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A 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. an athlete testing positive takes the drugs?

b. an athlete testing positive does not take the drugs?

c. an athlete testing negative takes the drugs?

d. an athlete testing negative does not take the drugs?

I was given a hint that I can use Bayes Theorem, but I have no idea how to do this and it looks complicated.. Can anyone help me to solve this..

tien lee
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  • p(E/F) denotes the probability that event E occurs given that event F has occurred. p(EF) denotes the probability that events E and F both occur. Bayes theorem : p(A)p(B/A) = p(AB) = p(B)p(A/B). Show (for example) work around question a, indicating where you are stumped, and you will receive a good hint. – user2661923 May 02 '18 at 23:23
  • If Bayes theorem (i.e. conditional probability) is a brand new topic, then you need to first explore simpler conditional probability problems, from your textbook, or getting help from a teacher or a friend. Do not try to attack the posted problems cold, if you are new to the topic. Also, the forum does not normally function as the first point of support for a math topic that is totally new to you. – user2661923 May 02 '18 at 23:28
  • Possibly helpful: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 – Ethan Bolker Sep 11 '18 at 22:29

2 Answers2

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Refer to the probability tree diagram (D-take drug, D'-do not take drug, P-positive, N-negative): enter image description here

a) $P(D|P)=\frac{P(D)P(P|D)}{P(D)P(P|D)+P(D')P(P|D')}=\frac{0.0288}{0.0288+0.0194}=0.5975.$

b) $P(D'|P)=\frac{P(D')P(P|D')}{P(D')P(P|D')+P(D)P(P|D)}=\frac{0.0194}{0.0194+0.0288}=0.4025.$

c) $P(D|N)=\frac{P(D)P(N|D)}{P(D)P(N|D)+P(D')P(N|D')}=\frac{0.0012}{0.0012+0.9506}=0.0013.$

d) $P(D'|N)=\frac{P(D')P(N|D')}{P(D')P(N|D')+P(D)P(N|D)}=\frac{0.9506}{0.9506+0.0012}=0.9987.$

farruhota
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1

If you are someone looking at these kind of probabilities for the first time, it helps to first convert the data into percentages of the whole population of athletes.

Test Positive/Take Drugs 2.88% (96% of 3%)

Test Positive/No Drugs 1.94% (2% of 97%)

Test Negative/Take Drugs 0.12% (4% of 3%)

Test Negative/No Drugs 95.06% (98% of 97%)

Answer to a. An athlete testing positive takes drugs.......

The conditional part here is, given someone tests positive what portion actually takes drugs. So the percentage that test positive is 2.88 + 1.94 = 4.82%. Of these 2.88% actually take drugs so the probability is 2.88/4.82 = .598

In proper notation $P(A/B) = P(B/A)P(A)/P(B)$

Where A is taking drugs and B is testing positive

$= \frac{.96*.03}{(.96*.03)+(.97*.02)} = \frac{.0288}{.0482} = .598$

Phil H
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