Here,
A=Event that 4 appears on first die.
A={(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}.
∴n(A)=6
Let B=Event that sum of the dice is 8 or more.
Then, B={(3,6),(6,3),(2,6),(6,2),(5,3),(3,5),(4,4),(4,5),(4,6)}
Now, (A∩B)={(4,4),(4,5),(4,6)}
⇒n(A∩B)=3
∴ Required probability , P(
B
A
)=
n(A∩B)
n(A)
=
3
6
=
1
2
