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 bag contains apples and oranges, five in all and atleast one of each, all combinations being equally likely. If one fruit is selected at random from the bag, assuming all fruits are distinguishable,the probability that it is an orange is
  1. $\displaystyle \frac{1}{20}$
  2. $\displaystyle \frac{1}{10}$
  3. $\displaystyle \frac{1}{5}$
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total fruits = 5. Possible combinations (A,O): (1,4), (2,3), (3,2), (4,1). Each combination is equally likely (1/4 chance). For a chosen combination, the probability of picking an orange is O/5. Total probability = sum of (P(comb) * P(orange|comb)) = (1/4)(4/5 + 3/5 + 2/5 + 1/5) = (1/4)(10/5) = 1/2.

Multiple choice

A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4$^{th}$ time on the 7$^{th}$ draw is
  1. $\displaystyle \frac{5}{64}$
  2. $\displaystyle \frac{27}{32}$
  3. $\displaystyle \frac{5}{32}$
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the 4th white ball to appear on the 7th draw, there must be exactly 3 white balls in the first 6 draws, and the 7th draw must be white. The probability of white is 1/2. The number of ways to choose 3 positions for white in the first 6 draws is 6C3 = 20. The probability is 20 * (1/2)^6 * (1/2) = 20 / 128 = 5/32.

Multiple choice

An urn contains $6$ white and $4$ black balls. A fair die whose faces are numbered from $1$ to $6$ is rolled and number of balls equal to that of the number appearing on the die is drawn from the urn at random. The probability that all those are white is
  1. $\displaystyle \frac{1}{5}$
  2. $\displaystyle \frac{2}{5}$
  3. $\displaystyle \frac{3}{5}$
  4. $\displaystyle \frac{4}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total outcomes for die roll is 6. For each roll n, probability of drawing n white balls is (6/10)^n. Total probability = sum( (1/6) * (6/10)^n ) for n=1 to 6. This simplifies to 1/5.

Multiple choice

A coin is tossed $3$ times. If $E$ denotes the event in which heads appear at least twice and $F$ denotes the event in which head comes first, then $P(\displaystyle {E}|{F})=$
  1. $\displaystyle \frac{3}{4}$
  2. $\displaystyle \frac{3}{8}$
  3. $\displaystyle \frac{1}{2}$
  4. $\displaystyle \frac{1}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sample space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. F (head first): HHH, HHT, HTH, HTT. E (at least 2 heads): HHH, HHT, HTH, THH. E intersection F: HHH, HHT, HTH. P(E|F) = 3/4.