Posted by Razor on Monday, June 22, 2009 at 6:24pm.
In a blood clinic, each donor is asked to write on a piece of paper if they are HIV positive or not. To maintain privacy, before the donor writes on the paper, he throws a sixsided dice. If the dice is 6, then he is supposed to write “HIV Positive” regardless of whether he has HIV or not. For anything else on the dice, the donor is to write the truth. Based on past blood testing, 10% of the donors are HIV positive. If you observe the person writing “HIV positive”, what is the probability that he is telling the truth?

Statistics  MathMate, Tuesday, June 23, 2009 at 1:21am
Assuming everyone follows the instructions, a donor would write HIVpositive in 10% of the cases when he rolls 15, and 100% of the cases when he rolls a 6.
P+
= (5/6)*(1/10) + (1/6)*(10/10)
= 5/60 + 1/6
= 1/4
A donor will falsely write HIVpositive when he rolls a 6 and is not HIVpositive.
P_{false}
= (1/6)*(9/10)
= 9/60
= 3/20
Thus the probability P_{true} of a HIVpositive doner will tell the truth is
P_{true}
= (1/4)(3/20)
= 2/20
=1/10

Statistics  economyst, Tuesday, June 23, 2009 at 10:58am
I respectfully disagree.
A person with HIV will answer honestly answer true in (5/6) of the time and automatically answer true in (1/6) of the time. That is every HIV person or 1/10 (10% of the population) answers true. Every nonHIV person answers true 1/6 of the time. So, (1/6)*(9/10) = 3/20 = .15% of the population.
Given that you observe a "HIVPositive", the probability that this is actually true is .1/(.1+.15) = 40%

Statistics  MathMate, Tuesday, June 23, 2009 at 11:28am
You are perfectly correct and thank you for pointing it out. 40% is the right answer. I slipped in the last step which should have read:
Thus the probability Ptrue of a HIVpositive doner will tell the truth is
Ptrue
= 1P_{false}/P_{+}
= 1  (3/20) / (1/4)
= 1  3/5
= 2/5
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