Practice Work

1

Let $S$ be the probability that a racer uses steroids. Let $P$ be the probability that a racer tests positive for steroids.

• $8\%$ of racers use steroids. $P(S)=0.08$ $P(\bar{S})=0.92$
• Bicyclist who uses steroids tests positive for steroids $96\%$ of the time. $P(P|S)=0.96$
• Bicyclist who does not use steroids tests positive $9\%$ of the time. $P(P|\bar{S})=0.09$

Find the probability that $P(S|P)$.

$$\begin{array}{rl}P(S|P)& =\frac{p(P|S)p(S)}{p(P|S)p(S)+p(P|\bar{S})p(\bar{S})}\\ & =\frac{0.96\times 0.08}{0.96\times 0.08+0.09\times 0.92}\\ & =0.48\end{array}$$

title: Correct answer.

2

Prove Bayes’ theorem.

$$\begin{array}{rl}\frac{p(P|S)p(S)}{p(P|S)p(S)+p(P|\bar{S})p(\bar{S})}& =\frac{\frac{p(P\cap S)}{p(S)}\cdot p(S)}{\frac{p(P\cap S)}{p(S)}\cdot p(S)+\frac{p(P\cap \bar{S})}{p(\bar{S})}\cdot p(\bar{S})}\\ & =\frac{p(P\cap S)}{p(P\cap S)+p(P\cap \bar{S})}\\ & =\frac{p(P\cap S)}{p(P\cap (S\cup \bar{S}))}\\ & =\frac{p(S\cap P)}{p(P)}\\ & =p(S|P)\end{array}$$

See conditional probability and union of events.

title: Correct answer.