A dice is tossed twice, what is the probability of getting the sum as (i) 9 (ii) 13.

The outcomes, When a dice is tossed twice are:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4) (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

(i) Probability of getting the sum as 9

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

The no. of favorable outcomes = 4

The no. of total outcomes = 36

As we know,

Probability =

⇒ Probability (Getting the sum as 9)

=

⇒

(ii) Probability of sum getting as 13

The no. of favorable outcomes = 0

The no. of total outcomes = 36

As we know,

Probability =

⇒ Probability (Getting the sum as 13)

=

⇒ 0