Question

In a simultaneous throw of a pair of dice, find the probability of getting neither 9 nor 1 1 as the sum of the numbers on the faces

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)

`barE` ⟶ event of getting neither 9 nor 11 as the sum of numbers on faces

E ⟶ getting either 9 or 11 as the sum of no’s on faces

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

Total no. of possible outcomes = 36

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

P(E) =6/36 =1/6
P`(barE)` = 1 − 𝑃 = 1 − 1/6 =5/6