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Percentage of athletes that use steroids $10\%=0.10$ Percentage of tests that are accurate when steroids are used $95\%=0.95$ Percentage of tests that are accurate when steroids are not used $97\%=0.97$

Formula used:$P\left(A\right)+P\left(B\right)=1$

$P\left(A\right)=1−P\left(B\right)$

Let S stand for the taking of a steroid. The letter NS denotes the absence of a steroid. The letter P stands for a positive outcome. The letter N stands for a negative outcome. Based on the information provided:$$\begin{gathered} P\left(S\right)=0.10 \\ P\left(P\right|S)=0.95 \\ P(N\left|NS\right)=0.97 \end{gathered}$$

Here,

$$\begin{gathered} P\left(P\right)=P(Sâˆ\copyright P)+P(NSâˆ\copyright P) \\ P(Sâˆ\copyright P)=P\left(S\right)×P\left(P\right|S) \\ P(NSâˆ\copyright P)=P(NS)×P(P\left|NS\right) \end{gathered}$$

The probabilities

$$\begin{gathered} P(Sâˆ\copyright P)=0.10×0.95=0.095 \\ P(NSâˆ\copyright P)=0.90×0.03=0.027 \end{gathered}$$

Adding the probabilities together to arrive at the required probability

$=0.095+0.027=0.122$

As a result, $0.122$is the required probability.

From the part b

In this case,

$P\left(S\right|P)=\frac{P(Sâˆ\copyright P)}{P\left(P\right)}$

Using the probabilities shown above,

$P\left(S\right|P)=\frac{0.095}{0.122}=0.7787$

As a result, $0.7787$ is the required probability.