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 fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is

  1. 2/36

  2. 2/6

  3. 5/12

  4. 1/2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\text{Total outcome are 36 out of which favorable outcomes are:} //// (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6); \\ (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) \text{which are 15.} \ \text{Thus} \hspace{1cm} P(E) = \frac{\text{No. of favourable outcomes}}{\text{No. of total outcomes}}=\frac{15}{36} = \frac{5}{12}$

Multiple choice

Balls marked 1 to 15 are mixed up and put in a bucket. Then one ball is drawn at random. What is the probability that the ball drawn has a number which is a multiple of 2 or 3?

  1. 1/3

  2. 2/3

  3. 1

  4. 4/3

  5. 5/3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here, total possibilities, S = {1, 2, 3, ...,15} or n(S) = 15

Let E = event of getting a multiple of 2 or 3 = {2,3,4, 6 ,8, 9,10, 12,14, 15}or n(E) = 10.

Then, the required probability, P(E) = n(E)/n(S) = 10/15 = 2/3.

Multiple choice

An urn contains 3 yellow, 2 green and 2 blue playing marbles. Two marbles are drawn at random. What is the probability that none of the marbles drawn is blue?

  1. 3/7

  2. 5/7

  3. 4/21

  4. 5/21

  5. 10/21
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Total number of marbles = (3 + 2 + 2) = 7 Let S be the sample space. Then, n(S) = Number of ways of drawing 2 marbles out of 7 = 7C2 = (7 x 6)/(2 x 1) = 21 Let E be the event of drawing 2 marbles, none of which is blue (i.e. either yellow or green marble is drawn).

Therefore, n(E) = Number of ways of drawing 2 marbles out of (2 green + 3 yellow) marbles = 5C2 = (5 x 4)/(2 x 1) = 10 P(E) = n(E)/n(S) = 10/21

Multiple choice

A dice is thrown twice. What is the probability of getting number 1 or 5 on the first throw, irrespective of the result of the second throw?

  1. 1/3

  2. 2/3

  3. 1

  4. 4/3

  5. 5/3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Probability of getting a 1 on the first throw = 1/6 Probability of getting a 5 on the first throw = 1/6 Therefore, using Addition theorem of probability, the probability of getting number 1 or 5 on the first throw irrespective of the result of second throw = (1/6) + (1/6) = 1/3

Multiple choice

A kid has 3 identical shaped cards in his pocket, two of them have T printed on one side and H on the other whereas the third one has H printed on both the sides. He takes one card out of his pocket randomly and finds H printed on one side of the card, what is the probability that the drawn card is the one with H printed on both the sides?

  1. 1/2

  2. 2/5

  3. 3/5

  4. 4/5

  5. 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If all the three cards are considered, then originally they had 4 Hs and 2 Ts. Since one H has already been revealed, now the probability of getting one more H out of 3 Hs and 2 Ts is to be found OR (3 Hs)/(3 Hs + 2 Ts) = 3/5

Multiple choice

In four independent tosses of an unbiased coin, what is the probability of getting at least one head?

  1. 1/4

  2. 3/4

  3. 15/16

  4. 11/16

  5. 3/16
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let S be the sample space, i.e. possible outcomes of four independent tosses of an unbiased coin. Then, n(S) = 2 x 2 x 2 x 2 = 16 Out of these 16 outcomes, there is only one possible outcome where one gets all the four tails. In all the remaining (16 - 1 = 15) cases, there will be at least one head. Therefore, Required probability = 15/16

Multiple choice

Three cards are drawn at random from a pack of 52 cards. What is the probability that all are the cards of diamond?

  1. 11/850

  2. 9/850

  3. 3/850

  4. 3/52

  5. 3/4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let S be the sample space and E be the event of selecting 3 cards of diamonds. Number of ways of drawing 3 cards out of 52 = n(S) = 52C3 Number of ways of drawing 3 cards of diamonds out of 13 = n(E) = 13C3 Then, required probability P(E) = n(E)/n(S) = 13C3 / 52C3 = 11/850

Multiple choice

An urn contains 80 blue and 3 yellow balls. Two balls are drawn without replacement. What is the probability that the second ball is yellow, if it is known that the first ball drawn was blue?

