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

1/6

The correct answer is Option (1) → 1/6

$\text{P(B) = P(4 on first throw) × P(5 on second throw)}$

$=\frac{1}{6}×\frac{1}{6}=\frac{1}{36}$

$\text{P(A|B)=P(6 on third throw)}$

$=\frac{1}{6}$

$P(A∩B)=P(B)×P(A|B)=\frac{1}{36}×\frac{1}{6}=\frac{1}{216}$

$⇒P(A|B)=\frac{P(A∩B)}{P(B)}=\frac{\frac{1}{216}}{\frac{1}{36}}=\frac{1}{6}$