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

Counters numbered $1,2,3$ are placed in a bag and one is drawn at random and replaced. The operation is being repeated three times. The probabolity of obtaining a total of $6$ is ?
  1. $\dfrac {7}{9}$
  2. $\dfrac {6}{27}$
  3. $\dfrac {7}{27}$
  4. $\dfrac {5}{27}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 3 * 3 * 3 = 27. Favorable outcomes for sum 6: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1), (2,2,2). There are 7 such outcomes. Probability = 7 / 27.

Multiple choice

A card is lost from a pack of $52$ playing cards. From the remainder of the pack, two cards are drawn and are found to be spade. The probability that the missing card is a spade is :
  1. $ \dfrac{11}{50} $
  2. $ \dfrac{13}{50} $
  3. $ \dfrac{17}{50} $
  4. $ \dfrac{23}{50} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let E be the event that two drawn cards are spades. Let M_s be the event that the missing card is a spade. P(M_s | E) = P(E | M_s) * P(M_s) / P(E). P(M_s) = 13/52 = 1/4. P(E | M_s) = (12/51) * (11/50). P(E | not M_s) = (13/51) * (12/50). Using Bayes' theorem, the result is 11/50.

Multiple choice

Two bags contains respectively 3 white and 2 red balls and 2 white and 4 red balls. One ball is drawn at random from the first bag and put it into the second; then a ball is drawn from the second is white is
  1. $\frac{13}{35}$
  2. $\frac{13}{25}$
  3. $\frac{12}{35}$
  4. $\frac{12}{25}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Bag 1: 3W, 2R (Total 5). Bag 2: 2W, 4R (Total 6). Case 1: White drawn from B1 (3/5), B2 becomes 3W, 4R (Total 7). Prob(W from B2) = 3/7. Case 2: Red drawn from B1 (2/5), B2 becomes 2W, 5R (Total 7). Prob(W from B2) = 2/7. Total prob = (3/5 * 3/7) + (2/5 * 2/7) = 9/35 + 4/35 = 13/35.

Multiple choice

A bag contains $3$ white, $3$ black, and $2$ red balls. One by one three balls are drawn without replacing them, then find the probability that the third ball is red.
  1. $ \displaystyle\frac{1}{4}$
  2. $\displaystyle \frac{3}{4}$
  3. $\displaystyle \frac{1}{3}$

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

In sampling without replacement, the probability of any specific ball appearing at any specific position is equal to the initial proportion of that ball in the bag. Total balls = 8. Red balls = 2. Probability = 2/8 = 1/4.

Multiple choice

An unbiased normal coin is tossed n times. Let $E_1$: event that both heads and tails are present in n tosses. $E_2$:event that the coin shows up heads at most once. The value of n for which $E_1$ and $E_2$ are independent is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

E1: At least one H and one T. P(E1) = 1 - (P(all H) + P(all T)) = 1 - (1/2^n + 1/2^n) = 1 - 2/2^n = 1 - 1/2^(n-1). E2: At most one H. P(E2) = P(0 H) + P(1 H) = (1 + n)/2^n. Independence requires P(E1 intersect E2) = P(E1) * P(E2). For n=3, P(E1)=6/8=3/4, P(E2)=4/8=1/2. P(E1 intersect E2) = P(exactly 1 H) = 3/8. Since 3/4 * 1/2 = 3/8, n=3 is correct.

Multiple choice

A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement til all the black balls are drawn. The probability that the test will come to an end at the 7th draw is
  1. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{6}}$
  2. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{6}}\times \frac{2}{7}$
  3. $\frac{^{10}c_{4}\times ^{3}c_{2}}{^{13}c_{7}}$
  4. $\frac{^{10}c_{5}\times ^{3}c_{2}}{^{13}c_{7}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To end at the 7th draw, the 3rd black ball must be drawn on the 7th attempt. This means in the first 6 draws, exactly 2 black balls and 4 white balls were drawn. The probability is (ways to pick 2 black from 3 AND 4 white from 10) / (total ways to pick 6 from 13) * (probability that the 7th is the remaining black ball). The formula provided in A correctly represents the selection of the first 6 balls.

Multiple choice

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :
  1. $\dfrac { 150 } { 6 ^ { 5 } }$
  2. $\dfrac { 175 } { 6 ^ { 5 } }$
  3. $\dfrac { 200 } { 6 ^ { 5 } }$
  4. $\dfrac { 225 } { 6 ^ { 5 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To end on the 5th throw, the 5th throw must be a 4, and the 4th throw must be a 4. The first 3 throws must not contain the sequence '44'. Total outcomes = 6^5. Favorable outcomes: (not 4, not 4, not 4, 4, 4) + (4, not 4, not 4, 4, 4) + (not 4, 4, not 4, 4, 4). Calculating these sequences: 5*5*5 + 1*5*5 + 5*1*5 = 125 + 25 + 25 = 175.

Multiple choice

The face cards are removed from a full pack. Out of the remaining $40$ cards, $4$ are drawn at random. What is the probability that they belong to different suits?
  1. $\dfrac {1000}{9139}$
  2. $\dfrac {1001}{9136}$
  3. $\dfrac {100}{913}$
  4. $\dfrac {1002}{9129}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total cards = 40. Choosing 4 cards = 40C4 = 91390. Choosing one from each suit = 10C1 * 10C1 * 10C1 * 10C1 = 10000. Probability = 10000 / 91390 = 1000 / 9139.