Question

Find the probability that the sum of the numbers showing on two dice is 8, given that at least one die does not show five.

Answer 1

Consider the given events.
A = At least one die does not show 5
B = The sum of the numbers on two dice is 8.

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

$$\begin{gathered} \mathrm{Now}, \\ A\cap B=\left\{\left(4,4\right),\left(6,2\right),\left(2,6\right)\right\} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(B/A\right)=\frac{n\left(A\cap B\right)}{n\left(A\right)}=\frac{3}{25} \end{gathered}$$