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 coin is tossed $100$ times and tail is obtained $10$ times. Now, if a coin is tossed at random, what is the probability of getting a head?
  1. $\dfrac{11}{10}$
  2. $\dfrac{9}{10}$
  3. $\dfrac{90}{10}$
  4. $\dfrac{100}{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of an event is the ratio of favorable outcomes to total trials. If a coin is tossed 100 times and tails occurred 10 times, the experimental probability of heads is (100-10)/100 = 90/100 = 9/10.

Multiple choice

Two different dice are tossed together. Find the probability (i) that the number on each die is even. (ii) that the sum of numbers appearing on the two dice is 5.
  1. i) $\cfrac 13$ and ii) $\cfrac 18$
  2. i) $\cfrac 12$ and ii) $\cfrac 17$
  3. i) $\cfrac 15$ and ii) $\cfrac 16$
  4. i) $\cfrac 14$ and ii) $\cfrac 19$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

i) Even numbers on two dice: (2,4,6) * (2,4,6) = 9 outcomes. 9/36 = 1/4. ii) Sum 5: (1,4), (2,3), (3,2), (4,1) = 4 outcomes. 4/36 = 1/9.

Multiple choice

A bag contains $9$ marbles, $3$ of which are red, $3$ of which are blue, and $3$ of which are yellow. If three marbles are selected from the bag at random, what is probability that they are all of different colors?
  1. $\dfrac {1}{15}$
  2. $\dfrac {9}{28}$
  3. $\dfrac {1}{19}$
  4. $\dfrac {1}{20}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total ways to choose 3 marbles from 9 is 9C3 = 84. Favorable ways to choose one of each color (1 red, 1 blue, 1 yellow) is 3C1 * 3C1 * 3C1 = 3 * 3 * 3 = 27. Probability = 27/84 = 9/28.

Multiple choice

A biased coin is tossed twice.The probability of head is twice the tail.The PDF of number of heads is x $0$ $1$ $2$ p(x) $\dfrac{a}{d}$ $\dfrac{b}{d}$ $\dfrac{c}{d}$ then values of $a,b,c,d$ are
  1. $a=1,b=2,c=3,d=4$
  2. $a=1,b=4,c=4,d=9$
  3. $a=1,b=4,c=4,d=10$
  4. $a=1,b=2,c=1,d=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let P(Tail) = p, then P(Head) = 2p. p + 2p = 1 => p = 1/3. P(H) = 2/3, P(T) = 1/3. For two tosses: P(0 heads) = P(TT) = 1/3 * 1/3 = 1/9. P(1 head) = P(HT, TH) = 2 * (2/3 * 1/3) = 4/9. P(2 heads) = P(HH) = 2/3 * 2/3 = 4/9. Thus a=1, b=4, c=4, d=9.

Multiple choice

When a pair of six faced fair dice are thrown, the probability that the sum of the numbers on the two dice is greater than $7$, is
  1. $\cfrac { 1 }{ 3 } $
  2. $\cfrac { 5 }{ 12 } $
  3. $\cfrac { 1 }{ 2 } $
  4. $\cfrac { 1 }{ 4 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total outcomes = 36. Sum > 7: sums of 8, 9, 10, 11, 12. Outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) [5]; (3,6), (4,5), (5,4), (6,3) [4]; (4,6), (5,5), (6,4) [3]; (5,6), (6,5) [2]; (6,6) [1]. Total = 5+4+3+2+1 = 15. Probability = 15/36 = 5/12.

Multiple choice

Bag $A$ contains three red and four white Balls; bag $B$ contains two red and three white balls. If one ball is drawn from bag $A$ and two balls from bag $B$, find the probability that: (i) One ball is red and two balls are white; (ii) All the three balls are of the same colour.
  1. i) $\cfrac {33}{70}$ and ii) $\cfrac {3}{14}$
  2. i) $\cfrac {22}{31}$ and ii) $\cfrac {1}{3}$
  3. i) $\cfrac 12$ and ii) $\cfrac 12$
  4. i) $\cfrac {8}{25}$ and ii) $\cfrac {2}{7}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bag A: 3R, 4W (Total 7). Bag B: 2R, 3W (Total 5). Drawing 1 from A, 2 from B. (i) 1R from A, 2W from B OR 1W from A, 1R+1W from B. P(i) = (3/7 * 3/5 * 2/4) + (4/7 * (2/5 * 3/4 + 3/5 * 2/4)) = 18/140 + 4/7 * 12/20 = 18/140 + 48/140 = 66/140 = 33/70. (ii) All red: 3/7 * 2/5 * 1/4 = 6/140. All white: 4/7 * 3/5 * 2/4 = 24/140. Total = 30/140 = 3/14.

Multiple choice

A garage mechanic keeps a box of good springs to use as replacements on customers cars. The box contains $5$ springs. A colleague, thinking that the springs are for scrap, tosses three faulty springs into the box. The mechanic picks two springs out of the box while servicing a car. Find the probability that the second spring drawn is faulty.
  1. $\dfrac{1}{8}$
  2. $\dfrac{2}{8}$
  3. $\dfrac{3}{8}$
  4. $\dfrac{4}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total springs = 5 + 3 = 8. Good = 5, Faulty = 3. Probability second is faulty = P(1st good, 2nd faulty) + P(1st faulty, 2nd faulty) = (5/8 * 3/7) + (3/8 * 2/7) = 15/56 + 6/56 = 21/56 = 3/8.