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

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting black face card
  1. $\displaystyle \frac{6}{49}$
  2. $\displaystyle \frac{3}{49}$
  3. $\displaystyle \frac{5}{49}$
  4. $\displaystyle \frac{4}{49}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Original pack has 52 cards. 3 spade face cards removed, leaving 49 cards. Black face cards are 6 (3 spades, 3 clubs). Since 3 spades are removed, 3 clubs remain. Probability = 3 / 49.

Multiple choice

A die is thrown once. What is the sample space?
  1. ${1, 2, 3}$
  2. ${6}$
  3. ${1, 2, 3, 4, 5, 6}$
  4. ${1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A standard die has six faces numbered 1 through 6. The sample space is the set of all possible outcomes of a single throw, which is {1, 2, 3, 4, 5, 6}.

Multiple choice

Two die are thrown. Find the probability that t he number on the upper face of the first dice is less than the number on the upper face of the second dice.
  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{7}{12}$
  3. $\displaystyle \frac{1}{3}$
  4. $\displaystyle \frac{5}{12}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

There are 36 equally likely ordered outcomes when two dice are thrown. Six outcomes have equal numbers, and the remaining 30 split equally between the first die being larger and the first die being smaller. Thus, the required probability is 15/36 = 5/12.

Multiple choice

Two die are thrown. Find the probability of the event that the sum of the numbers on their upper faces is multiple of $7$:
  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{1}{3}$
  3. $\displaystyle \frac{1}{9}$
  4. $\displaystyle \frac{2}{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The possible sums that are multiples of 7 are 7 and 14. Pairs summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totaling 6 outcomes. There are 36 total outcomes, so 6/36 = 1/6.