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

Each of the urns contains $4$ white and $6$ blackballs. The (n + 1)th urn contains $5$ white and $5$black balls. One of the (n+1) urns is chosen atrandom and two balls are drawn from it withoutreplacement and both the balls turn out to beblack. Then the probability that the (n+1)th urn was chosen to draw the balls is $1/16$, the value of n is
  1. $10$
  2. $11$
  3. $12$
  4. $13$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let n urns have 4W, 6B and 1 urn have 5W, 5B. Total urns = n+1. P(Urn_n+1 | 2B) = [P(2B | Urn_n+1) * P(Urn_n+1)] / [P(2B | Urn_n+1) * P(Urn_n+1) + P(2B | Urn_n) * P(Urn_n)]. P(2B | Urn_n+1) = 5C2 / 10C2 = 10/45 = 2/9. P(2B | Urn_n) = 6C2 / 10C2 = 15/45 = 1/3. P(Urn_n+1) = 1/(n+1), P(Urn_n) = n/(n+1). 1/16 = (2/9 * 1/(n+1)) / (2/9 * 1/(n+1) + 1/3 * n/(n+1)) = (2/9) / (2/9 + n/3) = 2 / (2 + 3n). 2 + 3n = 32, 3n = 30, n = 10.

Multiple choice

One bag contains 5 white and 3 black balls and an other contains 4 white, 5 red balls. Two balls are drawn from one of them choosing at random.The probability that they are of different colours is
  1. $\displaystyle \frac{15}{56}$
  2. $\displaystyle \frac{5}{18}$
  3. $\displaystyle \frac{275}{504}$
  4. $\displaystyle \frac{275}{624}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of choosing bag 1 is 1/2. P(diff colors | bag 1) = (5/8 * 3/7) + (3/8 * 5/7) = 30/56 = 15/28. Probability of choosing bag 2 is 1/2. P(diff colors | bag 2) = (4/9 * 5/8) + (5/9 * 4/8) = 40/72 = 5/9. Total probability = 1/2 * (15/28 + 5/9) = 1/2 * ((135 + 140) / 252) = 275 / 504.

Multiple choice

A bag contains $2$ white, $3$ black and $4$ green balls.Two balls are drawn one after another with replacement.The probability that the first one is white and the second one is black is
  1. $\displaystyle \frac{5}{27}$
  2. $\displaystyle \frac{1}{9}$
  3. $\displaystyle \frac{2}{27}$
  4. $\displaystyle \frac{4}{27}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total balls = 2 + 3 + 4 = 9. Probability of white = 2/9. Probability of black = 3/9 = 1/3. Since the draw is with replacement, the events are independent: (2/9) * (3/9) = 6/81 = 2/27.

Multiple choice

An urn $A$ contains $2$ white and $3$ black balls. Another urn $B$ contains $3$ white and $4$ black balls.Out of these two urns, one is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that I. It is from urn $A$ is $21/40$, II. It is from urn $B$ is $20/41$ Which of the following statements is correct

  1. I only

  2. II only

  3. both I and II

  4. neither I nor II
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Urn A: 2W, 3B (Total 5). Urn B: 3W, 4B (Total 7). P(A) = 1/2, P(B) = 1/2. P(B|A) = 3/5, P(B|B) = 4/7. P(Black) = (1/2 * 3/5) + (1/2 * 4/7) = 3/10 + 4/14 = 21/70 + 20/70 = 41/70. P(A|Black) = (3/10) / (41/70) = 21/41. P(B|Black) = (20/70) / (41/70) = 20/41. Statement II is correct.

Multiple choice

From a box containing 10 cards numbered 1 to10, four cards are drawn together. The probability that their sum is odd is
  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{11}{21}$
  3. $\displaystyle \frac{10}{21}$
  4. $\displaystyle \frac{1}{21}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total ways to choose 4 cards from 10 is 10C4 = 210. For the sum to be odd, we need an odd number of odd cards (1 or 3). Odd cards: {1,3,5,7,9} (5), Even cards: {2,4,6,8,10} (5). Case 1 (1 odd, 3 even): 5C1 * 5C3 = 5 * 10 = 50. Case 2 (3 odd, 1 even): 5C3 * 5C1 = 10 * 5 = 50. Total favorable = 100. Probability = 100/210 = 10/21.

Multiple choice

From 80 cards numbered 1 to 80, two cards are drawn at random. The probability that the cards have the numbers divisible by 4 is
  1. $\displaystyle \frac{15}{316}$
  2. $\displaystyle \frac{17}{316}$
  3. $\displaystyle \frac{19}{316}$
  4. $\displaystyle \frac{21}{316}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Numbers divisible by 4 between 1 and 80 are 80/4 = 20 numbers. The probability of picking two such numbers is (20/80) * (19/79) = (1/4) * (19/79) = 19/316.

Multiple choice

In a bag there are $N$ balls out of which $R$ are red.If $n$ balls are drawn at random from the bag, the probability that $r$ of them are red is
  1. $\displaystyle \frac{n_{C_{r}}}{^NC_{n}}$
  2. $\displaystyle \frac{R_{C_{r}}}{^NC_{n}}$
  3. $\displaystyle \frac{^{R}C_{r}\times ^{N-R}c_{n-r}}{^{N}C_{n}}$
  4. $\displaystyle \frac{^{N}C_{r}}{^{R}C_{n}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a classic hypergeometric distribution problem. The number of ways to choose r red balls from R is RCr, and the number of ways to choose the remaining (n-r) balls from (N-R) is (N-R)C(n-r). The total ways to choose n balls from N is NCn.

Multiple choice

From $100$ cards numbered $1$ to $100$, two cardsare drawn one by one with replacement. The probability that both are divisible by $5$ is
  1. $\displaystyle \frac{1}{5}$
  2. $\displaystyle \frac{1}{10}$
  3. $\displaystyle \frac{1}{25}$
  4. $\displaystyle \frac{1}{15}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Numbers divisible by 5 between 1 and 100 are 20. Probability of picking one is 20/100 = 1/5. Since replacement occurs, the probability for both is (1/5) * (1/5) = 1/25.

Multiple choice

If n coins are tossed then find the probability of not getting either all heads or all tails
  1. $\displaystyle 1-\frac{1}{2^{n}}$
  2. $\displaystyle \frac{1}{2^{n-1}}$
  3. $\displaystyle 1-\frac{1}{2^{n+1}}$
  4. $\displaystyle 1-\frac{1}{2^{n-1}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes = 2^n. Outcomes with all heads = 1. Outcomes with all tails = 1. Probability of (all heads or all tails) = 2 / 2^n = 1 / 2^(n-1). Probability of NOT getting these = 1 - 1 / 2^(n-1).