Question

Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Answer 1

Consider the given events.
A = 4 appears on second die
B = The sum of the numbers on two dice is 8.

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

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