excans:102b498704

Exercise

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]]