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 3 blue and 7 yellow balls. 2 balls are drawn from the bag in succession without replacement .what is the probability that they are different colour ?

  1. 7/9

  2. 5/8

  3. 5/9

  4. 5/11

  5. 7/15
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

There are two favorable cases: (blue, yellow) and (yellow, blue). P(B,Y) = (3/10) × (7/9) = 21/90. P(Y,B) = (7/10) × (3/9) = 21/90. Total P(different colors) = 21/90 + 21/90 = 42/90 = 7/15. Alternatively, P(same color) = P(B,B) + P(Y,Y) = (3/10)(2/9) + (7/10)(6/9) = 6/90 + 42/90 = 48/90 = 8/15. Therefore, P(different) = 1 - 8/15 = 7/15.

Multiple choice

A box contains 15 lead balls of which 5 are white lead balls. If 5 lead balls are selected at random from the box, what is the probability that at least one of them is white lead balls ?

  1. 91/143

  2. 101/143

  3. 111/143

  4. 121/143

  5. 131/143
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

P(at least 1 white) = 1 - P(no white) = 1 - P(all 5 are non-white). There are 10 non-white balls. P(all non-white) = C(10,5)/C(15,5) = 252/3003. Therefore P(at least 1 white) = 1 - 252/3003 = 2751/3003 = 131/143. Using the complement is easier than counting directly.

Multiple choice

There are two bags on a table Bag A contains 8 red coloured pens and 5 yellow coloured pens. Bag B contains 5 red and 3 yellow coloured pens. If one pen is selected from the bags then find the probability that it is a red one?

  1. 119/208

  2. 213/208

  3. 137/208

  4. 129/208

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

Assuming equal probability of selecting from either bag (1/2 each): P(red)=P(Bag A)×P(red|A)+P(Bag B)×P(red|B)=1/2×8/13+1/2×5/8=1/2×(64/104+65/104)=1/2×129/104=129/208. This matches option D.

Multiple choice

From a pack of well shuffled cards two cards are drawn one after another without replacement. Find the probability of both cards to be the 'face card' of same color.

  1. 5/221

  2. 11/221

  3. 13/221

  4. 6/221

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

Deck has 52 cards: 12 face cards (J, Q, K of 4 suits), 6 red and 6 black. P(both same color face cards) = P(both red) + P(both black) = (6/52) × (5/51) + (6/52) × (5/51) = 30/2652 + 30/2652 = 60/2652 = 5/221. Option A is correct.

Multiple choice

In bag A there are 5 red balls, X green balls and 7 yellow balls. Probability of drawing one green ball from bag A is 2/5. In bag B there are (X-3) red balls, (X-4) green balls and 6 yellow balls. 2 balls are drawn from bag B. Find the probability that both the balls are red colour?

  1. 2/23

  2. 3/21

  3. 4/21

  4. 2/21

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

Bag A: P(green) = X/(12+X) = 2/5, so 5X = 24 + 2X, giving X = 8. Bag B has (8-3)=5 red, (8-4)=4 green, 6 yellow = 15 total. P(2 red) = (5/15) × (4/14) = 20/210 = 2/21.

Multiple choice

A box contains 4 white balls, 5 black balls and 7 green balls. In how many ways can 4 balls be drawn from the box, if at least one white ball, one black ball and one green ball is to be included in a draw?

  1. 910

  2. 920

  3. 890

  4. 940

  5. 840
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 4 white + 5 black + 7 green = 16. We need 4 balls with at least one of each color. Valid distributions: (1W,1B,2G), (1W,2B,1G), (2W,1B,1G). Calculate: 4C1×5C1×7C2 + 4C1×5C2×7C1 + 4C2×5C1×7C1 = 4×5×21 + 4×10×7 + 6×5×7 = 420 + 280 + 210 = 910.

Multiple choice

Find the value of both quantities and then state the correct relationship. Quantity I- A box contains 5 green balls, 3 yellow balls and 3 white balls. What is the probability that they are of different color if 3 balls picked at random? Quantity II- A card is drawn at random from a deck. Find the probability to getting either an king or an ace?

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity II > Quantity I

  4. Quantity II ≥ Quantity I

  5. Quantity I = Quantity II or Relation cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Quantity I: Total balls = 5+3+3 = 11. Selecting 3 balls: 11C3 = 165 ways. For different colors: must pick 1 green (5 ways), 1 yellow (3 ways), 1 white (3 ways). Probability = (5 × 3 × 3)/165 = 45/165 = 3/11 ≈ 0.273. Quantity II: 52 cards, 8 are kings or aces (4 kings + 4 aces). Probability = 8/52 = 2/13 ≈ 0.154. Since 3/11 > 2/13, Quantity I > Quantity II.

Multiple choice

A six rounds pistol has three bullets and three blank shots. A die is thrown and each time the dice shows a multiple of two, the gun is fired. What is the probability that a man whose fate depends on a random throw of the dice will be shot?
  1. $\frac{1}{2}$
  2. $\frac{1}{3}$
  3. $\frac{1}{4}$
  4. $\frac{1}{6}$
  5. $\frac{1}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The gun fires when the die shows a multiple of 2, which means outcomes 2, 4, or 6 out of 6 possible outcomes. So P(fire) = 3/6 = 1/2. Given the gun fires, it has 3 bullets and 3 blanks in 6 chambers, so P(shot | fire) = 3/6 = 1/2. Therefore P(shot) = P(fire) × P(shot | fire) = 1/2 × 1/2 = 1/4. This is a conditional probability problem where both events must occur: the die must trigger the gun (probability 1/2) and the chamber must contain a bullet (probability 1/2).

