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

Cards marked with the numbers $2$ to $101$ are put in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number on the card is a n even prime number less than $16$.
  1. $\displaystyle\frac{5}{28}$
  2. $\displaystyle\frac{3}{50}$
  3. $\displaystyle\frac{3}{25}$
  4. $\displaystyle\frac{1}{100}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The only even prime number is 2, and it is less than 16. There are 100 cards from 2 through 101, so the probability is 1/100.

Multiple choice

A bag contains 50 tickets numbered 1, 2, 3, ..., 50 of which five are drawn at random and arranged in ascending order of magnitude $(x_1 < x_2 < x_3 < x_4 < x_5)$. The probability that $x_3=30$ is
  1. $\displaystyle \frac{^{20}C_2}{^{50}C_5}$
  2. $\displaystyle \frac{^{29}C_2}{^{50}C_5}$
  3. $\displaystyle \frac{^{20}C_2\times ^{29}C_2}{^{50}C_5}$

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

We choose 5 numbers out of 50. For x3 = 30, we must choose 2 numbers from {1, ..., 29} and 2 numbers from {31, ..., 50}. The number of ways is (29C2) * (20C2). Total ways is 50C5.

Multiple choice

Four die are thrown simultaneously. The probability that $4$ and $3$ appear on two of the die given that $5$ and $6$ have appeared on other two die is?
  1. $1/6$
  2. $1/36$
  3. $12/151$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The condition is given that two specific dice show 5 and 6. We need the probability that the remaining two dice show 4 and 3. There are 36 total outcomes for the remaining two dice. The favorable outcomes are (4,3) and (3,4), which is 2/36 = 1/18. Since 1/18 is not listed, 'None of these' is correct.

Multiple choice

A coin is tossed three times in succession. If $E$ is the event that there are at least two heads and $F$ is the event in which first throw is a head, then $P(E/F)$ is equal to:
  1. $3/4$
  2. $3/8$
  3. $1/2$
  4. $1/8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice

A fair coin is tossed $12$ times. Find the probability that two heads do not occur consecutively.
  1. Required probability is $\cfrac{375}{4096}$
  2. Required probability is $\cfrac{376}{4096}$
  3. Required probability is $\cfrac{377}{4096}$
  4. Required probability is $\cfrac{383}{4096}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The number of ways to toss a coin 12 times such that no two heads are consecutive is given by the (n+2)-th Fibonacci number, where n=12. F(14) = 377. Total outcomes = 2^12 = 4096. Probability = 377/4096.

Multiple choice

Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is K $(3\leq K \leq8)$ is
  1. $\displaystyle \frac { (k-1) (k-2) }{432 } $
  2. $\displaystyle \frac{k(k-1)}{432}$
  3. $\displaystyle \frac{k^2}{432}$

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

The number of ways to get a sum K with three dice is the coefficient of x^K in (x + x^2 + x^3 + x^4 + x^5 + x^6)^3. For 3 <= K <= 8, this is given by the formula (K-1)(K-2)/2. Dividing by the total outcomes 6^3 = 216 gives (K-1)(K-2)/432.