Question

In a simultaneous throw of a pair of dice, find the probability of getting 8 as the sum

Answer 1

In a throw of pair of dice, total no of possible outcomes = 36 (6 × 6) which 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) Let E be event of getting the sum as 8

No. of favorable outcomes = 5 {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

We know that, Probability P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`

P(E) =5/36