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

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well shufled pack of eleven cards numbered $2,3,4,...12$ is picked and the number on the card is noted. The probability that the noted number is either $7$ or $8$, is
  1. $\displaystyle \frac { 193 }{ 792 } $
  2. $\displaystyle \frac { 164 }{ 792 } $
  3. $\displaystyle \frac { 231 }{ 792 } $

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

P(Head) = 1/2. P(Sum 7 or 8 | Head) = P(Sum 7) + P(Sum 8) = 6/36 + 5/36 = 11/36. P(Tail) = 1/2. P(7 or 8 | Tail) = 2/11 (cards 2-12). Total probability = (1/2 * 11/36) + (1/2 * 2/11) = 11/72 + 1/11 = (121 + 72) / 792 = 193/792.

Multiple choice

Three groups of children contain 3 girls and one boy; 2 girl and 2 boys, one girl and 3 boys. One child is selected at random from each group. Show that the chance that the three selected consist of 1 girl and 2 boys is
  1. $\dfrac{13}{36}$
  2. $\dfrac{13}{32}$
  3. $\dfrac{15}{32}$
  4. $\dfrac{15}{36}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Probabilities of picking a girl from each group: G1=3/4, G2=2/4, G3=1/4. Probabilities of picking a boy: B1=1/4, B2=2/4, B3=3/4. We need 1 girl and 2 boys. Possible cases: (G,B,B) = 3/4 * 2/4 * 3/4 = 18/64; (B,G,B) = 1/4 * 2/4 * 3/4 = 6/64; (B,B,G) = 1/4 * 2/4 * 1/4 = 2/64. Sum = 26/64 = 13/32.

Multiple choice

Four whole numbers taken at random are multiplied together. What is the chance that the last digit in the product is $1,3,7$ or $9?$
  1. $\displaystyle \frac { 16 }{ 625 } $
  2. $\displaystyle \frac { 1 }{ 210 } $
  3. $\displaystyle \frac { 8 }{ 125 } $
  4. $\displaystyle \frac { 4 }{ 25 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The last digit of a product of four numbers depends only on the last digits of the individual numbers. There are 10 possibilities (0-9) for each digit. The probability that the last digit is 1, 3, 7, or 9 is 4/10 = 2/5 for one number. For four numbers, the probability is (2/5)^4 = 16/625.

Multiple choice

6 married couples are standing in a room. if 4 people are chosen at random, then the chance that exactly one married couple is among the 4 is-
  1. $\cfrac{16}{33}$
  2. $\cfrac{8}{33}$
  3. $\cfrac{17}{33}$
  4. $\cfrac{24}{33}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are 12 people total (6 couples). The total ways to choose 4 people is 12C4 = 495. To choose exactly one couple, we select 1 couple out of 6 (6C1) and then 2 people from the remaining 5 couples such that they are not a couple (5C2 * 2^2). This gives 6 * 10 * 4 = 240. The probability is 240/495 = 16/33.

Multiple choice

Five horses are in a race . A man selects two of them at random and bets on them . All the horses have equal chances of winning . The probability that the man selected the winning horse is
  1. $ \displaystyle \frac{3}{5}$
  2. $ \displaystyle \frac{1}{5}$
  3. $ \displaystyle \frac{2}{5}$
  4. $ \displaystyle \frac{4}{25}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total ways to select 2 horses from 5 is 5C2 = 10. The number of ways to select the winning horse is 4C1 = 4 (the winner plus one of the 4 losers). The probability is 4/10 = 2/5.

Multiple choice

$6$ ordinary dice are rolled. The probability that at least half of them will show at least $3$ is
  1. $\displaystyle 41 \times \frac {2^4}{3^6}$
  2. $\displaystyle \frac {2^4}{3^6}$
  3. $\displaystyle 20 \times \frac {2^4}{3^6}$

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

Probability of showing at least 3 is 4/6 = 2/3. Let p = 2/3, q = 1/3. We need at least 3 successes in 6 trials (Binomial distribution). P(X >= 3) = sum from k=3 to 6 of 6Ck * (2/3)^k * (1/3)^(6-k).

