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

The chance that a person with two dices, the faces of each being numbered $1$ to $6$, will throw aces exactly $4$ times in $6$ trials is
  1. ${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }$
  2. ${ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
  3. $_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 4 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 2 }$
  4. $_{ 4 }^{ 6 }{ C }{ \left( \cfrac { 1 }{ 36 } \right) }^{ 2 }{ \left( \cfrac { 35 }{ 36 } \right) }^{ 4 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Getting aces with two dice means rolling two sixes, which has probability 1/36. Exactly four successes in six trials have probability C(6,4)(1/36)^4(35/36)^2, so option C is correct.

Multiple choice

A card is drawn and replaced four times from an ordinary pack of $52$ playing cards. The probability that at least once heart is drawn
  1. $(\displaystyle \frac{3}{4})^{4}$
  2. $1-(\displaystyle \frac{1}{2})^{4}$
  3. $1-(\displaystyle \frac{3}{4})^{4}$
  4. $(\displaystyle \frac{1}{2})^{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The probability of drawing a heart is 1/4, so the probability of not drawing a heart is 3/4. The probability of not drawing a heart in 4 trials is (3/4)^4. The probability of drawing at least one heart is 1 - (3/4)^4.

Multiple choice

From a pack of cards, two are drawn, the first being replaced before the second is drawn. Find the probability that the first is a club and the second is a red card.
  1. $\dfrac{1}{52}$
  2. $\dfrac{1}{26}$
  3. $\dfrac{1}{8}$
  4. $\dfrac{4}{13}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of first card being a club = 13/52 = 1/4. Probability of second card being a red card (hearts or diamonds) = 26/52 = 1/2. Since the first is replaced, the events are independent. Total probability = (1/4) * (1/2) = 1/8.

Multiple choice

A bag contains $50$ coins and each coin is marked from $51$ to $100$. One coin is picked at random. The probability that the number of the coin is not a prime number, is ......
  1. $\dfrac {1}{5}$
  2. $\dfrac {3}{5}$
  3. $\dfrac {2}{5}$
  4. $\dfrac {4}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

There are 50 coins numbered 51 to 100. The prime numbers in this range are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (10 primes). The number of non-prime coins is 50 - 10 = 40. The probability is 40/50 = 4/5.

Multiple choice

A box contains $20$ balls bearing numbers, $1,2,3,4, ....., 20$. A ball is drawn at random from the box. What is the probability that the number on the ball is divisible by $2$ or $3$.
  1. $\dfrac{6}{10}$
  2. $\dfrac{11}{20}$
  3. $\dfrac{13}{20}$

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

Numbers divisible by 2: 10. Numbers divisible by 3: 6 (3, 6, 9, 12, 15, 18). Numbers divisible by both (6): 3 (6, 12, 18). Using inclusion-exclusion: 10 + 6 - 3 = 13. Probability = 13/20.

Multiple choice

An urn contains 8 blue and 4 green balls. Two balls are drawn at random. The probability that both balls are of the same colour (all balls are considered to be different sizes)
  1. $\displaystyle \frac{17}{33}$
  2. $\displaystyle \frac{1}{33}$
  3. $\displaystyle \frac{14}{33}$
  4. $\displaystyle \frac{1}{11}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total balls = 12. Ways to pick 2 = 12C2 = 66. Ways to pick 2 blue = 8C2 = 28. Ways to pick 2 green = 4C2 = 6. Total favorable = 28 + 6 = 34. Probability = 34/66 = 17/33.

Multiple choice

A coin is tossed '$2n$' times. The chance that the number of times one gets head is not equal to the number of times one gets tail is
  1. $\displaystyle \frac{(2n!)}{(n!)^{2}}. \left(\frac{1}{2} \right)^{2n}$
  2. $1-\displaystyle \frac{(2n!)}{(n!)^{2}}$
  3. $1-\displaystyle \frac{(2n!)}{(n!)^{2}}.\frac{1}{4^{n}}$
  4. $\displaystyle \frac{(2n!)}{(n!)^{2}}\frac{1}{4^{n}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total outcomes = 2^(2n). Outcomes with equal heads and tails = 2nCn. Probability of equal heads and tails = (2nCn) / 2^(2n). Probability of not equal = 1 - (2nCn) / 4^n.

Multiple choice

A fair coin is tossed if the result is a head, a pairof fair dice is rolled and the score is noted. If theresult is a tail, a card from a well shuffled pack ofeleven numbered 1,2,3,.......11 is picked and thenumber on the card is noted. Then the probability that the number noted is either 7 or 8 is
  1. $\displaystyle \frac{45}{512}$
  2. $\displaystyle \frac{193}{792}$
  3. $\displaystyle \frac{193}{1024}$
  4. $\displaystyle \frac{243}{792}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

P(Head) = 1/2. If Head, roll two dice (36 outcomes). Sum 7 or 8: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) [6] and (2,6),(3,5),(4,4),(5,3),(6,2) [5]. Total 11/36. P(Tail) = 1/2. If Tail, pick from 11 cards. 7 or 8: 2/11. Total P = (1/2 * 11/36) + (1/2 * 2/11) = 11/72 + 1/11 = (121 + 72) / 792 = 193/792.