Question

Two fair dice are thrown. Find the probability that number on the upper face of the first die is 3 or sum of the numbers on their upper faces is 6

Answer 1

When two dice are thrown, the sample space is

S = {(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 event A : The number on the upper face of the first die is 3.

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

∴ n(A) = 6

∴ P(A) = `("n"("A"))/("n"("S")) = 6/36`

Let event B : Sum of the numbers on their upper faces is 6

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

∴ n(B) = 5

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

Now, A ∩ B = {(3, 3)}

∴ n(A ∩ B) = 1

∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 1/36`

∴ Required probability = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B) 

= `6/36 + 5/36 - 1/36`

= `10/36`

= `5/18`.