both statements (1) and (2) together are not sufficient
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let A be apples, O be oranges, X be others. (1) A = 0.5O. (2) O = (1/3)Total. Neither alone gives the probability A/Total. Together, A = 0.5 * (1/3)Total = (1/6)Total. Probability = 1/6.

Multiple choice

A box contains 4 black and 5 red balls and another box contains 5 black and 4 red balls. One ball is to drawn from either of two boxes. What is the probability of drawing a black ball?

  1. 1/2

  2. 3/2

  3. 2/3

  4. 5/2

  5. 2/5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First box: 4 black, 5 red balls (total 9). Probability of black from box 1 = 4/9. Second box: 5 black, 4 red balls (total 9). Probability of black from box 2 = 5/9. Since one box is chosen at random (probability 1/2 for each box), total probability of black ball = (1/2)(4/9) + (1/2)(5/9) = 2/9 + 2.5/9 = 4.5/9 = 1/2.

Multiple choice

A basket contains 6 blue, 2 Red , 4 green and 3 yellow balls. If 4 balls are picked at random, what is the probability that 2 are red and 2 are green ?

  1. 4/15

  2. 5/57

  3. 1/3

  4. 2/455

  5. 6/455
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total balls = 6+2+4+3 = 15. We need to pick exactly 2 red and 2 green balls. Number of ways to choose 2 red from 2 = C(2,2) = 1. Number of ways to choose 2 green from 4 = C(4,2) = 6. Total favorable outcomes = 1 × 6 = 6. Total ways to choose any 4 balls from 15 = C(15,4) = 1365. Probability = 6/1365 = 2/455 after simplification (divide numerator and denominator by 3).

Multiple choice

Answer the following questions based on the statement given below: Two dice are rolled simultaneously. Find the probability of getting a doublet.
  1. $\frac{1}{2}$
  2. $\frac{1}{3}$
  3. $\frac{1}{4}$
  4. $\frac{1}{5}$
  5. $\frac{1}{6}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

When two dice are rolled, there are 6 × 6 = 36 total outcomes. A doublet means both numbers are the same: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) — that's 6 favorable outcomes. Probability = 6/36 = 1/6.

Multiple choice

Answer the following questions based on the statement given below: Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the two faces divisible by 3 or 4.

  1. 4/9

  2. 2/5

  3. 9/5

  4. 5/9

  5. 5/7
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes = 36. Sums divisible by 3: 3 (2 ways), 6 (5 ways), 9 (4 ways), 12 (1 way) = 12. Sums divisible by 4: 4 (3 ways), 8 (5 ways), 12 (1 way) = 9. But 12 counted twice. Total favorable = 12 + 9 - 1 = 20. Probability = 20/36 = 5/9. This tests counting outcomes and inclusion-exclusion principle.

Multiple choice

A basket contains 3 red balls, 5 blue balls and 2 yellow balls. If three balls are drawn at random then what is the probability that none is blue?

  1. 1/6

  2. 1/3

  3. 1/12

  4. 3/16

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We need 3 non-blue balls from 5 non-blue balls (3 red + 2 yellow) out of 10 total. Number of ways to choose 3 from 5 is C(5,3) = 10. Total ways to choose any 3 from 10 is C(10,3) = 120. Probability = 10/120 = 1/12. Option C is correct.

Multiple choice

Study the following information carefully to answer the question that follow. A box contains 2 blue balls, 5 green, 4 red and 1 yellow balls. If two balls are picked at random, what is the probability that both are blue?

  1. 7/66

  2. 3/132

  3. 1/66

  4. 5/67

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total balls = 2 + 5 + 4 + 1 = 12. Number of blue balls = 2. Probability of first ball being blue = 2/12. After removing one blue ball, probability of second ball being blue = 1/11. Combined probability = (2/12) × (1/11) = 2/132 = 1/66.

Multiple choice

From a pack of well suffled pack two cards are drawn one after another without repacement. Find the probability that both card are honour card of same colour?

  1. 28/663

  2. 10/221

  3. 8/221

  4. 29/663

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Honour cards are Ace, King, Queen, Jack of each suit - 16 total (8 red, 8 black). Probability both are red honours: 8/52 × 7/51 = 56/2652. Probability both are black honours: 8/52 × 7/51 = 56/2652. Total probability = 56/2652 + 56/2652 = 112/2652 = 28/663. Option B (10/221 = 30/663) and C (8/221 = 24/663) are close but incorrect.

Multiple choice

Messi strikes a goal three times in every eight chances given to him . If he strikes the ball three time then find the probability of hitting goal post only once? (Approx)

  1. 0.31

  2. 0.44

  3. 0.56

  4. 0.25

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is binomial probability with n=3 trials and success probability p=3/8. We need exactly 1 success: P(X=1) = ³C₁ × (3/8)¹ × (5/8)² = 3 × 0.375 × 0.390625 ≈ 0.439. Option A (0.31) underestimates, C (0.56) would require p ≈ 0.5.

Multiple choice

Two card are drawn from a pack of well shuffled cards one after another without replacement. Find the probability that only one card is of red colour.

  1. 13/51

  2. 26/51

  3. 51/221

  4. 84/221

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Probability of exactly one red = P(Red then Black) + P(Black then Red). P(Red,Black) = (26/52) × (26/51) = 26/51 × 1/2 = 13/51. P(Black,Red) = (26/52) × (26/51) = 13/51. Total = 13/51 + 13/51 = 26/51. This matches option B exactly. The answer is correct.