# Probability Questions and Answers updated daily – Aptitude

Probability Questions: Solved 93 Probability Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Probability Questions

61. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Correct Ans:(2/7)
Explanation:
Given,In a lottery, there are 10 prizes and 25 blanks.
So, Total number of sample space = 10 + 25 = 35
=> n(S) = 35

Event of getting prize = n(E) = 10

Required probability = P (getting a prize) = n(E) / n(S)
=10 / 35
= 2/7
Workspace

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

Correct Ans:(1/9)
Explanation:
In two throws of a dice, n(S) = (6 x 6) = 36.
Let E = event of getting a sum "9" = {(3, 6), (4, 5), (5, 4), (6, 3)}.
Therefore, P(E)= n(E)/n(S) = 4/36 = 1 / 9
Workspace

63. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Correct Ans:(1 / 3)
Explanation:
Total number of balls = (8 + 7 + 6) = 21. Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue. Therefore , n (E) = 7 Therefore P (E) = n (E) / n (S) = 7/21 = 1/3.
Workspace

64. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Correct Ans:(9/20)
Explanation:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

Therefore , P (E) = n (E) / n (s) = 9/20
Workspace

65. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Correct Ans:3/4
Explanation:
Solution is
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
=>n(E) = 27
P(E) = n(E) / n(S) = 27 / 36 = 3/4
Workspace

66. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Correct Ans:1/3
Explanation:
Solution is
Total no. of Balls = n(s) = 8 + 7 + 6 = 21
Let E = event that the ball drawn is neither green nor red =event that the ball drawn is blue
n(E) = 7
P(E) = n(E) / n(S) = 7/21 = 1/3
Workspace

67. Two coins are tossed. What is the probability of getting atleast one tail ?

Correct Ans:3/4
Explanation:
Solution is
Let H denote head and T denote tail.
When two coins are tossed, the possible cases are S = {HH, HT, TH , TT}
=> n(s) = 4

Let E be the event which denotes getting atleast one tail E = {HT , TH, TT}
=> n(E) = 3

Probability of getting atleast one tail,
P(E) = n(E) / n(S) = 3/4
Workspace

68. How many 3 digit numbers can be formed using the digits 1,3,4 and 8 ?

Correct Ans:24
Explanation:
Solution is
No. of digits given to us is 4.
Nothing is mentioned about repetition of digits, so we will take it granted, the digits are not suppose to repeat.
No. of ways that 3 digit numbers can be formed = 4 P3
= 4 ! = 4 x 3 x 2
= 24 numbers.
Workspace

69. In how many different ways can the letters of the word 'OFFICES' be arranged ?

Correct Ans:2520
Explanation:
Solution is
The word OFFICES contains O,F, F I, C, E, S contains 7 letters, in which 2 are identical.
No. of words that can formed = 7 ! / 2 !
= 7 * 6 * 5 * 4 * 3
= 2520
Workspace

70. In a box, there are 7 red, 6 blue and 5 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Correct Ans:1/3
Explanation:
Solution is
No. of Balls = n(s) = 7 + 6 + 5 = 18
Let E = event that the ball drawn is neither green nor red = event that the ball drawn is blue
n(E) = 6
P(E) = n(E)/n(S)
= 6/18
= 1/3
Workspace

71. An unbiased dice is thrown. What is the probability of getting an odd number ?

Correct Ans:1/2
Explanation:
Solution is
Sample Space , S = {1, 2, 3, 4, 5, 6}
=>n(S) = 6
E event of getting an odd number E = {1, 3, 5}
=> n(E)=3
P(E) = n(E) / n(S) = 3 / 6 = 1/2
Workspace

72. How many words can be formed with or without meaning by taking all the letters from the word COIN ?

Correct Ans:24
Explanation:
Solution is
There are totally 4 letters in the word
COIN (i.e., C, O, I and N)

No. Of Words formed = 4! = 4 x 3 x 2 x 1 = 24
Workspace

73. In a party there were totaly 20 people, each person shook his hands with the other person. How many hand shakes would have taken place ?

Correct Ans:190
Explanation:
Solution is
There are 20 people.
Every person has to shake hands with the other person,Which means we have to find the number of ways of choosing 2 people from the 20.
The number of ways it can happen = 20C2 = ( 20 x 19 ) / (1 x 2) = 190
Workspace

74. Two dice are thrown simultaneously. What is the probability that the sum of the two numbers appear on the top of the dice is 9?

Correct Ans:1/9
Explanation:
Solution is
Sample Space, S = { (1,1) , (1,2) , (1,3) , ... (6,6) }
=> n(S) = 6 x 6 = 36 possible cases.

