Probability questions are frequently asked in the school, college examinations and Engineering, GATE, CAT, SSC, IBPS, UPSC and other competitions. Probability also comes in the several Engineering and Medical entrance examination after school. So this is a very important topic to learn before going to exam. Let’s check your knowledge :
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Quiz Description :
Name : Probability mcq practice test for Competitions
Subject : Maths
Topic : Probability
Questions: 15 Objective type
Time Allowed : 20 Minutes
Important for : School students of Class 11th and 12th , B. Sc., M. Sc., BE/ B. Tech. College Students, GATE, IBPS, SBI, SSC, State level Competitions, Job interviews etc.
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 Question 1 of 15
1. Question
1 pointsIn a throw of coin what is the probability of getting tails
CorrectExplanation – In a coin there are 2 sides one is head and other one is tails when a coin is thrown so getting probability of head is 1/2.
IncorrectExplanation – In a coin there are 2 sides one is head and other one is tails when a coin is thrown so getting probability of head is 1/2.
UnattemptedExplanation – In a coin there are 2 sides one is head and other one is tails when a coin is thrown so getting probability of head is 1/2.
 Question 2 of 15
2. Question
1 points3 unbiased coins are tossed, what is the probability of getting at least 2 tails?
CorrectExplanation – Total cases are as follows :
TTT, HHH, TTH, THT, HTT, THH, HTH, HHT = 8
Favoured cases are : TTH, THT, HTT, TTT = 4
Hence, the probability is 4/8 or 1/2.
IncorrectExplanation – Total cases are as follows :
TTT, HHH, TTH, THT, HTT, THH, HTH, HHT = 8
Favoured cases are : TTH, THT, HTT, TTT = 4
Hence, the probability is 4/8 or 1/2.
UnattemptedExplanation – Total cases are as follows :
TTT, HHH, TTH, THT, HTT, THH, HTH, HHT = 8
Favoured cases are : TTH, THT, HTT, TTT = 4
Hence, the probability is 4/8 or 1/2.
 Question 3 of 15
3. Question
1 pointsIn a throw of dice what is the probability of getting number greater than 5
CorrectExplanation – In dice there are total 6 numbers, So no. Greater than 5 is only 6 which means the probability of getting no. Higher than 5 is 1. So, it is 1/6.
IncorrectExplanation – In dice there are total 6 numbers, So no. Greater than 5 is only 6 which means the probability of getting no. Higher than 5 is 1. So, it is 1/6.
UnattemptedExplanation – In dice there are total 6 numbers, So no. Greater than 5 is only 6 which means the probability of getting no. Higher than 5 is 1. So, it is 1/6.
 Question 4 of 15
4. Question
1 pointsWhat is the probability of getting the sum 9 from 2 throws of dice?
CorrectExplanation – Total no. Of cases are 36
Favoured cases are (3,6), (4,5), (6,3),(5,4) = 4
So, the probability is 4/36 or 1/9.
IncorrectExplanation – Total no. Of cases are 36
Favoured cases are (3,6), (4,5), (6,3),(5,4) = 4
So, the probability is 4/36 or 1/9.
UnattemptedExplanation – Total no. Of cases are 36
Favoured cases are (3,6), (4,5), (6,3),(5,4) = 4
So, the probability is 4/36 or 1/9.
 Question 5 of 15
5. Question
1 pointsIn a box, there are 8 green, 7 blue and 6 red balls. One ball is picked up randomly. What is the probability that it is neither blue nor red?
CorrectExplanation – Total no. Of balls = (8+7+6) = 21
Let (E) be the event
Therefore , n(E) = 8
P (E) = 8/21.
IncorrectExplanation – Total no. Of balls = (8+7+6) = 21
Let (E) be the event
Therefore , n(E) = 8
P (E) = 8/21.
UnattemptedExplanation – Total no. Of balls = (8+7+6) = 21
Let (E) be the event
Therefore , n(E) = 8
P (E) = 8/21.
 Question 6 of 15
6. Question
1 pointsA card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is
CorrectExplanation – Total no. Of cards are 52
Favorable card are 2
So, the probability is 2/52 or 1/26.
IncorrectExplanation – Total no. Of cards are 52
Favorable card are 2
So, the probability is 2/52 or 1/26.
UnattemptedExplanation – Total no. Of cards are 52
Favorable card are 2
So, the probability is 2/52 or 1/26.
 Question 7 of 15
7. Question
1 pointsFrom a pack of 52 cards, 1 card is drawn at random. Find the probability of a face card drawn
CorrectExplanation – Total no. Of cards are 52
Total no. Of face cards (favourable cards) are 16
So, probability is 16/52 or 4/13
IncorrectExplanation – Total no. Of cards are 52
Total no. Of face cards (favourable cards) are 16
So, probability is 16/52 or 4/13
UnattemptedExplanation – Total no. Of cards are 52
Total no. Of face cards (favourable cards) are 16
So, probability is 16/52 or 4/13
 Question 8 of 15
8. Question
1 pointsA bag contains 10 black and 20 white balls, one ball is drawn at random. What is the probability that the ball is white ?
CorrectExplanation – Total balls are 10+20 = 30
Total favourable balls are 20
So, the probability of white ball is 20/30
= 2/3.
IncorrectExplanation – Total balls are 10+20 = 30
Total favourable balls are 20
So, the probability of white ball is 20/30
= 2/3.
UnattemptedExplanation – Total balls are 10+20 = 30
Total favourable balls are 20
So, the probability of white ball is 20/30
= 2/3.
