A die is thrown three times. Events $A$ and $B$ are defined as below:
$A$ : 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.

$\frac{1}{6}$

The correct answer is Option (3) → $\frac{1}{6}$ ##

The sample space has 216 outcomes.

Now $A = \left\{ \begin{matrix} (1,1,4) & (1,2,4) & \dots & (1,6,4) & (2,1,4) & (2,2,4) & \dots & (2,6,4) \\ (3,1,4) & (3,2,4) & \dots & (3,6,4) & (4,1,4) & (4,2,4) & \dots & (4,6,4) \\ (5,1,4) & (5,2,4) & \dots & (5,6,4) & (6,1,4) & (6,2,4) & \dots & (6,6,4) \end{matrix} \right\}$

$B = \{(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)\}$

and $A \cap B = \{(6,5,4)\}$.

Now $P(B) = \frac{6}{216}$ and $P(A \cap B) = \frac{1}{216}$

Then $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}$