Mathematics

Probability

303 Questions

Probability measures the likelihood of an event occurring, such as rolling a specific number on a die or drawing a colored ball. Questions cover simple events, mutually exclusive outcomes, and dice or coin combinations. This topic is consistently asked in mathematics and reasoning sections of competitive exams.

dice probabilitycoin toss eventsdrawing balls probabilitiesplaying card problemsmutually exclusive events

Probability Questions

Multiple choice

In how many ways can one blue ball, one yellow ball, three green balls, two white balls, one purple ball and one violet ball be arranged (without missing any ball in the arrangement) in a line if the same coloured balls are indistinguishable among themselves?

  1. 30240

  2. 60480

  3. 15120

  4. 362880

  5. 120960
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have 1 blue,1 yellow, 3 green, 2 white, 1purple and 1 violet ball. That makes a total of 9 balls which can be arranged in 9! ways but since 3 green and 2 white balls are indistinguishable and themselves can be arranged in 3! and 2! ways respectively, so we will have to divide them with total to negate repetitions.

Therefore, total number of ways = (9!)/(3!2!) = 30240.

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 box contained 5 identical sized cubes out of which 3 were green and 2 were yellow. Two random draws of one cube each were made, such that the first cube was replaced into the box before making the second draw. What is the probability that first drawn cube was green and the second was yellow?

  1. 2/25

  2. 3/25

  3. 4/25

  4. 1/5

  5. 6/25
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The probability of the first drawn cube being green are 3/5. As the first cube is replaced back into the box, the probability of second drawn cube being yellow is 2/5. Since both the events are independent of each other, the combined probability, using Multiplication theorem of probability is (3/5)x(2/5) = 6/25.

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

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)