 Question 9 of 15
9. Question
1 pointsIn a lottery 10000 tickets are sold and 10 prizes are awarded. What is a probability of not getting a prize if you buy one ticket?
CorrectExplanation – Total outcomes out of those 1 is yours = 10000C1 = 10000
Favourable outcomes out of those 10 tickets one is yours 10C1 = 10
Hence, probability of winning is 10/10000
and probability of no winning is 110/10000
= 999/1000.
IncorrectExplanation – Total outcomes out of those 1 is yours = 10000C1 = 10000
Favourable outcomes out of those 10 tickets one is yours 10C1 = 10
Hence, probability of winning is 10/10000
and probability of no winning is 110/10000
= 999/1000.
UnattemptedExplanation – Total outcomes out of those 1 is yours = 10000C1 = 10000
Favourable outcomes out of those 10 tickets one is yours 10C1 = 10
Hence, probability of winning is 10/10000
and probability of no winning is 110/10000
= 999/1000.
 Question 10 of 15
10. Question
1 pointsTwo dice are tossed. The probability that the total score is a prime no. Is
CorrectExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
IncorrectExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
UnattemptedExplanation – Let (E) be the event of prime no.
n (S) = 36
then n (E) = 15
P (E) = 15/36 I.e 5/12.
 Question 11 of 15
11. Question
1 pointsA bag contains 4 white, 5 red and 6 blue balls. 3 balls are drawn at random from the bag. The probability that all of them are red is
CorrectExplanation – Let S be the sample space then, 3 balls out 15 are picked up So,
(15×14×13)/(3×2×1) = 455
Let (E) be the event of 3 red balls So,
n (E) = (5×4)/(2×1) = 10
P (E) = 10/455 = 2/91.
IncorrectExplanation – Let S be the sample space then, 3 balls out 15 are picked up So,
(15×14×13)/(3×2×1) = 455
Let (E) be the event of 3 red balls So,
n (E) = (5×4)/(2×1) = 10
P (E) = 10/455 = 2/91.
UnattemptedExplanation – Let S be the sample space then, 3 balls out 15 are picked up So,
(15×14×13)/(3×2×1) = 455
Let (E) be the event of 3 red balls So,
n (E) = (5×4)/(2×1) = 10
P (E) = 10/455 = 2/91.
 Question 12 of 15
12. Question
1 points3 unbiased coins are tossed. What is the probab of getting at most 2 heads?
CorrectExplanation – Total no. Of situations are (TTT, TTH, THT, HTT, THH, HTH, HHT, HHH)
= 8
favourable no. Are (TTT, TTH, THT, HTT, THH, HTH, HHT) = 7
So, p (E) = 7/8.
IncorrectExplanation – Total no. Of situations are (TTT, TTH, THT, HTT, THH, HTH, HHT, HHH)
= 8
favourable no. Are (TTT, TTH, THT, HTT, THH, HTH, HHT) = 7
So, p (E) = 7/8.
UnattemptedExplanation – Total no. Of situations are (TTT, TTH, THT, HTT, THH, HTH, HHT, HHH)
= 8
favourable no. Are (TTT, TTH, THT, HTT, THH, HTH, HHT) = 7
So, p (E) = 7/8.
 Question 13 of 15
13. Question
1 pointsTicket numbers 120 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a no. Which is a multiple of 3 or 5?
CorrectExplanation – Here S, (1,2,3,…..19,20)
Let (E) be the event of getting a multiple 3 or 5 = (3,6,9,12,15,18,5,10,20)
p (E) = 9/20.
IncorrectExplanation – Here S, (1,2,3,…..19,20)
Let (E) be the event of getting a multiple 3 or 5 = (3,6,9,12,15,18,5,10,20)
p (E) = 9/20.
UnattemptedExplanation – Here S, (1,2,3,…..19,20)
Let (E) be the event of getting a multiple 3 or 5 = (3,6,9,12,15,18,5,10,20)
p (E) = 9/20.
 Question 14 of 15
14. Question
1 pointsA bag contains 2 red balls , 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
CorrectExplanation – Total no. (2+3+2) = 7
n (S) = (7×6)/(2×1) = 21
n (E) = (5×4)/(2×1) = 10
So, p (E) = 10/21.
IncorrectExplanation – Total no. (2+3+2) = 7
n (S) = (7×6)/(2×1) = 21
n (E) = (5×4)/(2×1) = 10
So, p (E) = 10/21.
UnattemptedExplanation – Total no. (2+3+2) = 7
n (S) = (7×6)/(2×1) = 21
n (E) = (5×4)/(2×1) = 10
So, p (E) = 10/21.
 Question 15 of 15
15. Question
1 pointsIn a class there are 10 boys and 15 girls, 3 students are selected at random. The probability that 2 girls and 1 boy are selected is –
CorrectExplanation – n (S) = (25×24×23)/(3×2×1) = 2300
n (E) = 10× (15×14)/(2×1) = 1050
So, p (E) = 1050/2300
= 21/46.
IncorrectExplanation – n (S) = (25×24×23)/(3×2×1) = 2300
n (E) = 10× (15×14)/(2×1) = 1050
So, p (E) = 1050/2300
= 21/46.
UnattemptedExplanation – n (S) = (25×24×23)/(3×2×1) = 2300
n (E) = 10× (15×14)/(2×1) = 1050
So, p (E) = 1050/2300
= 21/46.