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

2K coins (K is an integer) each with probability P(O
  1. $\dfrac{K}{2K+1}$
  2. $\dfrac{K+1}{2K}$
  3. $\dfrac{K+1}{2K+1}$
  4. $\dfrac{2K}{K+1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For 2K tosses, equate the probabilities of K heads and K+1 heads. Cancelling the common binomial factor gives (1-P)/P = (K+1)/K. Solving yields P = (K+1)/(2K+1).

Multiple choice

A bag contain $(n + 1)$ coins. It is known that one of these has a head on both sides whereas the other coins are normal. One of these coins is selected at random & tossed. If the probability that the toss results in head is $\dfrac{7}{12}$, then the value of $n$ is
  1. $5$
  2. $6$
  3. $4$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the number of double-headed coins be 1 and normal coins be n. Total coins = n + 1. Probability of picking a double-headed coin is 1/(n+1), and a normal coin is n/(n+1). Probability of head = (1 * 1/(n+1)) + (0.5 * n/(n+1)) = 7/12. Solving (1 + 0.5n) / (n+1) = 7/12 gives 12 + 6n = 7n + 7, so n = 5.

Multiple choice

In a purse there are 10 coins, all 5 paise except one which is a rupee. In another purse, there are 10 coins, all 5 paise 9 coins are taken out from the former purse & put into the latter. Then 9 coins are taken out from the latter & put into the former. Then the chance that the rupee is still in the first purse is
  1. $\dfrac{9}{19}$
  2. $\dfrac{10}{19}$
  3. $\dfrac{4}{9}$

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

This is a classic probability problem. After the transfers, the probability that the rupee coin remains in the first purse is 10/19.

Multiple choice

Consider a bag containing $3$ normal and $6$ doubly headed coins. A coin is selected randomly and tossed $3$ times and found to fall head-wise on all the $3$ occasions. Probability that the selected coin was a normal coin is $\dfrac {1}{17}$.

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Assertion is incorrect but Reason is correct
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

There are ninety cards in a box numbered $10,11,12,\cdots,98, 99$. Three cards are drawn from the box one by one with replacement. The probability that product of the digits on the cards will be $12$ at least once is
  1. $1-\left(\displaystyle \frac{4}{5}\right)^{3}$
  2. $\left(\displaystyle \frac{43}{45}\right)^{3}$
  3. $1-\left(\displaystyle \frac{43}{45}\right)^{3}$
  4. $\displaystyle \frac{43}{45}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There are 90 cards (10-99). The product of digits is 12 for pairs like (2,6), (6,2), (3,4), (4,3). There are 4 such numbers out of 90. Probability of getting 12 in one draw is 4/90 = 2/45. Probability of NOT getting 12 is 1 - 2/45 = 43/45. Probability of not getting 12 in 3 draws is (43/45)^3. Probability of at least one is 1 - (43/45)^3.

Multiple choice

A fair coin is tossed $2n$ times. The probability of getting as many heads in the first $n$ tosses as in the last $n$ is
  1. $\displaystyle \frac{2n_{C_{n}}}{2^{2n}}$
  2. $\displaystyle \frac{2n_{C_{n-1}}}{2^{n}}$
  3. $\displaystyle \frac{n}{2^{n}}$
  4. $\displaystyle \frac{n^{2}}{2^{n}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total number of outcomes is 2^2n. The number of ways to have n heads in the first n tosses and n heads in the last n tosses is (nCn) * (nCn) = 1. However, the question implies having an equal number of heads in both halves, which is equivalent to choosing n positions out of 2n total positions for the heads, represented by 2nCn.

Multiple choice

5 cards are drawn at random from a well shuffled pack of 52 playing cards. If it is known that there will be at least 3 hearts, the probability that there are 4 hearts is
  1. $\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}+^{13}C_{4}+^{13}C_{5}}$
  2. $\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{39}C_{2}+^{13}c_{4}\times ^{39}C_{1}+^{13}C_{5}}$
  3. $\displaystyle \frac{^{13}C_{4}\times ^{39}C_{1}}{^{13}C_{3}\times ^{39}C_{2}+^{13}C_{4}\times ^{39}C_{1}+^{13}C_{5}}$
  4. $\displaystyle \frac{^{13}C_{4}}{^{13}C_{3}\times ^{13}C_{4}\times ^{13}C_{5}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is a conditional probability problem. Total ways to get at least 3 hearts is (13C3 * 39C2) + (13C4 * 39C1) + (13C5 * 39C0). The favorable outcome is getting exactly 4 hearts: (13C4 * 39C1). The ratio is the correct probability.

Multiple choice

A bag contains $n$ white, $n$ black balls. Pair of balls are drawn without replacement until the bag is empty. Then the probability that each pair consists of one white and one black ball is
  1. $\displaystyle \frac{n!}{(2n!)}$
  2. $\displaystyle \frac{(n!)^{2}}{(2n)!}$
  3. $\displaystyle \frac{n!.2^{2}}{(2n)!}$
  4. $\displaystyle \frac{(n!)^{2}2^{n}}{(2n)!}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total ways to arrange 2n balls is (2n)!. Ways to pick pairs such that each has one white and one black: there are n! ways to pair white balls with black balls, and 2^n ways to order the balls within each pair. The formula is (n! * 2^n) / (2n!/2^n) which simplifies to (n!)^2 * 2^n / (2n)!. This matches option D.

Multiple choice

A die is loaded so that the probability of face $i$ is proportional to $i,i=1,2,3,...,6$. The probability of an even number occurring when the die is rolled, is
  1. $\displaystyle \frac{2}{7}$
  2. $\displaystyle \frac{4}{7}$
  3. $\displaystyle \frac{3}{7}$

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

The probability of face i is k*i. Sum of probabilities = k(1+2+3+4+5+6) = 21k = 1, so k = 1/21. Even numbers are 2, 4, 6. Their probabilities are 2/21, 4/21, and 6/21. Sum = (2+4+6)/21 = 12/21 = 4/7.

Multiple choice

A bag contains 12 white balls and 3 black balls .Balls are drawn one by one without replacement till all the black balls are drawn . The probability that the procedure will come to an end at the $7^{th}$ draw is
  1. $\dfrac{27}{91}$
  2. $\dfrac{9}{91}$
  3. $\dfrac{3}{91}$

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

For the process to end at the 7th draw, the 7th draw must be the 3rd black ball, and exactly 2 black balls must have been drawn in the first 6 draws. Ways = (6C2 * 3C2 * 12C10) / 15C7. Calculation: (15 * 3 * 66) / 6435 = 2970 / 6435 = 3/91.