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

Two dice were thrown simultaneously. Then find the probability of g etting at least one outcome as multiple of 3.
  1. $\dfrac{2}{9}$
  2. $\dfrac{1}{6}$
  3. $\dfrac{5}{12}$
  4. $\dfrac{5}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Probability of at least one multiple of 3 = 1 - P(no multiple of 3). Multiples of 3 on a die are 3, 6 (2 outcomes). Non-multiples are 1, 2, 4, 5 (4 outcomes). P(no multiple) = (4/6) * (4/6) = 16/36 = 4/9. P(at least one) = 1 - 4/9 = 5/9.

Multiple choice

For $k=1, 2, 3$ the box ${ B }{ k }$ contains $k$ red balls and $\left( k+1 \right) $ white balls, let $P\left( { B }{ 1 } \right) =\dfrac { 1 }{ 2 } ,P\left( { B }{ 2 } \right) =1$ and $P\left( { B }{ 3 } \right) =\dfrac { 1 }{ 6 } $. A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box ${ B }_{ 2 }$ is
  1. $\dfrac { 35 }{ 78 } $
  2. $\dfrac { 14 }{ 39 } $
  3. $\dfrac { 10 }{ 13 } $
  4. $\dfrac { 12 }{ 13 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Alice, Bob and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice, Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcome of any other toss.)
  1. $\dfrac{1}{3}$
  2. $\dfrac{2}{9}$
  3. $\dfrac{5}{18}$
  4. $\dfrac{25}{91}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let p = 1/6 be the probability of rolling a 6. Carol wins if she rolls a 6 on her 1st turn (prob = (1-p)^2 * p), 2nd turn (prob = (1-p)^5 * p), etc. This is a geometric series with first term a = (5/6)^2 * (1/6) = 25/216 and common ratio r = (5/6)^3 = 125/216. Sum = a / (1-r) = (25/216) / (1 - 125/216) = (25/216) / (91/216) = 25/91.

Multiple choice

We roll a fair four-sided die. If the result is $1$ or $2$, we roll once more but otherwise, we stop. What is the probability that the sum total of our rolls is at least $4$?
  1. $\dfrac{1}{16}$
  2. $\dfrac{4}{16}$
  3. $\dfrac{7}{16}$
  4. $\dfrac{9}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Possible outcomes: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3), (4). Total outcomes = 10. Outcomes with sum >= 4: (1,3), (1,4), (2,2), (2,3), (2,4), (3), (4). There are 7 favorable outcomes out of 16 equally likely paths in the tree diagram (1/4 * 1/4 for each branch). The probability is 9/16.

Multiple choice

Let $u_1$ and $u_2$ be two urns such that $u_1$ contains 3 white, 2 red balls and $u_2$ contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn $u_1$ and put into $u_2$. However, if tail appears, then 2 balls are drawn at random from $u_1$ and put into $u_2$. Now, 1 ball is drawn at random from $u_2$. Then, probability of the drawn ball from $u_2$ being white is
  1. $\dfrac{13}{30}$
  2. $\dfrac{23}{30}$
  3. $\dfrac{19}{30}$
  4. $\dfrac{11}{30}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let H be the event of heads (p=0.5) and T be tails (p=0.5). If H, we move 1 ball from u1 to u2. If T, we move 2 balls. Calculate the probability of drawing a white ball from u2 by conditioning on the coin flip and the composition of the balls transferred.

Multiple choice

There are 4 urns. The first urn contains 1 white & 1 black ball, the second urn contains 2 white & 3 black balls, the third urn contains 3 white & 5 black balls & the fourth urn contains 4 white & 7 black balls. The selection of each urn is not equally likely. The probability of selecting $\displaystyle i^{th}$ urn is $\displaystyle \frac{i+1}{34}$ (i = 1, 2, 3, 4). If we randomly select one of the urn & draw a ball, then the probability of ball being white is
  1. $\displaystyle \frac{569}{1496}$
  2. $\displaystyle \frac{27}{56}$
  3. $\displaystyle \frac{8}{73}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Numbers 1, 2, 3, .... , 100 are written down on each of the cards A, B and C. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of a right angled triangle, is
  1. $\dfrac{4}{100^3}$
  2. $\dfrac{3}{50^3}$
  3. $\dfrac{3!}{100^3}$

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

A right triangle requires a^2 + b^2 = c^2. For numbers 1-100, the number of such triples is small (e.g., 3-4-5, 6-8-10, etc.). The total outcomes are 100^3. The count of valid triples is much higher than the options provided, making 'None of these' the correct choice.

Multiple choice

A die is rolled so that the probability of face $i$ is proportional to $i,\,\,{i=1,2,.....6}$. The probability of an even number occurring when the die is rolled, is
  1. $\dfrac { 7 }{ 4 } $
  2. $\dfrac { 4 }{ 7 } $
  3. $\dfrac { 5 }{ 7 } $

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

Probabilities are proportional to i, so P(i) = k*i. Sum of probabilities = k(1+2+3+4+5+6) = 21k = 1, so k = 1/21. Even numbers are 2, 4, 6. P(even) = P(2) + P(4) + P(6) = k(2+4+6) = 12k = 12/21 = 4/7.