When a pair of dice is rolled once, then the sample space
(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)
Let E : ‘the sum of the numbers on the dice is 4’ and F : ‘number appearing on the two dice are different’ then E = {(1, 3), (2, 2), (3, 1)} ⇒ P(E) =

3

36

F contains all points of S except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
This means that F contains 36 – 6 = 30 sample points ⇒ P(F) =

30

36

.
⇒ E ∩ F = {(1, 3), (3, 1)}
⇒P(E∩F)=

2

36

Hence, the required probability = P (E | F) =

P(E∩F)

P(F)

=

2∕36

30∕36

=

2

30

=

1

15

.