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 six faced fair dice is thrown until $1$ comes. The probability that $1$ comes in even no. of trails is
  1. $\displaystyle \frac{5}{11}$
  2. $\displaystyle \frac{5}{6}$
  3. $\displaystyle \frac{6}{11}$
  4. $\displaystyle \frac{1}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of rolling a 1 is p = 1/6, and not rolling a 1 is q = 5/6. The probability of getting the first 1 on an even trial is q*p + q^3*p + q^5*p + ... which is a geometric series with first term a = (5/6)*(1/6) = 5/36 and common ratio r = (5/6)^2 = 25/36. The sum is a/(1-r) = (5/36) / (1 - 25/36) = (5/36) / (11/36) = 5/11.

Multiple choice

$A$ and $B$ toss a fair coin each simultaneously $50$ times. The probability that both of them will not get tail at the same toss is
  1. $\displaystyle { \left( \frac { 3 }{ 4 } \right) }^{ 50 }$
  2. $\displaystyle { \left( \frac { 2 }{ 7 } \right) }^{ 50 }$
  3. $\displaystyle { \left( \frac { 1 }{ 8 } \right) }^{ 50 }$
  4. $\displaystyle { \left( \frac { 7 }{ 8 } \right) }^{ 50 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probability of getting a tail is 1/2. Probability of both NOT getting a tail = 1 - P(both get tails) = 1 - (1/2 * 1/2) = 3/4. For 50 tosses, it is (3/4)^50.

Multiple choice

Three faces of a fair dice are yellow, two faces red and one blue. The die is tossed 3-times. The probability that the colours yellow, red and blue appear in the first, second and the third tosses respectively is .............
  1. $1/36$
  2. $1/216$
  3. $1/72$
  4. $1/108$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probability of yellow = 3/6 = 1/2. Probability of red = 2/6 = 1/3. Probability of blue = 1/6. The probability of the sequence (yellow, red, blue) is (1/2) * (1/3) * (1/6) = 1/36.

Multiple choice

A man alternatively tosses a coin and throws a die. The probability of getting a head on the coin before he gets 4 on the die is
  1. $ \displaystyle \frac{6}{7} $
  2. $ \displaystyle \frac{2}{3} $
  3. $ \displaystyle \frac{3}{4} $
  4. $ \displaystyle \frac{1}{2} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let H be getting a head (p=1/2) and D be getting a 4 (p=1/6). The man wins if he gets H before D. The sequence of events is: H (win), D (continue), H (win), D (continue)... The probability of winning is P(H) + P(not H and not D) * P(H) + ... This is a geometric series. P(win) = (1/2) + (1/2 * 5/6) * (1/2) + (1/2 * 5/6)^2 * (1/2) = (1/2) / (1 - 5/12) = (1/2) / (7/12) = 6/7.

Multiple choice

A bag contains $n$ tickets marked $1,2,3,...,n$. If two tickets are drawn, then the chance that the difference of the numbers on the tickets exceed $m<(n-1)$ is
  1. $ \displaystyle \frac { \left( n-m \right) \left( n-m+1 \right) }{ n\left( n-1 \right) } $
  2. $ \displaystyle \frac { \left( n-m \right) \left( m-n+1 \right) }{ n\left( n-1 \right) } $
  3. $ \displaystyle \frac { \left( n-m \right) \left( n-m-1 \right) }{ n\left( n-1 \right) } $

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

Total ways to choose 2 tickets is n(n-1)/2. Favorable ways where difference > m: (n-m)(n-m-1)/2. Probability = [(n-m)(n-m-1)/2] / [n(n-1)/2] = (n-m)(n-m-1) / n(n-1).

Multiple choice

A fair die is tossed until a number greater than $4$ appear. The probability that an even number of tosses shall be required is:
  1. $ \displaystyle \frac{1}{2} $
  2. $ \displaystyle \frac{3}{5} $
  3. $ \displaystyle \frac{2}{5} $
  4. $ \displaystyle \frac{2}{3} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of success (number > 4) is 2/6 = 1/3. Probability of failure is 2/3. Even tosses means success on 2nd, 4th, 6th... toss. P = (2/3)(1/3) + (2/3)^3(1/3) + ... = (2/9) / (1 - 4/9) = (2/9) / (5/9) = 2/5.