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