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Question
A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.
Solution
Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and 5 on second throw
Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
\[\text{ Now } , \]
\[A \cap B = \left\{ \left( 6, 5, 4 \right) \right\}\]
\[ \therefore \text{ Required probability } = P\left( A/B \right) = \frac{n\left( A \cap B \right)}{n\left( B \right)} = \frac{1}{6}\]
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