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 box contains 'a' white and 'b' black balls. 'c' balls are drawn at random with replacement. The expected number of black balls drawn is
  1. $\displaystyle \frac{a}{a+b}$
  2. $\displaystyle \frac{ac}{a+b}$
  3. $\displaystyle \frac{bc}{a+b}$
  4. $\displaystyle \frac{c}{a+b}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The probability of drawing a black ball in one trial is b / (a + b). For c trials with replacement, the expected number is c times the probability of success in a single trial, which is c * (b / (a + b)).

Multiple choice

5 cards are drawn one after an other successively with replacement from a well shuffled pack of 52 cards. The probability that all the 5 cards are spades
  1. $\left(\displaystyle \frac{3}{4}\right)^{5}$
  2. 1-$\left(\displaystyle \frac{3}{4}\right)^{5}$
  3. $\left(\displaystyle \frac{1}{4}\right)^{5}$
  4. 1-$\left(\displaystyle \frac{1}{4}\right)^{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of drawing a spade in one draw is 13/52 = 1/4. Since draws are with replacement, the probability of drawing 5 spades in a row is (1/4) * (1/4) * (1/4) * (1/4) * (1/4) = (1/4)^5.

Multiple choice

A card is drawn and replaced three times from an ordinary pack of 52 playing cards. The probability that atleast one spade is drawn is
  1. $\displaystyle \frac{9}{64}$
  2. $\displaystyle \frac{1}{64}$
  3. $\displaystyle \frac{37}{64}$
  4. $\displaystyle \frac{3}{64}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of drawing a spade is 13/52 = 1/4. Probability of not drawing a spade is 3/4. Probability of no spades in 3 draws = (3/4)^3 = 27/64. Probability of at least one spade = 1 - 27/64 = 37/64.

Multiple choice

Twenty coins each with probability $P(0
  1. $\dfrac{9}{20}$
  2. $\dfrac{11}{20}$
  3. $\dfrac{11}{21}$
  4. $\dfrac{9}{21}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the binomial probability formula P(X=k) = C(n, k) * p^k * (1-p)^(n-k), we set C(20, 10) * p^10 * (1-p)^10 = C(20, 11) * p^11 * (1-p)^9. Simplifying gives C(20, 10) * (1-p) = C(20, 11) * p. Since C(20, 11) = C(20, 9), we get (1-p)/p = C(20, 11)/C(20, 10) = 10/11. Solving for p yields p = 11/21.

Multiple choice

$201$ coins each with probability $P(0
  1. $\dfrac{1}{4}$
  2. $\dfrac{1}{3}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For binomial distribution B(n, p), P(X=k) = P(X=k+1) implies nCk * p^k * (1-p)^(n-k) = nC(k+1) * p^(k+1) * (1-p)^(n-k-1). With n=201 and k=100, this simplifies to p = 1-p, so p = 0.5.

Multiple choice

A coin whose faces are marked 3 & 5 is tossed 4 times. The probability that the sum of the numbers thrown is greater than 15
  1. $\displaystyle\frac{11}{16}$
  2. $\displaystyle\frac{5}{16}$
  3. $\displaystyle\frac{5}{8}$
  4. $\displaystyle\frac{1}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total number of outcomes is 2^4 = 16. The possible sums are: 3+3+3+3=12, 3+3+3+5=14 (4 ways), 3+3+5+5=16 (6 ways), 3+5+5+5=18 (4 ways), 5+5+5+5=20 (1 way). Sums greater than 15 are 16, 18, and 20, which occur in 6+4+1 = 11 cases.

Multiple choice

A coin whose faces are marked 3 & 5 is tossed 4 times. The probability that the sum of the numbers thrown is 12 is
  1. $\displaystyle\frac{1}{16}$
  2. $\displaystyle\frac{5}{16}$
  3. $\displaystyle\frac{5}{8}$
  4. $\displaystyle\frac{1}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Possible sums with 4 tosses (each 3 or 5): 3+3+3+3=12. This is the only combination that sums to 12. The number of ways to get 3,3,3,3 is 1. Total outcomes = 2^4 = 16. Probability = 1/16.

Multiple choice

A symmetric die is thrown $(2n+1)$ times. The probability of getting a prime score on the upturned face at most $n$ times is
  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{1}{4}$
  4. $\displaystyle\frac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Prime scores on a die are 2, 3, 5 (3 outcomes). Probability of prime = 3/6 = 1/2. In (2n+1) trials, the probability of getting at most n successes in a symmetric binomial distribution is 1/2 due to symmetry.

Multiple choice

A bag contains $n+1$ coins. It is known that one of these coins has a head on both sides while the other coins are fair. One coins is selected at random and tossed. If head turns up, the probability that the selected coin was fair, is $\displaystyle \frac { n }{ n+2 } $

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Assertion is incorrect but Reason is correct
Reveal answer Fill a bubble to check yourself
A Correct answer