When a die is thrown three times the number of all favorable outcomes are n(S) = 216

Let E be the event that 4 occurs on the third toss, then the favorable outcomes will be

E ={(1, 1, 4),(1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4)

(2, 1, 4),(2, 2, 4), (2, 3,4), (2, 4, 4), (2, 5, 4), (2, 6, 4)

(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6,4)

(4, 1 ,4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)

(5, 1, 4), (5, 2,4), (5, 3, 4),(5,4, 4), (5, 5, 4), (5, 6, 4)

(6, 1, 4),(6, 2, 4), (6, 3, 4), (6,4, 4),(6, 5, 4), (6, 6, 4)}

n(E) = 36

So the probability that 4 appears on the third toss is

Let F be the event that 6 and 5 appear on first two tosses respectively, and then the favorable outcomes will be

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

n(F) = 6

So the probability that 6 and 5 appear on first two tosses respectively will be

So the favorable outcomes that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be

(E∩F) = {(6, 5, 4)}, n(E∩F) = 1

And the corresponding probability will be

So the conditional probability that if it is given that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be