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Exercise
ABy Admin
Apr 29'23
Answer
Solution: C
Let:
[math]S[/math] = Event of a smoker
[math]C[/math] = Event of a circulation problem
Then we are given that [math]\operatorname{P}[C] = 0.25[/math] and [math]\operatorname{P}[S | C] = 2 \operatorname{P}[S | C^c] [/math].
Now applying Bayes’ Theorem, we find that
[[math]]
\begin{align*}
\operatorname{P}[C | S] &= \frac{\operatorname{P}[ S | C ]\operatorname{P}[C ]}{\operatorname{P}[ S | C ]\operatorname{P}[C ] + \operatorname{P}[ S | C ]( \operatorname{P}[C^c ])}\\
&= \frac{2 \operatorname{P}[ S | C^c ]\operatorname{P}[C ]}{2 \operatorname{P}[ S |C^c ]\operatorname{P}[C ] + \operatorname{P}[ S |C^c ](1 − \operatorname{P}[C ])} \\
&= \frac{2(0.25)}{2(0.25) + 0.75} \\
&= \frac{2}{2+3} \\
&= \frac{2}{5}.
\end{align*}
[[/math]]