When a die is thrown three times, then the sample space contains 6 × 6 × 6 = 216 equally likely events.
The sample space is S = {(x, y, z) : x, y, z ∈ {1, 2, 3, 4, 5, 6}}.
E : ‘4 appears on the third toss’
i.e., E = {x, y, 4) : x, y ∈ {1, 2, 3, 4, 5,6}}
and F : ‘6 and 5 appears respectively on first two tosses’
i.e., F = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
⇒ E ∩ F = {(6, 5, 4)} ∴ P(E) =

36

216

,
P(F)=

6

216

=

1

36

and P(E∩F)=

1

216

∴P(E|F)=

P(E∩F)

P(F)

=

1 216

6 216

=

1

6