problem solving by conditional probability

Question

 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?

 

Answer 1

The fast way of solving this question is draw a flow chart:

 

                                         NFL Players

 

                 Users (3%)                             Non-users (100-3 = 97%)

 

positive(93%)         negative(7%)      negative(98%)                 positive (2%) 

 

 

From this, you can map out the question:

 

a. A player will test negative. (they didn't identify if it's an actual user or not so use both "users" and "non-users")

 

.03*.07 + 0.97*.98 = 0.9527 = 95.27%

 

b. "actually using drugs" so it's under "Users" and positive

 

.03*.93 = .0279 = 2.79% (Why is it too low? Imagine you are guessing twice. You need to guess from a pool of NFL players which one uses drugs and you have 3% to guess correctly. Then let's say you got lucky. The tester still has a chance to fail since it has 93% success so the 3% will be lowered further.)

Answer 2

Use a probability tree diagram.  Let A be the event that the NFL athlete is using drugs then A' is the event that the athlete is not using drugs.  Let B be the event that the test shows a positive result then B' is the event that the test shows a negative result.  This problems involves using Bayes' Theorem or Rule which is easily done using a probability tree diagram.  There are formulas but they may be somewhat confusing to use.

 

Note that the probabilities going vertical must add us to 1.  Use the multiplication rule to find the probability along a branch and us the total probability rule to find the probability of a single event.

Answer 3

I made a 'tree' emanating from a point at the left side of a sheet of paper.  From that point, I drew two branches,  heading at ±45^o relative to the horizontal. The upper one is labeled Y (uses drugs), .03; the lower one is labeled N, .97.

 

From the end point of the upper branch, I wrote two additional two branches, at ±45^o relative to the horizontal--labeled + .93 and - .07, representing the probabilities of + or - results of a drug test.

 

From the end point of the lower branch at the left side of the paper, I wrote two additional two branches, at ±45^o relative to the horizontal--labeled + .02 and - .98, representing the probabilities of + or - results of a drug test.

 

With this data, we can answer the (b) question, P(user|test is positive). This requires Bayes' Theorem; my answer is .03(.93) ÷ [.03(.93) + .97(.02)] = .5899