Answer:
93.25% probability that they have taken this steroid
Step-by-step explanation:
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
In this question:
Event A: Positive test
Event B: Taking the steroid.
Suppose the probability of an athlete taking a certain illegal steroid is 10%.
This means that [tex]P(B) = 0.1[/tex]
Given that the athlete has taken this steroid, the probability of a positive test result is 0.995.
This means that [tex]P(A|B) = 0.995[/tex]
Positive test:
99.5% of 10%(If the athlete has taken).
100-99.2 = 0.8% of 100-10 = 90%(Athlete has not taken)
Then
[tex]P(B) = 0.995*0.1 + 0.008*0.9 = 0.1067[/tex]
Given that a positive test result has been observed for an athlete, what is the probability that they have taken this steroid
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.1*0.995}{0.1067} = 0.9325[/tex]
93.25% probability that they have taken this steroid
