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

The probability of obtaining $4$ heads, $4$ tails when $8$ unbiased coins are tossed is :
  1. $\dfrac{1}{2}$
  2. $\dfrac{C_{4}^{8}}{2^{8}}$
  3. $\dfrac{C_{0}^{8}}{2^{8}}$
  4. $\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of getting exactly k successes in n Bernoulli trials is given by the binomial distribution formula: C(n, k) * p^k * (1-p)^(n-k). For 8 unbiased coins, n=8, k=4, and p=1/2, so the probability is C(8, 4) * (1/2)^4 * (1/2)^4, which simplifies to C(8, 4) / 2^8.

Multiple choice

Two uniform dice marked $1$ to $6$ are thrown together. The probability that the score on the two dice is at least seven is
  1. $\dfrac{5}{12}$
  2. $\dfrac{7}{12}$
  3. $\dfrac{3}{4}$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total outcomes = 36. Outcomes with sum < 7: (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1) = 15 outcomes. Outcomes with sum >= 7 = 36 - 15 = 21. Probability = 21/36 = 7/12.

Multiple choice

The probability that in the toss of two dice we obtain the sum 7 or 11 is
  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{1}{18}$
  3. $\displaystyle \frac{2}{9}$
  4. $\displaystyle \frac{23}{108}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 36. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) -> 6 outcomes. Sum 11: (5,6), (6,5) -> 2 outcomes. Total favorable = 8. Probability = 8/36 = 2/9.

Multiple choice

A bag contains 4 white and 5 black balls. A second bag contains 3 (identical) white and 6 (identical) black balls. One bag is chosen at random and a ball is drawn. Its colour is noted and the ball replaced. This is repeated four times. It was found that of these four, one was white and 3 were black. The $n$ the probability that the first bag was chosen is
  1. $\dfrac{162}{287}$
  2. $\dfrac{125}{287}$
  3. $\dfrac{250}{280}$
  4. $\dfrac{125}{574}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Two symmetrical dice are rolled. The probability that sum of the points on them is divisible by $5$ is
  1. $\dfrac{2}{9}$
  2. $\dfrac{4}{9}$
  3. $\dfrac{7}{36}$
  4. $\dfrac{5}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total outcomes when rolling two dice is 36. Sums divisible by 5 are 5 and 10. Pairs summing to 5 are (1,4), (2,3), (3,2), (4,1) [4 outcomes]. Pairs summing to 10 are (4,6), (5,5), (6,4) [3 outcomes]. Total favorable outcomes = 4 + 3 = 7. Probability = 7/36.