Revision as of 17:36, 29 April 2023 by Admin
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]]