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 dice are thrown. The probability of scoring a sum greater than $9$ or a doublet is
  1. $\dfrac{1}{6}$
  2. $\dfrac{1}{18}$
  3. $\dfrac{5}{18}$
  4. $\dfrac{2}{18}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 36. Sum > 9: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6) = 6 outcomes. Doublets: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. Union: { (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), (1,1), (2,2), (3,3), (4,4) } = 10 outcomes. 10/36 = 5/18.

Multiple choice

Six symmetrical dice are thrown simultaneously. The probability of having different points on them is
  1. $\displaystyle \frac{1}{6}$
  2. $\displaystyle \frac{1}{36}$
  3. $\displaystyle \frac {6!}{6^6}$
  4. $\displaystyle \frac{1}{18}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total number of outcomes when throwing 6 dice is 6^6. The number of ways to have different points on all six dice is the number of permutations of 6 distinct values, which is 6!. Thus, the probability is 6! / 6^6.

Multiple choice

From a bag containing $9$ distinct white and $9$ distinct black balls, $9$ balls are drawn at random one by one, the drawn balls being replaced each time. The probability that atleast four balls of each colour is in the draw is
  1. a little less than $\dfrac {1}{2}$
  2. a little greater than $\dfrac {1}{2}$
  3. $\dfrac {1}{2}$
  4. $\dfrac {1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

With 9 white and 9 black balls, the probability of drawing a white ball is 1/2. Drawing 9 balls with replacement follows a binomial distribution B(9, 0.5). The probability of at least four of each color is 1 - P(0, 1, 2, or 3 of one color). Due to symmetry, this is 1 - 2 * (P(0) + P(1) + P(2) + P(3)). Calculating this gives 1 - 2 * (1/512 + 9/512 + 36/512 + 84/512) = 1 - 2 * (130/512) = 1 - 260/512 = 252/512, which is slightly less than 1/2.

Multiple choice

An urn contains $25$ balls numbered $1$ through $25$. Suppose getting an odd number is a success. Two balls are drawn from the urn with replacement. The probability of getting at least one success is
  1. $\dfrac{481}{625}$
  2. $\dfrac{312}{625}$
  3. $\dfrac{144}{625}$
  4. $\dfrac{244}{625}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Numbers 1-25. Odd numbers: 1, 3, ..., 25 (13 numbers). P(success) = 13/25. P(failure) = 12/25. P(at least one success) = 1 - P(no success) = 1 - (12/25)^2 = 1 - 144/625 = 481/625.

Multiple choice

On a symmetrical die the numbers 1, -1, 2, -2, 3 and 0 are marked on its 6 faces. If such a die is thrown 3 times, the probability that the sum of points on them is 6 is
  1. $\dfrac{5}{27}$
  2. $\dfrac{5}{54}$
  3. $\dfrac{5}{108}$
  4. $\dfrac{3}{108}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 6^3 = 216. We need the sum of 3 faces to be 6. Possible combinations (order matters): (3,3,0)x3, (3,2,1)x6, (3,2,-1)x6, (2,2,2)x1, (2,2,0)x3, (2,1,1)x3, (3,1,2) is covered. Summing valid permutations leads to 10/216 = 5/108.

Multiple choice

$A$ and $B$ toss 3 coins. The probability that both obtain same number of tails is $p$ and the probability that both obtain same number of heads is $q$, then the value of $p+q$ is
  1. $\dfrac{5}{16}$
  2. $\dfrac{10}{16}$
  3. $\dfrac{12}{16}$
  4. $\dfrac{7}{16}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 3 coins, the total outcomes are 2^3 = 8. The probability of getting k tails is C(3,k)/8. Let T be number of tails and H be number of heads. P(same tails) = sum(P(A=k) * P(B=k)) = sum((C(3,k)/8)^2) = (1/64 + 9/64 + 9/64 + 1/64) = 20/64 = 5/16. Similarly, P(same heads) = 5/16. Total p+q = 10/16.

Multiple choice

A coin tossed $n$ times. If the probability that $4, 5, 6$ heads occur are in $A.P.$, then $n =$
  1. $14$
  2. $8$
  3. $15$
  4. $11$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probabilities of k heads: P(k) = C(n, k) * (1/2)^n. If P(4), P(5), P(6) are in AP, then 2 * P(5) = P(4) + P(6). This simplifies to 2 * C(n, 5) = C(n, 4) + C(n, 6). Using the formula C(n, k) = n! / (k!(n-k)!), we solve for n: 2 * (n! / (5!(n-5)!)) = n! / (4!(n-4)!) + n! / (6!(n-6)!). Dividing by n! and simplifying leads to n = 14.

Multiple choice

A card is drawn from a pack. The card is replaced and the pack is reshuffled. If this is done six times, the probability that $2$ herts, $2$ diamonds and $2$ black cards are drawn is
  1. $90.{\left( {\dfrac{1}{4}} \right)^6}$
  2. $\dfrac{{45}}{2}.{\left( {\dfrac{3}{4}} \right)^4}$
  3. $\dfrac{{90}}{{{2^{10}}}}$

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

This is a multinomial distribution problem. Probability of heart = 1/4, diamond = 1/4, black = 1/2. We want 2 of each in 6 trials. P = (6! / (2! 2! 2!)) * (1/4)^2 * (1/4)^2 * (1/2)^2 = 90 * (1/16) * (1/16) * (1/4) = 90 / 1024 = 45 / 512 = 90 / 2^10.

Multiple choice

A coin is tossed twice. Events E and F are defined as follows :E=heads on first toss, F= heads on second toss.Find the probability of $\displaystyle E\cup F.$
  1. Hence p $\displaystyle(E\cup F) =\frac{1}{4}$
  2. Hence p $\displaystyle(E\cup F) =\frac{3}{4}$
  3. Hence p $\displaystyle(E\cup F) =\frac{1}{2}$

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

P(E) = 1/2, P(F) = 1/2. Since events are independent, P(E intersect F) = 1/4. P(E union F) = P(E) + P(F) - P(E intersect F) = 1/2 + 1/2 - 1/4 = 3/4.

Multiple choice

There are $3$ bags each containing $5$ white balls and $2$ black balls and $2$ bags each containing $1$ white balls and $4$ black balls, a black ball having been drawn, find the chance that it came from the first group.
  1. $\displaystyle \frac{28}{43}$
  2. $\displaystyle \frac{15}{43}$
  3. $\displaystyle \frac{15}{28}$
  4. $\displaystyle \frac{13}{28}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let E be the event of drawing a black ball. P(E) = P(E|Group1)P(Group1) + P(E|Group2)P(Group2). Group 1: 3 bags, 5W 2B. P(E|G1) = 2/7. P(G1) = 3/5. Group 2: 2 bags, 1W 4B. P(E|G2) = 4/5. P(G2) = 2/5. P(E) = (2/7 * 3/5) + (4/5 * 2/5) = 6/35 + 8/25 = (30+56)/175 = 86/175. P(G1|E) = (6/35) / (86/175) = (6/35) * (175/86) = 30/86 = 15/43.