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 unbiased dice are drawn. Find the probability that neither a doublet nor a total of $8$ will appear.
  1. $\dfrac {15}{18}$
  2. $\dfrac {11}{18}$
  3. $\dfrac {13}{15}$
  4. $\dfrac {13}{18}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes for two dice are 36. Doublets are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) [6 outcomes]. Sum of 8 outcomes are (2,6), (3,5), (4,4), (5,3), (6,2) [5 outcomes]. Since (4,4) is counted in both, total unique outcomes to exclude are 6 + 5 - 1 = 10. Probability of neither is 1 - (10/36) = 26/36 = 13/18.

Multiple choice

A box contains $6$ red marbles numbered $1$ through $6$ and $4$ white marbles numbered from $12$ through $15$ . Find the probability that a marble drawn is Even numbered
  1. $\dfrac {1}{3}$
  2. $\dfrac {1}{4}$
  3. $\dfrac {1}{2}$
  4. $\dfrac {3}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are 6 red marbles (1-6) and 4 white marbles (12-15). The even numbers are {2, 4, 6} from the red set and {12, 14} from the white set, totaling 5 even marbles. The total number of marbles is 10, so the probability is 5/10 = 1/2.

Multiple choice

A box contains $10$ red marbles, $20$ blue marbles and $30$ green marbles, $5$ marbles are drawn at random. From the box, what is the probability that: At least one is green
  1. $\dfrac {4367}{4484}$
  2. $\dfrac {4671}{4484}$
  3. $\dfrac {467}{80004}$
  4. $\dfrac {437}{44684}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total marbles = 60. Probability of drawing no green marbles is C(30, 5) / C(60, 5). Probability of at least one green is 1 - (C(30, 5) / C(60, 5)). Calculating this yields 4367/4484.

Multiple choice

A box contains $10$ red marbles, $20$ blue marbles and $30$ green marbles, $5$ marbles are drawn at random from the box, what is the probability that a ll are blue?
  1. $\dfrac {34}{11977}$
  2. $\dfrac {34}{11966}$
  3. $\dfrac {68}{11977}$
  4. $\dfrac {68}{11966}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total marbles = 10 + 20 + 30 = 60. Probability of drawing 5 blue marbles = (20C5) / (60C5). 20C5 = (20*19*18*17*16) / (5*4*3*2*1) = 15504. 60C5 = (60*59*58*57*56) / (5*4*3*2*1) = 5461512. Ratio = 15504 / 5461512 = 34 / 11977.

Multiple choice

A bag contain tickets numbered from $1$ to $20$. Two tickets are drawn. Find the probability that On one there is a prime number and on the other there is a multiple of $4$.
  1. $\dfrac {4}{19}$
  2. $\dfrac {3}{19}$
  3. $\dfrac {5}{19}$

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

Total ways to choose 2 tickets from 20 is 20C2 = 190. Primes are {2, 3, 5, 7, 11, 13, 17, 19} (8 total). Multiples of 4 are {4, 8, 12, 16, 20} (5 total). Ways to pick one prime and one multiple of 4 = 8 * 5 = 40. Probability = 40/190 = 4/19.

Multiple choice

A fair coin is tossed $6$ times. What is the probability of getting at least $3$ heads?
  1. $\cfrac{11}{16}$
  2. $\cfrac{21}{32}$
  3. $\cfrac{1}{18}$
  4. $\cfrac{3}{64}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The probability of getting at least 3 heads in 6 tosses is calculated using the binomial distribution formula. The total outcomes are 2^6 = 64. The number of ways to get 3, 4, 5, or 6 heads is 20 + 15 + 6 + 1 = 42. Thus, the probability is 42/64 = 21/32.

Multiple choice

One lottery ticket is drawn at random from a bag containing $20$ tickets numbered from $1$ to $20$. Find the probability that the number drawn is either even or square of an integer:
  1. $\dfrac14$
  2. $\dfrac45$
  3. $\dfrac35$
  4. $\dfrac15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total number of possible outcomes is 20. The even numbers in this range are {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} (10 numbers), and the perfect squares are {1, 4, 9, 16}. Combining these two sets without duplication gives {1, 2, 4, 6, 8, 9, 10, 12, 14, 16, 18, 20}, which contains 12 numbers. Thus, the probability is 12/20 = 3/5.