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

Three faces of an ordinary dice are yellow, two faces are red and one face is blue. The dice is tossed $3$ times. The probability that yellow, red and blue faces appear in the first, second and third tosses respectively is
  1. $\displaystyle \frac {1}{36}$
  2. $\displaystyle \frac {1}{6}$
  3. $\displaystyle \frac {1}{30}$

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

P(Yellow) = 3/6 = 1/2. P(Red) = 2/6 = 1/3. P(Blue) = 1/6. The probability of the specific sequence (Yellow, Red, Blue) is (1/2) * (1/3) * (1/6) = 1/36.

Multiple choice

$A$ and $B$ draw two cards each, one after another, from a pack of well-shuffled pack of $52$ cards. The probability that all the four cards drawn are of the same suit is
  1. $\displaystyle \frac {44}{85 \times 49}$
  2. $\displaystyle \frac {11}{85 \times 49}$
  3. $\displaystyle \frac {13 \times 24}{17 \times 25 \times 49}$

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

Total ways to draw 4 cards = 52C4. Ways to draw 4 cards of same suit = 4 * 13C4. Probability = (4 * 13C4) / 52C4 = (4 * 13*12*11*10) / (52*51*50*49) = 44 / (17 * 5 * 49) = 44 / 4165. Simplifying 44 / (85 * 49) = 44 / 4165. Correct.

Multiple choice

One ticket is drawn at random from a bag containing 24 tickets numbered 1 to 24. Represent the sample space and the event of drawing a ticket containing number which is a prime also find the number of elements in them.
  1. $n(S)=24$ and $n(E)=8$
  2. $n(S)=24$ and $n(E)=7$
  3. $n(S)=24$ and $n(E)=9$
  4. $n(S)=24$ and $n(E)=10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Sample space S = {1, 2, ..., 24}, so n(S) = 24. Prime numbers between 1 and 24: {2, 3, 5, 7, 11, 13, 17, 19, 23}. Counting these, there are 9 primes. Thus, n(E) = 9.

Multiple choice

If a leap year is selected at random, what is the chance it will contain 53 Sunday?
  1. probability is $\cfrac{4}{7}$
  2. probability is $\cfrac{3}{7}$
  3. probability is $\cfrac{2}{7}$

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

A leap year has 366 days, which is 52 weeks and 2 days. The extra 2 days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), or (Sat, Sun). Out of these 7 possibilities, 2 contain a Sunday.

Multiple choice

A bag contains $12$ tickets numbered $1,2,3,...,12$.Five of them are drawn at random and arranged in the ascending order of their numbers . What is the probability that the third in the order is $5$?
  1. $\displaystyle \frac{7}{44}$
  2. $\displaystyle \frac{5}{12}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total number of ways to choose 5 tickets out of 12 is 12C5 = 792. For the third ticket to be 5, we must choose 2 tickets from the 4 smaller than 5 (1, 2, 3, 4) and 2 tickets from the 7 larger than 5 (6, 7, 8, 9, 10, 11, 12). The number of ways is 4C2 * 7C2 = 6 * 21 = 126. The probability is 126/792 = 7/44.

Multiple choice

Two dice each with faces marked $1,2,3,4,5,6$ are thrown simultaneously. Find the probability of the following events: (i) The sum of the numbers coming up $7$. (ii) The sum of the numbers coming up is a multiple of $4$.
  1. $(i)\displaystyle \frac{1}{6},(ii)\frac { 1 }{ 4 } $
  2. $(i)\displaystyle \frac{1}{3},(ii)\frac { 2 }{ 9 } $
  3. $(i)\displaystyle \frac{5}{6},(ii)\frac { 1 }{ 9 } $

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

(i) Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes. Prob = 6/36 = 1/6. (ii) Sum is multiple of 4 (4, 8, 12): (1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6) = 9 outcomes. Prob = 9/36 = 1/4.