E = {(x,y) / such that x+y=9}
E = {(4,5) , (5,4) , (3,6) , (6,3)}
=> n(E)=4

Required probability:
P(E) = n(E) / n(S)
= 4 / 36
= 1/9
Workspace

75. An unbiased dice is thrown. What is the probability of multiple of 3?

Correct Ans:1/3
Explanation:
Solution is :
Sample Space , S = {1, 2, 3, 4, 5, 6}
=>n(S) = 6,
E is the event of getting multiple of 3
E = {3, 6}
=> n(E)=2
Probability of getting multiple of 3 is
P(E) = n(E) / n(S)
= 2 / 6
= 1 / 3
Workspace

76. In how many ways 3 people can be made seated in a row containing 6 seats ?

Correct Ans:120
Explanation:
Solution is :
First person has got 6 ways to sit
Second person got 5 ways (since 1 seat occupied)
Third person got 4 ways.
Since all are independent
No. Of ways = 6 x 5 x 4 = 120
Workspace

77. Find the number of different ways of selecting 4 men from a group of 7 men ?

Correct Ans:35
Explanation:
Solution is :
Total number of men = 7
Number of men to select = 4
No.of Ways to select 4 men from 7 men = 7C4
= ( 7 x 6 x 5 x 4 ) / (4 x 3 x 2 x 1)
= 35
Workspace

78. Two coins are tossed. What is the probability of getting exactly one tail ?

Correct Ans:1/2
Explanation:
Solution is
Let H denote head and T denote tail.
When two coins are tossed, the possible cases are
S = {HH, HT, TH , TT}
=> n(s) = 4
Let E be the event which denotes getting exactly one tail
E = {HT , TH}
=> n(E) = 2
Probability of getting exactly one tail is
P(E) = n(E) / n(S)
= 2 / 4
= 1 / 2
Workspace

79. In a class there are some boys and 30 girls. The probability of selecting a girl is thrice the probaility of selecting a boy. Find the number of boys ?

Correct Ans:10
Explanation:
Solution is :
Let the number of boys be x
Given there are 30 girls
Total number of people are (boys + girls) = x+30
Probability of selecting a girl, P(G) = 30 / (x+30)
Probability of selecting a boy, P(B) = x / (x+30)
Given P(G) = 3 x P(B)
=> 30 / (x+30) = 3x / (x+30)
=> 30 = 3x
=> x = 10
There are totally 10 boys.
Workspace

80. How many words can be formed with or without meaning by taking all the letters from the word TAKEN ?

Correct Ans:120
Explanation:
Solution is :
There are totally 5 letters in the word TAKEN => T, A, K, E and N.
No. of Words formed = 5 !
= 5 x 4 x 3 x 2 x 1
= 120
Workspace

Are you seeking for good platform for practicing Probability questions in online. This is the right place. The time you spent in Fresherslive will be the most beneficial one for you.

## Online Test on Probability @ Fresherslive

This page provides important questions on Probability along with correct answers and clear explanation, which will be very useful for various Interviews, Competitive examinations and Entrance tests. Here, Most of the Probability questions are framed with Latest concepts, so that you may get updated through these Probability Online tests. Probability Online Test questions are granted from basic level to complex level.

## Why To Practice Probability Test questions Online @ Fresherslive?

Probability questions are delivered with accurate answer. For solving each and every question, very lucid explanations are provided with diagrams wherever necessary.
Practice in advance of similar questions on Probability may improve your performance in the real Exams and Interview.
Time Management for answering the Probability questions quickly is foremost important for success in Competitive Exams and Placement Interviews.
Through Fresherslive Probability questions and answers, you can acquire all the essential idea to solve any difficult questions on Probability in short time and also in short cut method.
Winners are those who can use the simplest method for solving a question. So that they have enough time for solving all the questions in examination, correctly without any tense. Fresherslive provides most simplest methods to answer any tough questions. Practise through Fresherslive test series to ensure success in all competitive exams, entrance exams and placement tests.

## Why Fresherslive For Probability Online Test Preparation?

Most of the job seekers finding it hard to clear Probability test or get stuck on any particular question, our Probability test sections will help you to success in Exams as well as Interviews. To acquire clear understanding of Probability, exercise these advanced Probability questions with answers.
You're Welcome to use the Fresherslive Online Test at any time you want. Start your beginning, of anything you want by using our sample Probability Online Test and create yourself a successful one. Fresherslive provides you a new opportunity to improve yourself. Take it and make use of it to the fullest. GOODLUCK for Your Bright Future.

## Online Test

Online Test for Aptitude
Online Test for Logical Reasoning
Online Test for Computer Knowledge
Online Test for General Knowledge
Online Test for Data Interpretation
Online Test for Verbal Ability
Online Test for C++
Online Test for Networking
Online Test for Java
Online Test for C Language
FreshersLive - No.1 Job site in India. Here you can find latest 2023 government as well as private job recruitment notifications for different posts vacancies in India. Get top company jobs for both fresher and experienced. Job Seekers can get useful interview tips, resume services & interview Question and answer. Practice online test free which is helpful for interview preparation. Register with us to get latest employment news/rojgar samachar notifications. Also get latest free govt and other sarkari naukri job alerts daily through E-mail...