sports association decides to implement a drug screening procedure to test its a

Question

sports association decides to implement a drug screening procedure to test its athletes for illegal performance enhancing drugs. A person who does not take the drugs will test positive with probability 0.02 and a person who does take the drugs will test negative with probability 0.04. Suppose that 3% of the athletes tested take performance enhancing drugs. What is the probability that a. (5 pt.) an athlete testing positive takes the drugs? b. (5 pt.) an athlete testing positive does not take the drugs? c. (5 pt) an athlete testing negative takes the drugs? d. (5 pt.) an athlete testing negative does not take the drugs?

Explanation / Answer

Answer:

a) P(tested positive)=P(takes the drug and tested positive+ not takes the drug and tested positive)

=0.03*(1-0.04) + (1-0.03)*0.02=0.0482

Hence P (takes the drug| tested positive)=P(takes the drug and tested positive)/P(tested positive)

=0.03*(1-0.04)/0.0482 =0.5975

b)

P (not takes drug|tested positive) =1-P(takes the drug|tested positive)=1-0.5975=0.4025

c)

P (tested negative) =1-P(tested positive) =1-0.0482=0.9518

P (takes the drug|tested negative)=P(takes the drug and tested negative)/P(tested negative)

=0.03*0.04/0.9518 =0.00126

D) P (not takes the drug| tested negative) =1-P(takes the drug|tested negative)=1-0.00126 =0.99874

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