Answer: (C) \(\frac{1}{3}\)
Let A: Event of getting a sum of 8 or greater than 8 in a throw of two dice
B: Event of getting 3 on the first dice.
Then, A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}.
⇒ n(A) = 13
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)}.
⇒ n(B) = 6
A ∩ B = {(3, 5), (3, 6)} ⇒ n(A ∩ B) = 2.
∴ P(Event of getting sum = 8 or > 8 when 3 appears on the first dice)
⇒ The occurrence of event A on the satisfaction of condition B
= P(A/B) = \(\frac{P(A\,\cap\,B)}{P(A)}\) = \(\frac{\frac{n(A\,\cap\,B)}{n(S)}}{\frac{n(B)}{n(S)}}\) = \(\frac{n(A\,\cap\,B)}{n(B)}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
