We known that when a die is thrown twice, then the sample space has 36 possible outcomes.
Let A = event that 4 appears at least once, and
`B=` event that the sum of the numbers appearing is 6.
Then,
`A{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4) , (6, 4)}`
and `B={(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}`.
`:. A nn B={ (2, 4), (4, 2)}`.
So, `P(A)=(n(A))/(n(S))=11/36, P(B)=(n(B))/(n(S))=5/36`
and `P(A nn B)=(n(A nn B))/(n(S))=2/36=1/18`.
Suppose B has already occurred and then A occurs.
so, we have to find `P(A//B)`.
Noe, `P(A//B)=(P(A nn B))/(P(B))=((1//18))/((5//36))=(1/18xx36/5)=2/5`
Hence, the required probability is `2/5`.
