A die is thrown three times. Events A and B are defined as follows: A:4 on the third throw, B:6 on the first and 5 on the second throw. Find the probability of A given that B has already occurred.

Explanation

total otcomes `= 6*6*6 = 216`

`P(A) = 36/216`

`P(B)= 6/216`

`P(A nn B)=1/216`

`P(A|B) = (P(A nn B))/(P(B))`

`=(1/216)/(6/216)= 1/6`