Given that the two numbers appearing …

Consider the given events
A = Numbers appearing on two dice are different
B = The sum of the numbers on two dice is 4

Clearly,
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
                     (2, 1), (2, 3), (2, 4), (2, 5), (2, 6),
                     (3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
                     (4, 1), (4, 2), (4, 3), (4, 5), (4, 6),
                     (5, 1), (5, 2), (5, 3), (5, 4), (5, 6),
                     (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = {(1, 3), (3, 1) and (2, 2)}

 Now ,

A∩B={(1,3) and (3,1)}

∴ Required probability =P(B/A) = n(A∩B) / n(A) = 2 / 30 =1 / 15

When two dice are thrown, there are 36 possibilities.
However, we are given, every time both dice contain different numbers means doublet is not a possibility.
As, number of doublets are 6, so, sample space, S=36-6=30
Number of possibilities of the event when sum of dice is 4, (1,3),(3,1).
Here, we won′t take (2,2) as both dice contain different numbers.So,
E=2
So, P(E)=2/30=1/15