Aptitude - Probability

Exercise :: Probability - General Questions

6. 

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

A.
1
2
B.
3
4
C.
3
8
D.
5
16

Answer: Option B

Explanation:

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) = 27 = 3 .
n(S) 36 4


7. 

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A.
21
46
B.
25
117
C.
1
50
D.
3
25

Answer: Option A

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25
= 25C3 `
= (25 x 24 x 23)
(3 x 2 x 1)
= 2300.

n(E) = (10C1 x 15C2)
= 10 x (15 x 14)
(2 x 1)
= 1050.

P(E) = n(E) = 1050 = 21 .
n(S) 2300 46


8. 

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

A.
1
10
B.
2
5
C.
2
7
D.
5
7

Answer: Option C

Explanation:

P (getting a prize) = 10 = 10 = 2 .
(10 + 25) 35 7

9. 

From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

A.
1
15
B.
25
57
C.
35
256
D.
1
221

Answer: Option D

Explanation:

Let S be the sample space.

Then, n(S) = 52C2 = (52 x 51) = 1326.
(2 x 1)

Let E = event of getting 2 kings out of 4.

n(E) = 4C2 = (4 x 3) = 6.
(2 x 1)

P(E) = n(E) = 6 = 1 .
n(S) 1326 221


10. 

Two dice are tossed. The probability that the total score is a prime number is:

A.
1
6
B.
5
12
C.
1
2
D.
7
9

Answer: Option B

Explanation:

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
      (5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E) = 15 = 5 .
n(S) 36 12





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