Exercise


Apr 30'23

Answer

Solution: C

Let event A be the selection of the die with faces (1,2,3,4,5,6), event B be the selection of the die with faces (2,2,4,4,6,6) and event C be the selection of the die with all 6’s. The desired probability is, using the law of total probability,

[[math]] \begin{align*} \operatorname{P}(6, 6) &= \operatorname{P}(6, 6 | A) \operatorname{P}( A) + \operatorname{P}(6, 6 | B) \operatorname{P}( B) + \operatorname{P}(6, 6 | C ) \operatorname{P}(C ) \\ &= (1/ 36)(1/ 2) + (1/ 9)(1/ 4) + 1(1/ 4) \\ &=1/ 72 + 2 / 72 + 18 / 72 = 21/ 72 \\ &= 0.292. \end{align*} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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