Probability

Question 1

A number p is chosen at random from the numbers [-3, 3]. What is the probability that value of p < 2

• A. 3/7
• B. 5/7
• C. 4/7
• D. 1/7

Answer: (B)

The required favourable sample space is -3,-2,-1,0,1. Therefore the required answer is 5/7.

Question 2

A box contains 6 black, 5 brown and 2 yellow balls. If 2 balls are selected at random, what is the probability that both are black?

• A. 4/23
• B. 5/26
• C. 7/26
• D. 8/15

Answer: (B)

There are total 13 balls in the sample space Probability of choosing any two of them is 13C2. There are total 6 black balls and probability of choosing any two of them is 6C2 Therefore the required answer is 6C2/13C2 = 5/26.

Question 3

4 dices are thrown simultaneously. What is the probability that all the dices show the same number.

• A. 1/216
• B. 1/36
• C. 4/216
• D. 4/36

Answer: (A)

The total sample space is 6⁴=1296 The favourable sample space is 6 as there are 6 numbers in a dice= 6 The required answer is 6/1296=1/216.

Question 4

An urn contains 12 red and 18 white marbles. Two marbles are drawn without replacement one after another. What is the probability that first is red and second is white?

• A. 16/145
• B. 9/32
• C. 36/145
• D. 5/32

Answer: (C)

Two marbles are drawn without replacement one after another, then the probability that first is red and second is white,

= (12/30)*(18/29)
= (2/5)*(18/29)
= (2*18) / (5*29)
= 36 / 145 

So, option (C) is correct.

Question 5

What is the probability that a number selected from numbers [1,30] is prime number?

• A. 1/3
• B. 2/7
• C. 5/9
• D. 5/30

Answer: (A)

The number of prime number from 1 to 30 are 10 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29) Therefore the required answer is 10/30=1/3

Question 6

A bag contains 21 balls numbered 1 to 21. two balls are drawn one after another without replacement. Find the probability that both balls are even numbered.

• A. 3/14
• B. 2/3
• C. 6/9
• D. 1/8

Answer: (A)

Total numbers in sample space is 21 Probability of choosing any two of them is 21C2. There are total 10 even numbers from 1 to 21 and probability of choosing any two of them is 10C2 Therefore the required answer is 10C2/21C2 = 3/14

Question 7

A 5 digit number is formed by using digits 1, 2, 3, 4 and 5 with no two digits same. What is the probability that the formed number is divisible by 4?

• A. 1/5
• B. 4/5
• C. 3/5
• D. 1/4

Answer: (A)

To be divisible by 4, the number has to end with the last two digits forming a number divisible by 4. Since there are only 5 digits: 1, 2, 3, 4, 5. So, possible last two digits according to divisibility by 4: 12 24 32 52 For each of them we now have to put the remaining numbers in front: 3x2x1, therefore 6 combinations for each number set = 6x4 (we have 4 sets from above) = 24. So, required probability is:P = Number of desired outcomes / number of possible outcomes = 24 /(5*4*3*2*1) = 24 / (5!) = 24 / 120 = 1/5So, option (A) is correct.

Question 8

A box having 5 black and 3 brown flags. Another box having 4 black and 6 brown flags. If one flag is drawn from each box. Find the probability that both flags are of different color.

• A. 21/40
• B. 10/19
• C. 3/8
• D. 3/10

Answer: (A)

Box 1 consists of 5 black and 3 brown flags. Box 2 consists of 4 black and 6 brown flags. So if we select black flags from box 1 we have to select brown flags from box2 according to the question and probability of that is P1= 5/8 * 6/10 If we select brown flags from box 1 we have to select black flags from box2 according to the question and probability of that is P2=3/8 * 4/10 So the answer is P1+P2= 21/40

Question 9

Ram speaks truth in 3/4 of cases and Shyam in 4/5 of cases. In what percent of cases while narrating the same event are they likely to contradict each other?

• A. 35%
• B. 25%
• C. 20%
• D. 40%

Answer: (A)

They will contradict when 1 is says truth and other lies.

p(ram_truth)=3/4

p(shyam_truth)=4/5

p(ram_lies)=1-3/4= 1/4 
p(shyam_lies)=1-4/5= 1/5 

contradict %= [p(ram_truth) * p(shyam_lies) + p(shyam_truth) * p(ram_lies) ]*100

=[3/4 * 1/5 + 4/5 * 1/4]*100

=35%

Question 10

What is the probability of getting a sum of 9 from two throws of a dice?

• A. 1/3
• B. 1/6
• C. 1/9
• D. 1/12

Answer: (C)

Total number of outcomes = 6 * 6 = 36. Total number of favorable outcomes = (3, 6), (4, 5), (5, 4) (6, 3) = 4. Therefore, required probability = 4/36 = 1/9.