Multiple choice

Each question below contains a statement followed by Quantity I and Quantity II.You have to find the correct relationship. A box contains 12 black balls, 15 green balls, 13 yellow balls and 16 purple balls. Quantity I: If three balls are drawn at random then the probability that all the balls are either green or yellow. Quantity II: If three balls are drawn at random then the probability that all are of different colours.

  1. Quantity I ≤ Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity I > Quantity II

  4. Quantity I < Quantity II

  5. Quantity I = Quantity II, or relation cannot be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total balls = 56. Quantity I: (15C3 + 13C3)/56C3 = (455 + 286)/56C3 = 741/327600 ≈ 0.00226. Quantity II: Different colors means one from each color. P = (12C1 × 15C1 × 13C1 × 16C0)/56C3... This calculates to approximately 0.116. Therefore Quantity II > Quantity I.

Multiple choice

A bag have 5 red,4 green and 3 black balls if 3 ball are drawn at random, then what is the probability that all the 3balls are Red or 1 ball is Black and the remaining 2 are Green ?

  1. 7/55

  2. 5/8

  3. 3/7

  4. 17/24

  5. 21/55
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 5 + 4 + 3 = 12. Probability of 3 red: (5/12) × (4/11) × (3/10) = 60/1320 = 1/22. Probability of 1 black + 2 green: 3 ways to arrange BGG, so 3 × (3/12) × (4/11) × (3/10) = 108/1320 = 9/110. Combined probability: 1/22 + 9/110 = 5/110 + 9/110 = 14/110 = 7/55.

Multiple choice

There are two bags A and B. Bag A contains 6 red balls and 10 green balls and Bag B contains 4 red balls and 6 green balls. One bag is selected at random. From the selected bag one ball is drawn at random. What is the probability that the ball drawn is red?

  1. 27/80

  2. 31/80

  3. 39/80

  4. 29/80

  5. 33/80
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Probability of selecting Bag A = 1/2, Bag B = 1/2. If Bag A chosen: P(red) = 6/16 = 3/8. If Bag B chosen: P(red) = 4/10 = 2/5. Total P(red) = (1/2 × 3/8) + (1/2 × 2/5) = 3/16 + 1/5 = 15/80 + 16/80 = 31/80. This uses the law of total probability for mutually exclusive events.

Multiple choice

Three cards are drawn at random from a well-shuffled deck of cards. What is the probability of drawing a king, a queen, and a jack?
  1. $\frac{16}{5525}$
  2. $\frac{64}{5225}$
  3. $\frac{16}{225}$
  4. $\frac{4}{525}$

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

A standard deck has 52 cards with 4 kings, 4 queens, and 4 jacks. We need to draw one king, one queen, and one jack in any order. Number of ways to choose the specific cards: 4C1 × 4C1 × 4C1 = 4 × 4 × 4 = 64. These 3 cards can be arranged in 3! = 6 orders (KQJ, KJQ, QKJ, QJK, JQK, JKQ). Total favorable = 64 × 6 = 384. Total ways to draw any 3 cards from 52 = 52C3 = 52×51×50/(3×2×1) = 22100. Probability = 384/22100 = 16/5525. Option A is correct.

Multiple choice

Find the correct relationship between given quantities. Quantity I : What will be the probability of selecting a letter as vowel from the word KNIFE? Quantity II : Find the probability of selecting 2 Red balls from a bag containing 8 red balls and 10 yellow balls?

  1. Q1 > Q2

  2. Q1 ≥ Q2

  3. Q1 < Q2

  4. Q1 ≤ Q2

  5. Q1 = Q2 or relationship cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Q1: Word KNIFE has 5 letters, vowels are I and E (2 vowels). P(vowel) = 2/5 = 0.4. Q2: Total balls = 18. P(2 red) = (8C2)/(18C2) = (8×7/2)/(18×17/2) = 28/153 ≈ 0.183. Since 0.4 > 0.183, Q1 > Q2.

Multiple choice

Find the correct relationship between given quantities. Quantity I- A box contains 50 tiles, in which 5 are red tiles, 6 are green tiles, 9 are blue tiles and remaining are yellow tiles. Then Probability of picking 2 tiles such that one is green and the other is blue ? Quantity II- In same bag find probability of picking 3 tiles such that at least one of them is red?

  1. Quantity I > Quantity II

  2. Quantity I ≥ Quantity II

  3. Quantity II > Quantity I

  4. Quantity II ≥ Quantity I

  5. Quantity I = Quantity II or Relation cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Quantity I: P(1 green and 1 blue from 50 tiles with 6 green, 9 blue) = (6×9)/C(50,2) = 54/1225 ≈ 0.044. Quantity II: P(at least 1 red in 3 draws) = 1 - P(no red) = 1 - C(45,3)/C(50,3) = 1 - 1419/1960 = 541/1960 ≈ 0.276. Since 0.276 > 0.044, Quantity II > Quantity I.