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)}

\[\text{ Now } , \]

\[A \cap B = \left\{ \left( 4, 4 \right), \left( 6, 2 \right), \left( 2, 6 \right) \right\}\]

\[ \therefore \text{ Required probability } = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{3}{25}\]