Multiple choice

A man throws a die and tosses a coin alternately, starting with the coin. Find the probability that he gets head before $5$ or $ 6$ on the die.
  1. $\displaystyle\frac{3}{4}$
  2. $\displaystyle\frac{3}{7}$
  3. $\displaystyle\frac{1}{2}$

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

The probability of getting a head is 1/2. The probability of getting 5 or 6 on a die is 2/6 = 1/3. The man wins if he gets a head (H) or if he fails to get 5/6 on the die (D') and then gets a head on the next turn. The probability is P(H) + P(D')P(T)*P(H) + ... which is a geometric series: (1/2) + (2/3)(1/2)(1/2) + (2/3)(1/2)(2/3)(1/2)(1/2) + ... = (1/2) / (1 - (2/3)(1/2)) = (1/2) / (2/3) = 3/4.

Multiple choice

A die is thrown $7$ times. What is the chance that an odd number turns up ? (i) exactly $4$ times (ii) at least $4$ times.
  1. (i)$\displaystyle \frac{35}{128}$, (ii)$\displaystyle \frac{1}{2}$
  2. (i)$\displaystyle \frac{45}{128}$, (ii)$\displaystyle \frac{1}{2}$
  3. (i)$\displaystyle \frac{35}{128}$, (ii)$\displaystyle \frac{1}{4}$

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

Probability of odd number (p) = 1/2. (i) P(X=4) = 7C4 * (1/2)^4 * (1/2)^3 = 35 * (1/128) = 35/128. (ii) P(X>=4) = P(4)+P(5)+P(6)+P(7) = (35+21+7+1)/128 = 64/128 = 1/2.

Multiple choice

Three identical dice are thrown. (i) What is the probability of getting a total of $15$? (ii) What is the probability that an odd number is obtained on each die given that the sum obtained is $15$ ?
  1. $\displaystyle \frac{5}{108}$, $\displaystyle \frac{1}{10}$
  2. $\displaystyle \frac{5}{108}$, $\displaystyle \frac{3}{10}$
  3. $\displaystyle \frac{5}{54}$, $\displaystyle \frac{3}{10}$

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

Total outcomes = 6^3 = 216. Sum of 15: (6,6,3)x3, (6,5,4)x6, (5,5,5)x1 = 10 outcomes. Prob = 10/216 = 5/108. For (ii), odd numbers summing to 15: (5,5,5) is 1 way. Total odd-sum combinations: (5,5,5) is the only one. Prob = 1/10.

Multiple choice

A bag has $13$ red, $14$ green and $15$ white balls. ${ p }{ 1 }$ is the probability of drawing exactly $2$ white balls when four balls are drawn. Then the number of balls of each color are doubled. Let ${ p }{ 2 }$ be the probability of drawing $4$ balls when $8$ balls are drawn, then
  1. ${ p }_{ 1 }={ p }_{ 2 }$
  2. ${ p }_{ 1 }>{ p }_{ 2 }$
  3. ${ p }_{ 1 }<{ p }_{ 2 }$

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

p1 is the probability of 2 white balls from 42 total (13R, 14G, 15W). p2 is the probability of 4 white balls from 84 total (26R, 28G, 30W) when 8 are drawn. This is a comparison of hypergeometric distributions. As the population size increases while maintaining proportions, the probability distribution changes. Calculation confirms p1 > p2.

Multiple choice

A bag contains three white, two black and four red balls. If four balls are drawn at random with replacement, the probability that the sample contains just one white ball is
  1. $\displaystyle \frac { 32 }{ 81 } $
  2. $\displaystyle \frac { 11 }{ 2 } $
  3. $\displaystyle \frac { 32 }{ 93 } $

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

Total balls = 9. P(White) = 3/9 = 1/3. P(Not White) = 2/3. For 4 draws with replacement, P(1 white) = 4C1 * (1/3)^1 * (2/3)^3 = 4 * (1/3) * (8/27) = 32/81.