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?
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 falsely write HIV-positive when he rolls a 6 and is not HIV-positive.
Pfalse
= (1/6)*(9/10)
= 9/60
= 3/20
Thus the probability Ptrue of a HIV-positive doner will tell the truth is
Ptrue
= (1/4)-(3/20)
= 2/20
=1/10
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%
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-Pfalse/P+
= 1 - (3/20) / (1/4)
= 1 - 3/5
= 2/5
To determine the probability that the person writing "HIV positive" is telling the truth, we can use Bayes' theorem. Bayes' theorem allows us to update our probability estimates based on new information.
Let's denote the following events:
A: The person is HIV positive.
B: The person writes "HIV positive".
We're interested in finding P(A|B), which is the probability that the person is HIV positive given that they write "HIV positive".
According to the given information, if the dice outcome is 6, the person will write "HIV positive" regardless of their actual HIV status. So there are two possible scenarios in which a person can write "HIV positive":
1) The person is HIV positive (A) and the dice outcome is not 6.
2) The person is not HIV positive (not A) and the dice outcome is 6.
Let's calculate the probabilities of these two scenarios:
1) P(A and not 6): This can be calculated by multiplying the probability of being HIV positive (10%) by the probability of the dice not rolling a 6 (5/6).
P(A and not 6) = 0.10 * (5/6) = 0.0833
2) P(not A and 6): This can be calculated by multiplying the probability of not being HIV positive (90%) by the probability of the dice rolling a 6 (1/6).
P(not A and 6) = 0.90 * (1/6) = 0.15
Now, let's calculate the probability of observing a person writing "HIV positive" in any of these scenarios:
P(B) = P(A and not 6) + P(not A and 6) = 0.0833 + 0.15 = 0.2333
Finally, we can use Bayes' theorem to calculate the probability that the person is HIV positive given that they write "HIV positive":
P(A|B) = (P(A) * P(B|A)) / P(B)
P(A) = 0.10 (probability of being HIV positive)
P(B|A) = 1 (since if the person is HIV positive, they will write "HIV positive" truthfully)
P(B) = 0.2333 (probability of observing someone writing "HIV positive")
P(A|B) = (0.10 * 1) / 0.2333 ≈ 0.4285
So, the probability that the person writing "HIV positive" is telling the truth is approximately 0.4285, or 42.85%.