All athletes at the Olympic Games (OG) are tested for performance-enhancing ster

Question

All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (AST) test gives positive results (indicating drug use) for 90% of all steroidusers but also (and incorrectly) for 2% of those who do not use steroids. Suppose that 5% of all registered athletes use steroids. Based on this information compute the following probabilities.
(a) What is the probability that an athlete uses steroids and the AST test fails to detect it?

(b) What is the probability that a randomly selected athlete will get a positive AST?

(c) If an athlete has tested negative, what is the probability that he/she does not use steroids?

(d) Is ”testing positive” statistically independent of ”using steroids”? Clearly prove your answer.

(e) [Bonus 1pts] What is the probability that the athlete will either use steroids or she/he will get a positive result?

Explanation / Answer

We are given that: The basic Anabolic Steroid Test (AST) test gives positive results (indicating drug use) for 90% of all steroidusers . Therefore

P( AST positive | steroids ) = 0.9 and therefore P( AST negative | steroidusers ) = 1 - 0.9 = 0.1

Also we are given that:

P( AST positive | no steroids ) = 0.02

Also we are given that P( steroids ) = 0.05 because 5% of all registered athletes use steroids

a) Probability that an athlete uses steroids and the AST test fails to detect it is computed as:

= P( AST positive | steroids ) P( steroids ) = 0.9*0.05 = 0.045

Therefore 0.045 is the required probability.

b) Now the probability that a randomly selected athlete will get a positive AST is computed as:

( law of total probability )

P( AST positive ) = P( AST positive | steroids ) P( steroids ) + P( AST positive | no steroids )P( no steroids )

P( AST positive ) = 0.9*0.05 + 0.02*(1 - 0.05) = 0.045 + 0.019 = 0.064

Therefore 0.064 is the required probability.

c) Given that the athlete is tested negative, the probability that he/she does not use steroids is computed as:

= P( AST negative | no steroids )P( no steroids ) / [ 1- P(positive ) ]

= ( 1- 0.02)*(0.95) / [ 1 - 0.064 ]

= 0.9947

Therefore 0.9947 is the required probability.

d) P( positive ) = 0.064

P( steroids ) = 0.05

Therefore, P( positive )P( steroids ) = 0.064*0.05 = 0.0032

P( positive and steroids ) = 0.9*0.05 = 0.045

Therefore, P( positive )P( steroids ) is not equal to P( positive and steroids )

Therefore the 2 events are not independent.

e) Probability that the athlete will either use steroids or she/he will get a positive result is computed as:

= P( positive ) + P( negative | steroids ) = 0.064 + 0.1 = 0.164

Therefore 0.164 is the required probability.

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