  1. 80C3

  2. 83C2

  3. 3/80

  4. 3/82

  5. 3/83
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If all the 83 balls are considered, then originally there were 80 blue and 3 yellow balls. Since one of the blue ball has already been revealed, now the probability of getting one yellow ball out of 3 yellow and remaining (80 - 1) or 79 blue balls = 3/(3+79) = 3/82

Multiple choice

In a draw of lottery, tickets numbered 1 to 50 were thoroughly shuffled and put in an urn. Then three tickets were drawn randomly from the urn, what is the probability that there is at least one ticket with one digit number in the drawn tickets?

  1. 41C3

  2. 50C3

  3. 1 - (41C3)/(50C3)

  4. (41C3)x(50C3)

  5. 1 - (41C3)x(50C3)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Number of ways of drawing 3 tickets out of 50 = 50C3 Number of two-digit tickets = 50 - 9 = 41 (since number of one-digit tickets, i.e. tickets marked 1 to 9 = 9)
Number of ways of drawing 3 tickets out of 41 = 41C3 Therefore, probability of getting two-digit numbered ticket = (41C3)/(50C3) Probability of getting at least 1 one-digit numbered ticket = 1 - probability of getting two-digit numbered ticket So, required probability = 1 - (41C3)/(50C3)

Multiple choice

Five cards are drawn simultaneously from a pack of 52 playing cards. What is the probability of drawing 2 cards of hearts and 3 black cards?

  1. 13C2x26C3

  2. (13C2)x(26C3)/(52C5 )

  3. 26C5

  4. 52C5

  5. (13C2)x(52C5 )/(26C3)x(52C5 )
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As the cards are drawn simultaneously, there is no replacement made. Let S be the sample space and E1 be the event of selecting 2 hearts and E2 be the event of selecting 3 black cards. Number of ways of drawing 5 cards out of 52= n(S) = 52C5 Number of ways of drawing 2 hearts out of 13 = n(E1) = 13C2 Number of ways of drawing 3 black out of 26 = n(E2) = 26C3 Since both the events are independent of each other, then the probability of drawing 2 hearts and 3 black cards = P(E) = n(E1)x n(E2)/n(S) = (13C2)x(26C3) / (52C5 )

Multiple choice

In a box, there are 5 plastic balls and 10 wooden balls. Out of the 5 plastic balls, 3 are red and 2 are green. Similarly, the 10 wooden balls have 3 red and 7 green. What is the probability that a ball drawn from the box is either a plastic ball or red in color?

  1. 3/15

  2. 3/5

  3. 6/15

  4. 11/15

  5. 8/15
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The event of interest is; “the drawn ball should be either a plastic ball or red in color”. There are 5 plastic balls. Number of red balls is 6; 3 plastic and 3 wooden. Plastic red balls are already counted in 5 plastic balls. But the 3 wooden red balls should be added to get the number of balls that are either red or of plastic. Thus number of such balls is 5+3 = 8. Since, there are total 15 balls; required probability is 8/15.

Multiple choice

A card is drawn from a standard pack of playing cards. It is known that this card is red. What is the probability that it is a King? (A standard pack of playing cards has 52 cards. They are in 2 colors – red and black. There are 4 suits Spades, Hearts, Diamonds and Clubs. Each suit has 13 cards: Ace, 2, 3, --- 9, 10, Jack, Queen and King. Hearts and Diamonds are red in color while Spades and Clubs are black.)

  1. 1/13

  2. 2/52

  3. 4/26

  4. 1/52

  5. 1/2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There are 26 cards in red color out of which 2 are Kings. Hence, the required probability is 2/26 = 1/13.

This is correct answer.