probability aptitude questions

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

 

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