OverviewExercise"In the middle of difficulty lies opportunity."
 Albert Einstein

6. 
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Answer: Option
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:

Answer: Option
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 

= ^{25}C_{3} ` 

= 
(25 x 24 x 23) 
(3 x 2 x 1) 


= 2300. 
n(E) 
= (^{10}C_{1} x ^{15}C_{2}) 

= 

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?

Answer: Option
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? 
Answer: Option
Explanation:
Let S be the sample space.
Then, n(S) = ^{52}C_{2} = 
(52 x 51) 
= 1326. 
(2 x 1) 
Let E = event of getting 2 kings out of 4.
n(E) = ^{4}C_{2} = 
(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:

Answer: Option
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 


