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

Let a red die, a blue die, a green die and a white die are rolled once, the dice being fair. The outcomes on the red, blue, green and white denote the numbers, $a,b,c$ and $d$ respectively. Let $E$ denotes the event that absolute value of $(a-1)(b-2)(c-3)(d-6)=1$, then $P(E)$ is:
  1. $\dfrac{1}{324}$
  2. $\dfrac{1}{648}$
  3. $\dfrac{2}{324}$
  4. $\dfrac{1}{162}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The condition |(a-1)(b-2)(c-3)(d-6)| = 1 means each factor must be 1 or -1. For each die, count the values that satisfy this. For (a-1), a=2 or 0 (impossible). So a=2. For (b-2), b=3 or 1. For (c-3), c=4 or 2. For (d-6), d=5 (d=7 impossible). Total favorable outcomes = 1 * 2 * 2 * 1 = 4. Total outcomes = 6^4 = 1296. P(E) = 4 / 1296 = 1 / 324.

Multiple choice

Three dice are rolled simultaneously. The probability of getting the total sum of more than $5$ is
  1. $\dfrac{5}{54}$
  2. $\dfrac{67}{72}$
  3. $\dfrac{103}{108}$
  4. $\dfrac{53}{54}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes for three dice = 6^3 = 216. The sum is 5 or less for combinations: (1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1). There are 10 such outcomes. Probability of sum <= 5 is 10/216 = 5/108. Probability of sum > 5 is 1 - 5/108 = 103/108.

Multiple choice

The six faces of a die are marked A,B,C,D,E and F. The event M is getting a vowel on the upper face of the die when it is tossed. The probability of the event M is

  1. 1

  2. 1/2

  3. 1/3

  4. 2/3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The six faces of the die are marked with the letters A, B, C, D, E, and F. Among these letters, only A and E are vowels, making a total of 2 favorable outcomes. Since there are 6 possible outcomes in total, the probability of getting a vowel is 2/6, which simplifies to 1/3.

Multiple choice

A certain coin is tossed with probability of showing head being 'p'. Let Q denotes probability that when the coin is tossed two times, the number of heads obtained is odd. Then
  1. there is no value of p, if $Q = \frac { 1 } { 4 }$
  2. there is exactly one value of p, if $Q = \frac { 3 } { 4 }$
  3. there is exactly two value of p, if $Q = \frac { 3 } { 5 }$
  4. there is no value of p, if $Q = \frac { 4 } { 5 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Probability of head = p, tail = 1-p. Two tosses: HH (p^2), HT (p(1-p)), TH ((1-p)p), TT ((1-p)^2). Odd number of heads means HT or TH. Q = 2p(1-p). If Q = 4/5, 2p - 2p^2 = 0.8, 2p^2 - 2p + 0.8 = 0, p^2 - p + 0.4 = 0. Discriminant = 1 - 4(0.4) = 1 - 1.6 = -0.6. No real value for p.