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 three coins, one of which has head on both sides, another is a biased coin that shows up heads $90\%$ of the time and the third one is an unbiased coin. A coin is taken out from the bag at random and tossed. If it shows up a head, then the probability that it is the unbiased coin, is:
  1. $\cfrac { 5 }{ 12 } $
  2. $\cfrac { 5 }{ 24 } $
  3. $\cfrac { 1 }{ 3 } $
  4. $\cfrac { 3 }{ 8 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let H be the event of getting heads. The probabilities of picking each coin are 1/3. P(H) = (1/3 * 1) + (1/3 * 0.9) + (1/3 * 0.5) = (1 + 0.9 + 0.5) / 3 = 2.4 / 3 = 0.8. The probability it is the unbiased coin given heads is P(Unbiased and H) / P(H) = (1/3 * 0.5) / 0.8 = 0.5 / 2.4 = 5/24.

Multiple choice

In a bag there are $6$ red, $4$ black balls. From it $2$ balls (without replacement) are drawn. If the first drawn ball is known to be red, the probability for the second drawn ball is also red is
  1. $\dfrac {5}{9}$
  2. $\dfrac {5}{18}$
  3. $\dfrac {1}{3}$
  4. $\dfrac {2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initially there are 6 red and 4 black balls (total 10). If one red ball is removed, 5 red and 4 black balls remain (total 9). The probability that the second ball is red is 5/9.

Multiple choice

If from each of the three boxes containing 3 white and 2 black, 2 white and 3 black and 1 white and 4 black balls, one ball is drawn at random, then the probability that two white and one black ball will be drawn is...
  1. $\dfrac { 37 }{ 125 } $
  2. $\dfrac { 6 }{ 125 } $
  3. $\dfrac { 1 }{ 125 } $
  4. $\dfrac { 36 }{ 125 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probabilities of drawing white from boxes are 3/5, 2/5, 1/5. Probabilities of black are 2/5, 3/5, 4/5. We need 2 white and 1 black: (W1, W2, B3) + (W1, B2, W3) + (B1, W2, W3) = (3/5*2/5*4/5) + (3/5*3/5*1/5) + (2/5*2/5*1/5) = 24/125 + 9/125 + 4/125 = 37/125.

Multiple choice

A die is thrown $2n + 1$ times. The probability of getting $1$ or $3$ or $4$ atmost $n$ times is
  1. $\dfrac12$
  2. $\dfrac1n$
  3. $\dfrac{n}{(2n +1)}$
  4. $\dfrac1{(2n + 1)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a binomial distribution problem with p = 3/6 = 1/2. For a symmetric distribution with 2n+1 trials, the probability of success in at most n trials is exactly 1/2 because the probability of success in at most n trials equals the probability of success in at least n+1 trials.

Multiple choice

Bag $A$ contains $6$ red and $4$ black balls and bag $B$ contains $4$ red and $6$ black balls. One ball is drawn at random from bag $B$ and placed in bag $A$. If one ball is now drawn from bag $A$. The probability that it is found to be red is
  1. $\dfrac {32}{55}$
  2. $\dfrac {33}{55}$
  3. $\dfrac {32}{63}$
  4. $\dfrac {25}{66}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are two cases: drawing a red ball from B (prob 4/10) or a black ball from B (prob 6/10). If red is moved, Bag A becomes 7 red, 4 black (prob 7/11). If black is moved, Bag A becomes 6 red, 5 black (prob 6/11). Total probability = (4/10 * 7/11) + (6/10 * 6/11) = 28/110 + 36/110 = 64/110 = 32/55.

Multiple choice

Two players $A$ and $B$ toss a coin alternatively, with $A$ beginning the game. The players who first throw a head is deemed to be the winner. $B$'s coin is fair and $A$'s is biased and has a probability $p$ showing a head. Find the value of $p$ so that the game is equiprobable to both the players.
  1. $\dfrac {1}{2}$
  2. $\dfrac {1}{3}$
  3. $1$
  4. $\dfrac {1}{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let p be the probability of A getting a head and 1/2 be the probability of B getting a head. The probability of A winning is p + (1-p)(1/2)p + (1-p)(1/2)(1-p)(1/2)p + ... which is a geometric series summing to p / (1 - (1-p)/2) = 2p / (1+p). Setting this equal to 1/2 gives 4p = 1+p, so 3p = 1, p = 1/3.