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)}

\[\text{ Now } , \]
\[A \cap B = \left\{ \left( 4, 4 \right) \right\}\]
\[ \therefore \text{ Required probability } = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{1}{6}\]