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Question
Two dice are thrown simultaneously. Find the probability of getting: the sum as a prime number
Solution
n(s) = 36 i.e.
(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)}
Event = { sum as a prime number }
i.e., = {(1, 1), (1, 2), (2, 1), (1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5) and (5, 6)}
n(E) = 15
P(E) = ?
∴ P(E) = `"n(E)"/"n(s)" = 15/36 = 5/12`
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