# Probability

• Last Updated : 16 Sep, 2021

 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
Probability
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Question 1 Explanation:
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
Probability
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Question 2 Explanation:

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
Probability
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Question 3 Explanation:

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
Probability
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Question 4 Explanation:
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
Probability
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Question 5 Explanation:
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
Probability
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Question 6 Explanation:

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
Probability
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Question 7 Explanation:
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
Probability
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Question 8 Explanation:

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%
Probability
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Question 9 Explanation:

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
Probability
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Question 10 Explanation:
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.
There are 19 questions to complete.
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