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Exercise


ABy Admin
Apr 29'23

Answer

Solution: B

[[math]] \begin{align*} \operatorname{P}[ D] &= \operatorname{P}[ H ]\operatorname{P}[ D | H ] + \operatorname{P}[ M ]\operatorname{P}[ D | M ] + \operatorname{P}[ L]\operatorname{P}[ D | L] \\ 0.009 &= \operatorname{P}[ H ]\operatorname{P}[ D | H ] + \operatorname{P}[M] \left( \frac{1}{2} \operatorname{P}[D | H] \right) + \operatorname{P}[L] \left( \frac{1}{2} \frac{1}{3} \operatorname{P}[D | H] \right) \\ 0.009 &= 0.20 \operatorname{P}[ D | H ] + 0.35 \left( \operatorname{P}[ D | H ] \right) + 0.45 \left( \operatorname{P}[ D | H ] \right) = 0.45 \operatorname{P}[ D | H ] \\ \operatorname{P}[ D | H ] &= 0.009/ 0.45 = 0.02 \end{align*} [[/math]]

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

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