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

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets a 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability of getting a 1, 2, 3 or 4 on the die?

  1. 8/11

  2. 1/11

  3. 3/11

  4. 2/3

  5. 4/11
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let E be the event of getting exactly one head. P(E) = P(E|5,6)P(5,6) + P(E|1,2,3,4)P(1,2,3,4). P(E|5,6) = 3C1 * (0.5)^3 = 3/8. P(E|1,2,3,4) = 1C1 * 0.5 = 1/2. P(E) = (3/8 * 2/6) + (1/2 * 4/6) = 1/8 + 1/3 = 11/24. P(1,2,3,4|E) = (1/2 * 4/6) / (11/24) = (1/3) / (11/24) = 8/11.

Multiple choice

Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are red or both are jacks?

  1. 330/1225

  2. 1/1326

  3. 332/1326

  4. 55/221

  5. 331/1326
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the inclusion-exclusion principle, P(A or B) = P(A) + P(B) - P(A and B). P(both red) = (26/52) * (25/51) = 325/1326. P(both jacks) = (4/52) * (3/51) = 12/1326. P(both red and both jacks) = P(both red jacks) = (2/52) * (1/51) = 2/1326. Total probability = (325 + 12 - 2) / 1326 = 335/1326. This simplifies to 55/221.

Multiple choice

A and B throw a dice in succession to win a bet with A starting first. Whoever throws '1' first wins an amount of Rs. 110. What are the respective expectations of A and B?

  1. Rs. 70 and Rs. 40

  2. Rs. 60 and Rs. 50

  3. Rs. 75 and Rs. 35

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

Prob(A wins) = 1/6 + (5/6)(5/6)(1/6) + ... = (1/6) / (1 - 25/36) = (1/6) / (11/36) = 6/11. Prob(B wins) = 1 - 6/11 = 5/11. Expectation A = 6/11 * 110 = 60. Expectation B = 5/11 * 110 = 50.

Multiple choice

There are 7 gold rings, 5 silver rings and 10 diamond rings in a bag. In how many ways can a person pick 3 rings from the bag such that at least one of them is a diamond ring?

  1. 1280

  2. 1320

  3. 1410

  4. 1480

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total rings = 7+5+10 = 22. Total ways to pick 3 rings = 22C3 = (22*21*20)/(3*2*1) = 1540. Ways to pick 3 rings with no diamonds (only gold or silver) = 12C3 = (12*11*10)/(3*2*1) = 220. Ways with at least one diamond = 1540 - 220 = 1320.

Multiple choice

An unbiased dice is rolled four times. The probability that the minimum number on any roll is not less than 2 and maximum number on any roll is not greater than 5 is

  1. 16/81

  2. 1/81

  3. 65/81

  4. 80/81
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Each roll has 6 outcomes. The condition 'not less than 2' and 'not greater than 5' means the outcome must be 2, 3, 4, or 5. There are 4 favorable outcomes out of 6 per roll. Probability for one roll = 4/6 = 2/3. For four rolls, probability = (2/3)^4 = 16/81.