Question

Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.

Answer 1

Sample space = {(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)}

n(S) = 36

Let A be the event of getting an even number on the first time

A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

n(A) = 18

P(A) = `("n"("A"))/("n"("S")) = 18/36`

Let B be the event of getting a total of face sum 8.

B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

n(B) = 5

P(B) = `("n"("B"))/("n"("S")) = 5/36`

A ∩ B = {(2, 6) (4, 4) (6, 2)}

n(A ∩ B) = 3

P(A ∩ B) = `3/36`

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= `18/36 + 5/36 - 3/36`

= `(18 + 5 - 3)/36`

= `20/36`

= `5/9`

The required probability = `5/9`