conditional probability
Professional athletes who play in the NLF are subject to random drug tests. Suppose the probability that the test shows a positive result given that the athlete is using drugs is 0.93 and the probability that the test shows a negative result given that the athlete is not using drugs is 0.98. Assume that three percent of the NFL players use drugs.
a) What is the probability that a NFL player chosen at random will test negative for drug use?
b) What is the probability that the player is actually using drugs given that he has a positive test?
Could you please help me with this question by given me any hint?
Best Answer
Let event A=The player is using the drug
B=The player is not using the drug
P=The test result is positive.
N=The test result is negative.
We have $Pr(A)=0.03, Pr(B)=0.97$ and $Pr(P)+Pr(N)=1$
We also know $Pr(P|A)=0.93$ and $Pr(N|B)=0.98$.
(a)
By Law of Total Probability, we have
$Pr(N)=Pr(N|A)Pr(A)+Pr(N|B)Pr(B)=(1-Pr(P|A))Pr(A)+Pr(N|B)Pr(B)=(1-0.93)\times 0.03+0.98\times 0.97=0.9527$
(b)
By Bayes' Rule, we get
$Pr(A|P)=\frac{Pr(P|A)Pr(A)}{Pr(P)}=\frac{0.93\times 0.03}{1-0.9527}=0.5899$