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

The sample space has 216 outcomes.
so possible outcome`=216`
Now `A = (1,1,4) (1,2,4) ... (1,6,4) (2,1,4) (2,2,4) ... (2,6,4) (3,1,4) (3,2,4) ... (3,6,4) (4,1,4) (4,2,4) ...(4,6,4) (5,1,4) (5,2,4) ... (5,6,4) (6,1,4) (6,2,4) ...(6,6,4)`
` B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} `
and` A ∩ B = {(6,5,4)}`.
Now, `P(B) = 6/216` and `P(A ∩ B) = 1/216 `
Then `P(A/B) = (P(A ∩ B))/P(B)`
` = (1/216)/(6/216) = 1/6 `