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Question
Determine P(E|F).
A die is thrown three times,
E: 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses
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Solution
A die is thrown three times:
E: Number 4 appears on the third toss
= {(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),
(3, 1, 4), (2, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4),
(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4),
(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4),
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
= 36 results
F: 6 and 5 appearing on the first two tosses respectively
= {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
= 6 results
E ∩ F = {6, 5, 4}
P(E ∩ F) = `1/216`
∵ The number of possible outcomes from three dice
= 6 × 6 × 6
= 216
P(F) = `6/216`
`P(E/F) = (P(E ∩ F))/(P(F))`
`= (1/216)/(6/216)`
`= 1/6`
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