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

If a fair coin is tossed 15 times, what is the probability of getting head as many times in the first ten throws as in the last five?
  1. Hence the required probability=$\displaystyle \frac{3003}{32768.}$
  2. Hence the required probability=$\displaystyle \frac{1001}{32768.}$
  3. Hence the required probability=$\displaystyle \frac{3003}{65536.}$
  4. Hence the required probability=$\displaystyle \frac{1001}{65536.}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let H1 be heads in first 10 tosses and H2 be heads in last 5. We want P(H1 = H2). This occurs if H1=0, H2=0; H1=1, H2=1; H1=2, H2=2; H1=3, H2=3; H1=4, H2=4; H1=5, H2=5. Summing the probabilities using combinations (10Ck * 5Ck) / 2^15 gives 3003 / 32768.

Multiple choice

A purse contains four coins; two coins having been drawn are found to be 2 Rupee coins; find the chance (i) that all coins are 2 Rupee coins (ii) that if the coins are replaced, another drawing will give a 2 Rupee coin.
  1. $\displaystyle \dfrac{2}{5}$,$\displaystyle \dfrac{7}{8}$
  2. $\displaystyle \dfrac{3}{5}$,$\displaystyle \dfrac{1}{8}$
  3. $\displaystyle \dfrac{3}{5}$,$\displaystyle \dfrac{7}{8}$
  4. $\displaystyle \dfrac{2}{5}$,$\displaystyle \dfrac{1}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice

In a multiple choise question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if be ticks all the correct answers.The candidate decides to tick answers at random. If he is allowed upto three chances to answer the question, find the probability that he will get marks in the question.
  1. $\displaystyle \dfrac{4}{5}$
  2. $\displaystyle \dfrac{1}{5}$
  3. $\displaystyle \dfrac{2}{5}$
  4. $\displaystyle \dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

A die is thrown three times, and sum of three numbers thrown is 15. Find the chance that the first throw was a four.
  1. $\displaystyle \dfrac{1}{2}.$
  2. $\displaystyle \dfrac{1}{5}.$
  3. $\displaystyle \dfrac{1}{15}.$
  4. $\displaystyle \dfrac{1}{13}.$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total outcomes for sum 15 with 3 dice: (6,6,3)x3, (6,5,4)x6, (5,5,5)x1 = 10 outcomes. Outcomes where first die is 4: (4,5,6), (4,6,5) = 2 outcomes. Probability = 2/10 = 1/5.

Multiple choice

In shuffling a pack of cards; four are accidently dropped; find the chance that the missing cards should be one from each suit.
  1. $\displaystyle 1-\frac{\left ( 13 \right )^{4}}{^{52}C_{4}}$
  2. $\displaystyle \frac{\left ( 13 \right )^{4}}{^{52}C_{4}}$
  3. $\displaystyle \frac{\left ( 12 \right )^{4}}{^{52}C_{4}}$

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

The total number of ways to choose 4 cards from 52 is 52C4. The number of ways to choose one card from each of the 4 suits is 13 * 13 * 13 * 13 = 13^4.

Multiple choice

The probability that in the toss of two dice we obtain an even sum or a sum less than 5 is
  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle \frac{1}{6}$
  3. $\displaystyle \frac{2}{3}$
  4. $\displaystyle \frac{5}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes for two dice are 36. Even sums are 2, 4, 6, 8, 10, 12 (18 outcomes). Sums less than 5 are 2, 3, 4. Combining these sets (avoiding double counting), we have sums 2, 3, 4, 6, 8, 10, 12. Counting these: 2(1), 3(2), 4(3), 6(5), 8(5), 10(3), 12(1) = 20 outcomes. 20/36 = 5/9.

Multiple choice

Six faces of an unbiased die are numbered with $2, 3, 5, 7, 11$ and $13$. If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is
  1. $\dfrac5{18}$
  2. $\dfrac5{36}$
  3. $\dfrac{13}{18}$
  4. $\dfrac{25}{36}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The faces are {2, 3, 5, 7, 11, 13}. There are 5 odd numbers and 1 even number. Sum is odd if one die is even and the other is odd. Probability = (P(E)*P(O) + P(O)*P(E)) = (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36 = 5/18.

Multiple choice

Two dice are thrown. The probability of getting a sum of less than $7$ points or a double is :
  1. $\frac{1}{6}$
  2. $\frac{1}{3}$
  3. $\frac{2}{3}$
  4. $\frac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Total outcomes = 36. Sum < 7: (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1) = 15 outcomes. Doubles: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. Union: 15 + 6 - 3 (overlaps: 1,1; 2,2; 3,3) = 18 outcomes. Probability = 18/36 = 1/2.

Multiple choice

A bag contains three tickets numbered $1, 2$ and $3$. A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is
  1. $\dfrac {40}{81}$
  2. $\dfrac {41}{81}$
  3. $\dfrac {14}{27}$
  4. $\dfrac {13}{81}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let p be the probability of drawing an even number (2) in one trial, p = 1/3. Let q be the probability of drawing an odd number (1 or 3), q = 2/3. The sum is even if we draw an even number of odd-numbered tickets (0, 2, or 4). Using the binomial expansion (p+q)^4 and (p-q)^4, the sum of even terms is 1/2 * ((p+q)^4 + (p-q)^4). Here p=1/3, q=2/3, so 1/2 * (1^4 + (-1/3)^4) = 1/2 * (1 + 1/81) = 1/2 * (82/81) = 41/81.