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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’.
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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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