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March 28, 2015

March 28, 2015

Posted by **Razor** on Monday, June 22, 2009 at 6:24pm.

- Statistics -
**MathMate**, Tuesday, June 23, 2009 at 1:21amAssuming 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**falsely**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

- Statistics -
**economyst**, Tuesday, June 23, 2009 at 10:58amI 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%

- Statistics -
**MathMate**, Tuesday, June 23, 2009 at 11:28amYou 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

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