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If a fair die (with 6 faces) is cast twice, what is the probability that the two
numbers obtained differ by 2?
(A) 1/12
(B) 1/6
(C) 2/9
(D) 1/2
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2/9
1
1

1 Answer

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$probability=\frac{favorable\ cases}{sample space}=\frac{(1,3)(3,1)(2,4)(4,2)(3,5)(5,3)(4,6)(6,4)}{total \ 36 \ pairs}=\frac{8}{36}=\frac{2}{9}$

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