Question

A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once?

Answer 1

Consider the given events.
A = 5 appears on the die at least once
B = The sum of the numbers on two dice is 8.

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

$$\begin{gathered} \mathrm{Now}, \\ A\cap B=\left\{\left(3,5\right),\left(5,3\right)\right\} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A/B\right)=\frac{n\left(A\cap B\right)}{n\left(B\right)}=\frac{2}{5} \end{gathered}$$