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 bag contains $17$ tickets numbered from $1$ to $17$. A ticket is drawn at random, then another ticket is drawn without replacing the first one. The probability that both the tickets may show even numbers is?
  1. $\dfrac{7}{34}$
  2. $\dfrac{8}{17}$
  3. $\dfrac{7}{16}$
  4. $\dfrac{7}{17}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total tickets = 17. Even numbers are {2, 4, 6, 8, 10, 12, 14, 16} (8 total). Probability of first even = 8/17. Probability of second even = 7/16. Total probability = (8/17) * (7/16) = 56 / 272 = 7/34.

Multiple choice

A die is thrown four times. The probability of getting perfect square in at least one throw is
  1. $\cfrac { 16 }{ 81 } $
  2. $\cfrac { 65 }{ 81 } $
  3. $\cfrac { 23 }{ 81 } $
  4. $\cfrac { 58 }{ 81 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Perfect squares on a die are 1 and 4. Probability of getting a perfect square in one throw = 2/6 = 1/3. Probability of not getting one = 2/3. Probability of not getting one in 4 throws = (2/3)^4 = 16/81. Probability of at least one = 1 - 16/81 = 65/81.

Multiple choice

A coin is tossed three times. If $X$ denotes the absolute difference between the number of heads and the number of tails then $P(X=1)=$
  1. $\cfrac { 1 }{ 2 } $
  2. $\cfrac { 2 }{ 3 } $
  3. $\cfrac { 1 }{ 6 } $
  4. $\cfrac { 3 }{ 4 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Possible outcomes for 3 tosses: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. X is the absolute difference between heads and tails. For 3 tosses, the counts are (3,0) or (0,3) where X=3, and (2,1) or (1,2) where X=1. There are 3 outcomes with 2 heads (HHT, HTH, THH) and 3 with 1 head (HTT, THT, TTH). Total outcomes = 8. P(X=1) = 6/8 = 3/4.

Multiple choice

Two dice are thrown simultaneously What is the probability of getting two numbers whose product is even?
  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{3}{4}$
  3. $\displaystyle \frac{3}{8}$
  4. $\displaystyle \frac{5}{16}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The product of two numbers is odd only if both numbers are odd. The probability of one die being odd is 1/2. The probability of both being odd is 1/2 * 1/2 = 1/4. The probability of the product being even is 1 - P(odd) = 1 - 1/4 = 3/4.

Multiple choice

A bag contains $3$ red, $4$ white and $5$ blue balls. If two balls are drawn at random, then the probability that they are of different colours, is
  1. $\cfrac { 47 }{ 66 } $
  2. $\cfrac { 23 }{ 33 } $
  3. $\cfrac { 47 }{ 132 } $
  4. $\cfrac { 47 }{ 33 } $
  5. $\cfrac { 70 }{ 33 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 12. Total ways to pick 2 = 12C2 = 66. Ways to pick same color: (3C2 + 4C2 + 5C2) = 3 + 6 + 10 = 19. Ways to pick different colors = 66 - 19 = 47. Probability = 47/66.

Multiple choice

A pair of $12$-sided fair dice with faces numbered $1, 2, 3,..., 12$ is rolled. The probability that the sum of the numbers appearing has remainder $2$ when divided by $9$ is?
  1. $\displaystyle\frac{7}{72}$
  2. $\displaystyle\frac{5}{48}$
  3. $\displaystyle\frac{11}{144}$
  4. $\displaystyle\frac{1}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes = 12 * 12 = 144. We want sum S such that S mod 9 = 2. Possible sums: 2, 11, 20. Sum = 2: (1,1) [1 way]. Sum = 11: (1,10), (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2), (10,1) [10 ways]. Sum = 20: (8,12), (9,11), (10,10), (11,9), (12,8) [5 ways]. Total = 1 + 10 + 5 = 16. Probability = 16/144 = 1/9.

Multiple choice

From a well shuffled pack of $52$ playing cards two cards drawn at random. The probability that either both are red or both are kings is:
  1. $\dfrac{{\left( {^{26}{C_2} + \,{\,^4}{C_2}} \right)}}{{^{52}{C_2}}}$
  2. $\dfrac{{\left( {^{26}{C_2} + {\,^4}{C_2} - {\,^2}{C_2}} \right)}}{{^{52}{C_2}}}$
  3. $\dfrac{{^{30}{C_2}}}{{^{52}{C_2}}}$
  4. $\dfrac{{^{39}{C_2}}}{{^{52}{C_2}}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

P(A or B) = P(A) + P(B) - P(A and B). A = both red (26C2), B = both kings (4C2). A and B = both red kings (2C2). Total ways = 52C2. Probability = (26C2 + 4C2 - 2C2) / 52C2.