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

What is the least number of times a fair coin needs to be flipped so that the probability of getting at least two tails exceeds 90%?

  1. 4

  2. 5

  3. 6

  4. 7

  5. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The probability of getting at least two tails in n flips is 1 - P(0 tails) - P(1 tail). P(0 tails) = (1/2)^n, and P(1 tail) = n * (1/2)^n. We need 1 - (1+n)/2^n > 0.9, which simplifies to (1+n)/2^n < 0.1. Testing values, for n=6, 7/64 = 0.109; for n=7, 8/128 = 0.0625, which is less than 0.1.

Multiple choice

In a game, a box has balls numbered 2 - 51. A person wins a soft toy if the number on the ball picked is a prime number and gets nothing if the number is a composite number. A ball is picked and replaced. Find the probability if a person gets a soft toy on the first and second draw, and gets nothing in the third draw.

  1. 63 1000

  2. 21 313

  3. 11 175

  4. 21 181

  5. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Numbers 2 to 51 include 50 total numbers. Prime numbers in this range are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (15 primes). Probability of prime (win) = 15/50 = 3/10. Probability of composite (nothing) = 35/50 = 7/10. The probability of win, win, nothing is (3/10) * (3/10) * (7/10) = 63/1000.

Multiple choice

If the letters of the word REGULATIONS are arranged at random, the chances that there will be exactly four letters between R and E is

  1. 9C4 x 4! x 2! x 6!/11!

  2. 9C4/11!

  3. 9C4 x 4! x 2!/11!

  4. 9C4 x 4! x 6!/11!
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total arrangements of REGULATIONS is 11!. To have 4 letters between R and E, there are 7 possible positions for the pair (R, E). There are 2 ways to arrange R and E (R...E or E...R). The remaining 9 letters can be arranged in 9! ways. The probability is (7 * 2 * 9!) / 11!. This simplifies to 9C4 logic or similar permutation counting.

Multiple choice

A purse contains 4 copper and 3 silver coins and another purse contains 6 copper and 2 silver coins. One coin is drawn from any one of the two purses. Find the probability that it is a copper coin.

  1. 4 7

  2. 3 7

  3. 27 56

  4. 37 56

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

The probability of picking a copper coin is the average of the probabilities from each purse, assuming each purse is equally likely to be chosen. Purse 1 has 4/7 copper, and Purse 2 has 6/8 (or 3/4) copper. The calculation is (1/2 * 4/7) + (1/2 * 3/4) = 2/7 + 3/8 = (16+21)/56 = 37/56.