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

A six-faced fair dice is rolled five times. The probability (in %) of obtaining 'ONE' at least four times is

  1. 33.3

  2. 3.33

  3. 0.33

  4. 0.0033
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of getting 'ONE' in one roll is 1/6. For 5 rolls, getting 'ONE' at least 4 times follows binomial distribution: P(X=4) + P(X=5). P(X=4) = 5C4 * (1/6)^4 * (5/6)^1 = 5 * (1/1296) * (5/6) = 25/7776. P(X=5) = 5C5 * (1/6)^5 = 1/7776. Total = 26/7776 approx 0.00334. As a percentage, this is 0.334%.

Multiple choice

An urn contains 6 different black balls and 4 different blue balls. In how many ways can 6 balls be selected so that there are at least two balls of each colour?

  1. 185 ways

  2. 120 ways

  3. 125 ways

  4. 124 ways

  5. 123 ways
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total ways to select 6 balls from 10 is 10C6 = 210. We need at least two of each color. Possible combinations: (2B, 4b) or (3B, 3b) or (4B, 2b). 6C2*4C4 + 6C3*4C3 + 6C4*4C2 = (15*1) + (20*4) + (15*6) = 15 + 80 + 90 = 185.

Multiple choice

A bag contains 10 black marbles, some red marbles and some green marbles. If the number of red marbles is 5 times the number of green marbles and twice the number of black marbles, then what is the probability of getting at least one red marble when two marbles are drawn from the bag?

  1. 480/531

  2. 470/561

  3. 460/501

  4. None of these

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Black = 10. Red = 2 * Black = 20. Red = 5 * Green, so Green = 4. Total = 34. Probability of no red = (14/34) * (13/33) = 182 / 1122 = 91 / 561. Probability of at least one red = 1 - (91/561) = 470/561.

Multiple choice

A bag contains 10 black marbles, some red marbles and some green marbles. If the number of red marbles is 3 more than twice the number of green marbles and the total number of marbles in the bag is 31 (there are only three colour marbles in bag), then what is the probability of picking one red and one green marble when 2 marbles are drawn from the bag?

  1. 6/31

  2. 5/31

  3. 4/31

  4. None of these

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total = 31. Black = 10. Red + Green = 21. R = 2G + 3. (2G + 3) + G = 21 => 3G = 18 => G = 6, R = 15. Probability of one red and one green = (15C1 * 6C1) / 31C2 = (15 * 6) / (31 * 30 / 2) = 90 / 465 = 6/31.

Multiple choice

Rosy and Mary play a game. They roll a dice each simultaneously. If both faces bear the same number, it is called a doublet. In three rolls of the pair of dice, what is the probability of getting a doublet at least once?

  1. 21/216

  2. 91/136

  3. 22/197

  4. 91/216

  5. 1/2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Probability of a doublet in one roll = 6/36 = 1/6. Probability of no doublet in one roll = 5/6. Probability of no doublet in three rolls = (5/6)^3 = 125/216. Probability of at least one doublet = 1 - 125/216 = 91/216.

Multiple choice

There are three piles of identical yellow, black and green balls and each pile contains at least 20 balls. The number of ways of selecting 20 balls, if the number of black balls to be selected is twice the number of yellow balls, is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y, b, g be the number of yellow, black, and green balls. We have y + b + g = 20 and b = 2y. Substituting gives y + 2y + g = 20, so 3y + g = 20. Since g must be at least 0, 3y <= 20, so y can be 0, 1, 2, 3, 4, 5, or 6. Each value of y determines a unique g, resulting in 7 possible combinations.