Question

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 six-sided 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?

Answer 1

Assuming everyone follows the instructions, a donor would write HIV-positive in 10% of the cases when he rolls 1-5, 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 <b>falsely</b> write HIV-positive when he rolls a 6 and is not HIV-positive.
P_false
= (1/6)*(9/10)
= 9/60
= 3/20

Thus the probability P_true of a HIV-positive doner will tell the truth is
P_true
= (1/4)-(3/20)
= 2/20
=1/10

Answer 2

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 non-HIV person answers true 1/6 of the time. So, (1/6)*(9/10) = 3/20 = .15% of the population.

Given that you observe a "HIV-Positive", the probability that this is actually true is .1/(.1+.15) = 40%

Answer 3

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 HIV-positive doner will tell the truth is
Ptrue
= 1-P_false/P₊
= 1 - (3/20) / (1/4)
= 1 - 3/5
= 2/5