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 deck of 8 cards contains 3 red cards, 3 black cards, and 2 blue cards. A player randomly picks 2 cards. Quantity A Probability that both selected cards are the same color Quantity B Probability that the selected cards are of different colors

  1. Quantity A is greater.

  2. Quantity B is greater.

  3. The two quantities are equal.

  4. The relationship cannot be determined from the information given.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total ways to pick 2 cards = 8C2 = 28. Same color: (3C2 + 3C2 + 2C2) = 3 + 3 + 1 = 7. P(Same) = 7/28 = 1/4. Different color: 1 - 1/4 = 3/4. Quantity B (3/4) is greater than Quantity A (1/4).

Multiple choice

A bag is selected at random from two available bags. Bag 1 contains 6 red balls and 4 green balls, while Bag 2 contains 5 red balls and 5 green balls. The probability of selecting Bag 1 is 0.4, and the probability of selecting Bag 2 is 0.6. After selecting a bag, one ball is drawn at random from it. What is the probability that the ball drawn is red?

  1. 0.48

  2. 0.54

  3. 0.58

  4. 0.62

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

P(Red) = P(Bag1)*P(Red|Bag1) + P(Bag2)*P(Red|Bag2) = 0.4 * (6/10) + 0.6 * (5/10) = 0.24 + 0.30 = 0.54.

Multiple choice

Three bags contain 3 red, 7 black; 8 red, 2 black, and 4 red & 6 black balls respectively. One of the bags is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the third bag.

  1. 1/15

  2. 2/15

  3. 3/15

  4. 4/15
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Bayes' Theorem: P(Bag3|Red) = P(Red|Bag3)P(Bag3) / [P(Red|Bag1)P(Bag1) + P(Red|Bag2)P(Bag2) + P(Red|Bag3)P(Bag3)]. Probabilities: 3/10, 8/10, 4/10. Calculation: (4/10 * 1/3) / ((3/10 + 8/10 + 4/10) * 1/3) = 4/15.

Multiple choice

Three perfect dice D1, D2 and D3 are rolled. Let x, y and z represent the numbers on D1, D2 and D3, respectively. What is the number of possible outcomes such that x < y < z?

  1. 20

  2. 18

  3. 14

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We need to choose 3 distinct numbers from {1, 2, 3, 4, 5, 6} such that x < y < z. This is equivalent to choosing any 3 distinct numbers, as there is only one way to arrange them in increasing order. The number of ways is 6C3 = (6*5*4)/(3*2*1) = 20.

Multiple choice

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

  1. 3/7

  2. 7/3

  3. 1/7

  4. 1/3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are 15 balls total. The probability of drawing the first black ball is 10/15 = 2/3. After drawing one black ball, 9 black balls and 14 total balls remain, so the probability of the second being black is 9/14. The total probability is (2/3) * (9/14) = 18/42 = 3/7.

Multiple choice

In a city, there are three popular gyms: Gym X, Gym Y, and Gym Z. Gym X attracts 50% of the fitness enthusiasts, Gym Y attracts 30%, and Gym Z attracts the remaining 20%. It is known that 10% of the members at Gym X forget their access cards, 15% at Gym Y forget their access cards, and 5% at Gym Z forget their access cards. Now, if a randomly selected gym member is known to have forgotten their access card, what is the probability that the member is from Gym X?

  1. 0.38

  2. 0.42

  3. 0.48

  4. 0.52
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using Bayes' Theorem: P(X|Card) = P(Card|X)P(X) / [P(Card|X)P(X) + P(Card|Y)P(Y) + P(Card|Z)P(Z)]. P(X|Card) = (0.10 * 0.50) / (0.10 * 0.50 + 0.15 * 0.30 + 0.05 * 0.20) = 0.05 / (0.05 + 0.045 + 0.01) = 0.05 / 0.105 = 50/105 = 10/21 approx 0.476, which rounds to 0.48.

Multiple choice

There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?

  1. 6/14

  2. 2/7

  3. 6/35

  4. 6/29

  5. 6/42
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let R be the set of balls with red color and G be the set of balls with green color. Given |R intersect G| / |R| = 2/7 and |R intersect G| / |G| = 3/7. This implies |R| = 3.5 * |R intersect G| and |G| = 2.33 * |R intersect G|. Using the principle of inclusion-exclusion, the total number of balls is 58. Solving the system yields the probability.