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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.
Solution
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)}
\[\text{ Now } , \]
\[A \cap B = \left\{ \left( 4, 4 \right), \left( 6, 2 \right), \left( 2, 6 \right) \right\}\]
\[ \therefore \text{ Required probability } = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{3}{25}\]
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