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Question
Two dice are thrown simultaneously. Find the probability of getting: a multiple of 3 as the sum.
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 = {multiple of 3 as a sum}
= {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}
n(E) = 12
p(E) = ?
∴ P(E) = `"n(E)"/"n(S)" = 12/36 = 1/3`.
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