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6. Athletes are tested for doping by administering a drug test for testosterone.

ID: 3305812 • Letter: 6

Question

6. Athletes are tested for doping by administering a drug test for testosterone. Suppose 75% of "dopers" test postive for testosterone, whereas 3% of nonusers test postive for testosterone use (this is called a false positive), and further suppose that 5% of all tested athletes are doping with testosterone. o) I'an athlete i elected at random, what is the prodability dat they are a 1 (b) If an athlete is selected at random, what is the probability that they test positive for 3 (c) An athlete is found to have tested positive for doping and is subsequently banned from further competitions. What is the (conditional) probability that they actually were testosterone use? doping? Does this seem like a fair test? EC Replace the "3%" in the above description of the test with a generic probability p. How small must p be in order to ensure the probability you calculated in part (c) is at least 2 95%?

Explanation / Answer

Let D shows the event that athlete actually a doper. And G shows the event that athlete is non user. So

P(D) = 0.05, P(G) =1 -P(D) = 1-0.05= 0.95

Let P shows the event test results positive and N shows the event that test gives negative result. So

P(P|D) = 0.75, P(P|G) = 0.03

By the complement rule

P(N|D) = 1 - P(P|D) = 1 - 0.75 =0.25, P(N|G) =1 -P(P|G) = 1 -0.03 = 0.97

(a)

The probability that athelet is a non uses is P(G) = 1 -P(D) = 0.95

(b)

By the law of total probability,

P(P) = P(P|D)P(D) +P(P|G)P(G) = 0.75 *0.05 + 0.03 * 0.95 = 0.0375 + 0.0285 = 0.066

(c)

The required probability is

P(D|P ) = [P(P|D)P(D)] / P(P) = 0.0375 / 0.066 = 0.5682

No it does not seem that test is effective.

(EC)

P(P) = P(P|D)P(D) +P(P|G)P(G) = 0.75 *0.05 + p * 0.95 = 0.0375 + 0.95p

And

P(D|P ) = [P(P|D)P(D)] / P(P) = 0.0375 / (0.0375+0.95p)

We need this probability such that

0.0375 / (0.0375+0.95p) >= 0.95

0.0375 >= 0.95 * (0.0375+0.95p)

0.0375 >= 0.035625 + 0.9025p

0.001875 >= 0.9025p

0.0021 > = p

That is p must be less than equal to 0.21%.

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