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Question
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution
When a pair of dice is rolled once, then the sample space
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4); (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Let E: 'the sum of the numbers on the dice is 4' and F: 'numbers appearing on the two dice are different'
F contains all points of 5 except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). This means that F contains 36 - 6 = 30 sample points
⇒ `P (F) = 30/36`
⇒ E ∩ F = {(1, 3), (3, 1)}
⇒ `P (E cap F) = 2/36`
Hence, the required probability = P (E|F)
`= (P (E cap F))/(P (F)) = (2/36)/(30/36)`
`= 2/30 = 1/15`
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