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

A special dice is so constructed that the probabilities of throwing $1,2,3,4,5$ and $6$ are $\dfrac{1-k}{6},\dfrac{1+2k}{6},\dfrac{1-k}{6},\dfrac{1+k}{6},\dfrac{1-2k}{6}$ and $\dfrac{1+k}{6}$ respectively. If two such dice are thrown and the probability of getting a sum equal to $9$ lies between $\dfrac19$ and $\dfrac29$, find the set of integral values of $k$.
  1. Set of integral values of $k={0}$
  2. Set of integral values of $k={1}$
  3. Set of integral values of $k={0,1}$
  4. Set of integral values of $k=\phi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of probabilities must be 1. Sum = (1-k + 1+2k + 1-k + 1+k + 1-2k + 1+k)/6 = 6/6 = 1. The condition 1/9 < P(sum=9) < 2/9 must be satisfied. Calculating P(sum=9) for different k values shows that only k=0 satisfies the condition.

Multiple choice

There are n cards in a pack numbered from 1 to n. A and B are two individuals who draw a card from the pack. Find the probability that the card drawn by A is higher than that drawn by B. A
  1. $\dfrac1n$
  2. $\dfrac{n-1}{2n}$
  3. $\dfrac{n+1}{2n}$
  4. $\dfrac{n-1}{n}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

There are n(n - 1) ordered outcomes in which the two cards are different. Exactly half of these have A's card higher than B's, so favorable outcomes equal n(n - 1)/2. The probability is therefore (n - 1)/(2n).

Multiple choice

An unbiased die with faces $1,2,3,4,5,6$ is tossed $4$ times out of four face values obtained the probability that minimum face value is not less than $2$ and the maximum face value is not greater than $5$ is
  1. $ \displaystyle \frac{1}{81} $
  2. $ \displaystyle \frac{80}{81} $
  3. $ \displaystyle \frac{64}{81} $
  4. $ \displaystyle \frac{16}{81} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Each toss has 6 outcomes. We want faces in {2, 3, 4, 5}. There are 4 such values. Probability for one toss is 4/6 = 2/3. For 4 tosses, the probability that all are in {2, 3, 4, 5} is (2/3)^4 = 16/81.

Multiple choice

A bag contains $n$ red and $n$ white balls. Two balls at a time are drawn at random from the bag till all the ball are drawn. Find the probability that in each draw there is one white and one red ball.
  1. $\displaystyle \dfrac{(2^{n}){n!}}{^{2n}C_{n}}$
  2. $\displaystyle \dfrac{4^{n}}{^{2n}C_{n}}$
  3. $\displaystyle \dfrac{2^{n}}{^{2n}C_{n}}$
  4. $\displaystyle \dfrac{(2^{n})/n!}{^{2n}C_{n}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total number of ways to draw 2n balls in pairs is (2n)! / (2^n). The number of ways to draw one red and one white in each of n draws is n! * n!. The probability is (n! * n!) / ( (2n)! / 2^n ) = (2^n * n! * n!) / (2n)!. This simplifies to 2^n / (2nCn).

Multiple choice

A card is drawn and replaced in an ordinary pack of playing cards. The number of times a card must be drawn so that the probability of getting atleast a club card is greater than $\displaystyle \frac{3}{4}$ is
  1. $7$
  2. $6$
  3. $5$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of not getting a club in one draw is 3/4. Probability of not getting a club in n draws is (3/4)^n. We want 1 - (3/4)^n > 3/4, so (3/4)^n < 1/4. For n=4, (3/4)^4 = 81/256 = 0.316. For n=5, (3/4)^5 = 243/1024 = 0.237. Since 0.237 < 0.25, n=5 is the smallest integer.

Multiple choice

A purse contains $2$ six-sided dice. One is a normal fair die,while the other has two $1's, $ two $ 3' s$ and two $5'$, A die is picked up and rolled. Because of some secret magnetic attration of the unfair die, there is 75% chance of picking the unfair die and a 25% chance of picking a fair die. The die is rolled and shows up the face $3$ The probability that a fair die was picked up is
  1. $1/7$
  2. $1/4$
  3. $1/6$
  4. $1/24$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

P(Fair) = 0.25, P(Unfair) = 0.75. P(3|Fair) = 1/6. P(3|Unfair) = 2/6 = 1/3. Using Bayes' Theorem: P(Fair|3) = [P(3|Fair) * P(Fair)] / [P(3|Fair) * P(Fair) + P(3|Unfair) * P(Unfair)] = [(1/6 * 1/4)] / [(1/6 * 1/4) + (1/3 * 3/4)] = (1/24) / (1/24 + 3/12) = (1/24) / (1/24 + 6/24) = (1/24) / (7/24) = 1/7.

Multiple choice

An unbiased die, with faces numbered $1, 2, 3, 4, 5, 6$ is thrown n times and the list of n numbers showing up is noted. Then the probability that, among the numbers $1,2, 3,4, 5,6$ only three numbers appear in this list ?
  1. $\displaystyle \frac{[3^{n}-3(2^{n})+3]}{6^{n}}$
  2. $\displaystyle \frac{^{6} C_{3}[3^{n}-3(2^{n})+3]}{6^{n}}$
  3. $\displaystyle \frac{^{6} C_{3}[3^{n}-3(2^{n})]}{6^{n}}$

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

This is a combinatorics problem involving the inclusion-exclusion principle for surjective mappings. The number of ways to choose 3 numbers out of 6 is 6C3. The number of ways to map n throws to exactly 3 specific numbers is 3^n - 3(2^n) + 3.