Question

In a simultaneous throw of a pair of dice, find the probability of getting a sum more than 7

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)

E ⟶ event of getting a sum more than 7

No. of favorable outcomes = 15 {(2, 6) (3, 5) (3, 6) (4, 4) (4, 5) (4, 6), (5, 3) (5, 4)

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

Total no. of possible outcomes = 36

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

P(E) =15/36 =5/12