Answer
0.30
Work Step by Step
Let $E_{1}$ be the event of soccer players using steroids, $E_{2}$ be the event of soccer players not using steroids and A be the event of testing positive for steroids.
Then, $P(E_{1})=\frac{5}{100}$
Also, $P(E_{2})=\frac{95}{100}$,
$P(A|E_{1})=\frac{98}{100}$ and $P(A|E_{2})=\frac{12}{100}$.
By using Baye's theorem, we get
$P(E_{1}|A)=\frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1})+P(E_{2})P(A|E_{2})}=\frac{\frac{5}{100}\times\frac{98}{100}}{\frac{5}{100}\times\frac{98}{100}+\frac{95}{100}\times\frac{12}{100}}=0.30$
