We have, n(S) = 36
1. E = Event of 5 on black die.
E = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
P(E) = \(\frac{6}{36} = \frac{1}{6}\)
F = Getting a sum greater than 9.
F = {(4, 6), (5, 5), (6, 4)(5, 6), (6, 5), (6, 6)}
⇒ E ∩ F = {(5,5), (5,6)}
P(E ∩ F) = \(\frac{2}{36} = \frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E ∩ F)}{P(F)}\)= \(\frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{3}\)
2. E = Event of a number less than 4 on red die.
E = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3),
(3, 1), (3, 2), (3, 3), (3, 1), (3, 2), (3, 3),
(5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
P(E) = \(\frac{18}{36}= \frac{1}{2}\)
F = Getting a sum 8.
F = {(4, 4), (5, 3), (3, 5)(2, 6), (6, 2), (6, 6)}
⇒ E ∩ F = {(5, 3),(6, 2)}
P(E ∩ F) = \(\frac{2}{36} = \frac{1}{18}\)
Therefore the required probability
P(E/F) = \(\frac{P(E ∩ F)}{P(F)}\)= \(\frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{9}\).