Solve the underneath given Class 10 Maths MCQ Questions of Probability with Answers and check your answers with the detailed responses here. Likelihood has an enormous degree later on for higher examinations. In the event that this chapter has to understand by understudies, they can without much of a stretch solve the next-level issues, in view of this concept. It is recommended that students attempt to solve these questions first and afterward check with the appropriate answer. This practice will assist to gain problem-solving abilities and build their confidence level.

Students are encouraged to solve the MCQ Questions of Probability to know various ideas. Practicing the MCQ question for Class 1O Maths of Probability with answers will support your confidence subsequently assisting you with scoring great in the exam. Explore various MCQ Questions for Class 10 Maths with answers provided with detailed solutions by looking underneath.

Practice Class 10 Maths MCQ Questions of Probability

1. The probability of getting exactly one head in tossing a pair of coins is

(a) 0
(b) 1
(c) 1/3
(d)1/2

2. The probability of event equal to zero is called;

(a) Unsure event
(b) Sure Event
(c) Impossible event
(d) Independent event

3. The probability that cannot exist among the following:

(a) 2/3
(b) -1.5
(c) 15%
(d) 0.7

4. If P(E) = 0.07, then what is the probability of ‘not E’?

(a) 0.93
(b) 0.95
(c) 0.89
(d) 0.90

5. A bag has 3 red balls and 5 green balls. If we take a ball from the bag, then what is the probability of getting red balls only?

(a) 3
(b) 8
(c) 3/8
(d) 8/3

6. A dice is thrown. Find the probability of getting an even number.

(a) 2/3
(b) 1
(c) 5/6
(d) 1/2

7. Two coins are thrown at the same time. Find the probability of getting both heads.

(a) 3/4
(b) 1/4
(c) 1/2
(d) 0

8. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:

(a) 1/10
(b) 3/10
(c) 1/9
(d) 4/9

9. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.

(a) 3/4
(b) 27/50
(c) 1/4
(d) 29/100

10. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:

(a) 5
(b) 10
(c) 15
(d) 20

11. A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

(a) 143/150
(b) 147/150
(c) 1/25
(d) 1/50

12. The total number of events of throwing 10 coins simultaneously is

(a) 1024
(b) 512
(c) 100
(d) 10

13. The total events to throw three dice simultaneously is

(a) 6
(b) 18
(c) 81
(d) 216

14. Which of the following cannot be the probability of an event?

(a) 1.5
(b) 35
(c) 25%
(d) 0.3

15. A bag has 5 white marbles, 8 red marbles and 4 purple marbles. If we take a marble randomly, then what is the probability of not getting purple marble?

(a) 0.5
(b) 0.66
(c) 0.08
(d) 0.77

16. A dice is thrown in the air. The probability of getting odd numbers is

(a) 1/2
(b) 3/2
(c) 3
(d) 4

17. If we throw two coins in the air, then the probability of getting both tails will be:

(a) 1/2
(b) 1/4
(c) 2
(d) 4

18. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.

(a) 9/100
(b) 1/10
(c) 3/10
(d) 19/100

19. What is the probability of getting 53 Mondays in a leap year?

(a) 1/7
(b) 53/366
(c) 2/7
(d) 7/366

20. If two dice are thrown in the air, the probability of getting sum as 3 will be

(a) 2/18
(b) 3/18
(c) 1/18
(d) 1/36

Explaination: S = [HH, HT, TH, TT] = 4

Explanation: The probability of an event that cannot be happened or which is impossible is equal to zero.

Explanation: The probability lies between 0 and 1. Hence it cannot be negative.

Explanation: P(E) + P(not E) = 1

Since, P(E) = 0.05

So, P(not E) = 1 – P(E)

Or, P(not E) = 1 – 0.07

∴ P(not E) = 0.93

Explanation: Number of red balls = 3

Number of green balls = 5

Total balls in bag = 3+5 = 8

Probability of getting red balls = number of red balls/total number of balls

= 3/8

Explanation: Total number of cases = 6 (1,2,3,4,5,6)

There are three even numbers 2,4,6

Therefore probability of getting an even number is:

P (even) = 3/6

⇒ P (even) = 1/2

Explanation: Since two coins are tossed, therefore total number of cases = 22 = 4

Therefore, probability of getting heads in both coins is:

Explanation: Total cases = 36

Total cases in which sum of 9 can be obtained are:

(5, 4), (4, 5), (6, 3), (3, 6)

∴ P (9) = 4/36 = 1/9

Explanation: Total prime numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

That means 25 out of 100

So probability is:

P (prime) = 25/100

⇒ P (prime) = 1/4

Explanation: Let the number of blue balls be x

Then total number of balls will be 5 + x.

According to question,

x/(5 + x) = 2 X (5/5+x)

⇒ x = 10

Explanation: P (non-defective bulb) = 1 – P (Defective bulb)

= 1 – (12/600)

= (600 – 12)/600

= 588/600

= 147/150

Explaination: Reason: Total events 210 = 1024

Explanation: Total cards = (6)3= 216

Explanation: ∵ Probability of any event cannot be more than 1.

∴ 1.5 can not be the probability of any event.

Explanation: Total number of purple marbles = 4

Total number of marbles in bag = 5 + 8 + 4 = 17

Probability of getting purple marbles = 4/17

Hence, the probability of not getting purple marbles = 1-4/17 = 0.77

Explanation: A dice has six faces having values 1, 2, 3, 4, 5 and 6.

There are three odd numbers and three even numbers.

Therefore, the probability of getting only odd numbers is = 3/6 = 1/2

Explanation: When two coins are tossed, the total outcomes will be = 2 x 2 = 4

Hence, the probability of getting both tails = 1/4

Explanation: The perfect square numbers between 2 to 101 are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Total numbers from 2 to 101 =100

So probability of getting a card with perfect square number is:

P (perfect square) = 10/100

⇒ P (perfect square) = 1/10

Explanation: With 366 days, the number of weeks in a year is

366/7 = 52 (2/7)

i.e., 52 complete weeks which contains 52 Mondays,

Now 2 days of the year are remaining.

These two days can be

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are 7 pairs, in which Monday occurs in 2 pairs,

So probability is:

P (53 Monday) = 2/7