Question

First 6 faced die which is numbered 1 through 6 is thrown then a 5 faced die which is numbered 1 through 5 is thrown. What is the probability that the sum of the numbers on the upper faces of the dice is divisible by 2 or 3?

Answer 1

When a 6 faced die and a 5 faced die are thrown, the sample space is

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

∴ n (S) = 30

Let event A: The sum of the numbers on the upper faces of the dice is divisible by 2.

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

∴ n(A) = 15

∴ P(A) = `("n"("A"))/("n"("S")) = 15/30`

Let event B: Sum of the numbers on the upper faces of the dice is divisible by 3.

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

∴ n(B) = 10

∴ P(B) = `("n"("B"))/("n"("S")) = 10/30`

Now,

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

∴ n(A ∩ B) = 5

∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 5/30` 

∴ Required probability = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B) 

= `15/30 + 10/30 - 5/30`

= `20/30`

= `2/3`