Multiple choice

Four tickets marked $00,01,10$ and $11$ respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the tickets is $15$, is
  1. $\displaystyle \frac { 3 }{ 1024 } $
  2. $\displaystyle \frac { 5 }{ 1024 } $
  3. $\displaystyle \frac { 7 }{ 1024 } $

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

Each ticket is a binary number: 0, 1, 2, 3. We draw 5 times with replacement. Total outcomes = 4^5 = 1024. We want the sum of 5 draws to be 15. This is equivalent to finding the number of integer solutions to x1+x2+x3+x4+x5 = 15 where 0 <= xi <= 3. Using generating functions, the coefficient of x^15 in (1+x+x^2+x^3)^5 is 5.

Multiple choice

A die is thrown $ 2n $ times. The probability that the number greater than 4 appears at least once in $ 2n $ throws is
  1. $ \displaystyle \left ( \frac{1}{3} \right )^{2n} $
  2. $ 1-\displaystyle \left ( \frac{1}{3} \right )^{2n} $
  3. $ \displaystyle \frac{3^{2n}-\:2^{2n}}{3^{2n}} $

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

The probability of getting a number greater than 4 (i.e., 5 or 6) in one throw is 2/6 = 1/3. The probability of not getting a number greater than 4 is 1 - 1/3 = 2/3. In 2n throws, the probability of never getting a number greater than 4 is (2/3)^(2n). Thus, the probability of getting at least one is 1 - (2/3)^(2n) = 1 - (2^(2n) / 3^(2n)) = (3^(2n) - 2^(2n)) / 3^(2n).

Multiple choice

Three six faced dice are thrown together. The probability that sum of the numbers appearing on the dice is equal to $ \displaystyle k\left ( 3\leq k\leq 4\right ) $ is
  1. $ \displaystyle \frac{k^{2}}{108} $
  2. $ \displaystyle \frac{k\left ( k-1 \right )}{216} $
  3. $ \displaystyle \frac{\left ( k-1 \right )\left ( k-2 \right )}{2\times 6^{3}} $

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

For three dice, the total outcomes are 6^3 = 216. The number of ways to get a sum k is the coefficient of x^k in (x + x^2 + x^3 + x^4 + x^5 + x^6)^3. For k=3, ways = 1. For k=4, ways = 3. The formula (k-1)(k-2)/2 gives 1 for k=3 and 3 for k=4. Thus, the probability is (k-1)(k-2) / (2 * 6^3).

Multiple choice

A box contains 10 green, 20 red, 30 Black and 40 orange marbles. One marble is drawn from the box at random. The probability that marble is neither green nor orange is
  1. $ \displaystyle \frac{1}{2} $
  2. $ \displaystyle \frac{7}{10} $
  3. $ \displaystyle \frac{3}{5} $

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

Total marbles = 10 + 20 + 30 + 40 = 100. Marbles that are neither green nor orange are red and black. Total red and black = 20 + 30 = 50. Probability = 50 / 100 = 1/2.

Multiple choice

If $ \displaystyle n $ integers taken at random are multiplied together, the chance (probability) that the last digit of the product is 2,4,6 or 8 is
  1. $ \displaystyle \dfrac{5^{n}-4^{n}}{10^{n}} $
  2. $ \displaystyle \dfrac{2^{n}}{5^{n}} $
  3. $ \displaystyle \dfrac{4^{n}-2^{n}}{5^{n}} $

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

The last digit of a product is 2, 4, 6, or 8 if the product is even but not a multiple of 5. The probability that a single integer is not a multiple of 5 is 4/5, and the probability that the product of n integers is not a multiple of 5 is (4/5)^n. The probability that the product is even is 1 - (1/2)^n. Combining these, the probability that the product is even and not a multiple of 5 is (4/5)^n - (2/5)^n, which simplifies to (4^n - 2^n) / 5^n.