1
$\begingroup$

I've been having some difficulty with this homework problem related to conditional probabilities. I think part a is correct but the answer to part b seems wrong to me.

Professional athletes who play in the NFL 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.

Let $S=\{TP, TN, FP, FN \}$ where TP stands for True Positive, TN fo True Negative, FP for False Positive, and FN for false negative. All of these events are mutually exclusive.

a) What is the probability that a NFL player chosen at random will test negative for drug use?

$P(TN\cup FN)=P(TN)+P(FN)=0.97*0.98+0.03*0.07=0.9527$

b) What is the probability that the player is actually using drugs given that he has a positive test?

First, note that the probability of a positive test is $1-0.9527=0.0473$ since this event is the complement of the event in problem a. Also the probability that a player is using drugs is given at 0.03.

By the conditional probability formula: $$P(\text{using drugs }\vert \text{ positive test})=P(TP)\backslash P(\text{positive test})=P(TP)\backslash 0.0473$$ To find $P(TP)$, note that: $$P(\text{positive test }\vert \text{ using drugs})=P(TP)\backslash P(\text{using drugs})=P(TP)\backslash 0.03$$ This was give to be 0.93. Solving yields $P(TP)=0.0279$.

Hence the solution is $0.0279\backslash 0.0473=0.5899$

I am encouraged that my solution is between 0 and 1, but it seems off to me. Can anyone tell me where I made a mistake?

$\endgroup$
1
$\begingroup$

If the athlete has a positive test, he falls either into TP or into FP.

Given that

$$P(\text{positive test}|\text{using drugs}) = 0.93$$

we know

$$P(\text{negative test}|\text{using drugs}) = 0.07$$.

I assumed your (a) is correct, I check only (b) on a different way.

We need to find $P(\text{using drugs}|\text{positive test})$. So, we use the Bayes-theorem:

$P(\text{using drugs}|\text{positive test}) = P(\text{positive test}|\text{using drugs}) \cdot \frac{P(\text{using drugs})}{P(\text{positive test})} = 0.93 \cdot \frac{0.03}{1-P(\text{negative test})} = 0.93 \cdot \frac{0.03}{1-0.9527} = 0.589852$.

So, your calculation is correct.


What you've found, is a known surprising result of the probability theory. It is caused by the fact, that after a positive result, the human thinking is often minded to ignore the fact, that only a small part of the athletes are using drugs.

Thus, between the actual positive results, we will have much more false positives.

$\endgroup$
  • $\begingroup$ This "paradox" has also a name, unfortunately I forget it, but it is a well-known phenomenon. $\endgroup$ – peterh May 21 '18 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.