1.

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?

A.

1

2

B.

2

5

C.

8

15

D.

9

20

Answer & Explanation

Answer: Option D

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}.

P(E) = n(E) = 9 .

n(S) 20

2.

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A.

10

21

B.

11

21

C.

2

7

D.

5

7

Answer & Explanation

Answer: Option A

Explanation:

Total number of balls = (2 + 3 + 2) = 7.

Let S be the sample space.

Then, n(S) = Number of ways of drawing 2 balls out of 7

= 7C2 `

= (7 x 6)

(2 x 1)

= 21.

Let E = Event of drawing 2 balls, none of which is blue.

n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls.

= 5C2

= (5 x 4)

(2 x 1)

= 10.

P(E) = n(E) = 10 .

n(S) 21

3.

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?

A.

1

3

B.

3

4

C.

7

19

D.

8

21

E.

9

21

Answer & Explanation

Answer: Option A

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.

n(E) = 7.

P(E) = n(E) = 7 = 1 .

n(S) 21 3

4.

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

A.

1

6

B.

1

8

C.

1

9

D.

1

12

Answer & Explanation

Answer: Option C

Explanation:

In two throws of a die, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) = n(E) = 4 = 1 .

n(S) 36 9

5.

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A.

3

4

B.

1

4

C.

3

8

D.

7

8

Answer & Explanation

Answer: Option D

Explanation:

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) = n(E) = 7 .

n(S) 8