Question

A die is thrown three times. Find P (A/B) and P (B/A), if
A = 4 appears on the third toss, B = 6 and 5 appear respectively on first two tosses.

Answer 1

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw

Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),
. .
. .
. .
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

$$\begin{gathered} \mathrm{Now}, \\ A\cap B=\left\{\left(6,5,4\right)\right\} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{1}{6} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(B/A\right)=\frac{n\left(A\cap B\right)}{n\left(A\right)}=\frac{1}{36} \end{gathered}$$