Exercise


Apr 29'23

Answer

Solution: D

Let

[math]H[/math] = Event of a heavy smoker

[math]L[/math] = Event of a light smoker

[math]N [/math] = Event of a non-smoker

[math]D [/math] = Event of a death within five-year period

Now we are given that [math] \operatorname{P}[ D | L ] = 2 \operatorname{P}[ D N ] [/math] and [math]\operatorname{P}[ D | L ] = \frac{1}{2} \operatorname{P}[ D | H ] [/math]. Therefore, upon applying Bayes’ Formula, we find that

[[math]] \begin{align*} \operatorname{P}[ H | D ] &= \frac{\operatorname{P}[ D | H ] \operatorname{P}[ H ]}{\operatorname{P}[ D | N ] \operatorname{P}[ N ] + \operatorname{P}[ D | L ] \operatorname{P}[ L ] + \operatorname{P}[ D | H ] \operatorname{P}[ H ]} \\ &= \frac{2 \operatorname{P}[ D | L ] ( 0.2 )}{\frac{1}{2} \operatorname{P}[D |L ](0.5) + \operatorname{P}[D|L](0.3) + 2\operatorname{P}[D|L](0.2)} \\ &= \frac{0.4}{0.25 + 0.3 + 0.4} \\ &= 0.42. \end{align*} [[/math]]

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

00