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I have this problem, and I have solved it but I'm confused about some points.

HIV infected approximately 1,500 people in country X last year1. Mark heard about HIV, so he made a test for the HIV. The test correctly identifies the presence of HIV in 80% of cases and only gives false positives in 1/10,0000 cases. In Mark's case, the test was positive for HIV and Mark is very concerned. Assume that in the population of X country there are 20 million people susceptible to HIV.

  1. What is the probability that Mark has HIV?
  2. The HIV virus is fatal in 4% of the cases. What is the probability that Mark will die this year? Assume a fatality rate of any cause (car accident, etc.) of 0.15% that is independent of whether or not Mark has HIV.

I believe that:
probability that Mark has HIV, P(HIV) = 1,500/20,000,000 = 0.000075

Hence, P(~HIV) = 1- 0.000075 = 0.999925

let Pos is test is positive

P(Pos|HIV) = 0.80

P(Pos|~HIV)= 1/10,0000 = 0.00001

For question 1, I believe it wants me to find P(Mark has HIV|test is positive)?

My solution:

P(Mark has HIV|test is positive) = P(HIV|Pos)

P(HIV|Pos) = P(Pos|HIV).P(PHIV) / P(Pos)

= P(Pos|HIV).P(HIV) / P(Pos,HIV) + P(Pos,~HIV)

= P(Pos|HIV).P(HIV) / P(Pos|HIV).P(HIV) + P(Pos|~HIV).P(~HIV)

= 0.80 * 0.000075 / (0.80 * 0.000075 + 0.00001 * 0.999925)

= 0.857

For question 2:

the probability that Mark will die this year should be as follows: P(Mark has HIV|test is positive) * [P(Mark dies|Mark has HIV) + P(Mark dies|Accident)] + (1 - P(Mark has HIV|test is positive)) * P(Mark dies|Accident)

= 0.857 * (0.4 + 0.0015) + (1- 0.857) * 0.0015

= 0.3443

My Questions:
For Question 1, I feel that I miss something. Am I?

For Question 2, my question is that do I have to calculate the probability of both cases if Mark has HIV and If he Doesn't have HIV.

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  • $\begingroup$ @ssdecontrol, it is best not to add the [self-study] tag for the OP, but to ask the to do it themselves & to read its wiki (you can find example text here). That way, there is a better chance they will be familiar with our policy. If the OP doesn't add the tag w/i a suitable time, we can close. (Note that this Q does meet our requirements otherwise, though.) $\endgroup$ – gung Jan 22 '16 at 13:21
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    $\begingroup$ @gung indeed, that's what I should've done (and usually do). I jumped to conclusions here because the question seemed like it was copied directly out of the textbook $\endgroup$ – shadowtalker Jan 22 '16 at 13:23
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    $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung Jan 22 '16 at 13:25
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    $\begingroup$ "Mark heard about HIV, so he made a test for the HIV". I think that deserves some thought. $\endgroup$ – Scortchi Jan 22 '16 at 13:52
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Your steps in question one look right, but I'm also skimming on my phone so I could have missed something (I also didn't check your math). Also, the answer seems sensible.

You're overthinking question two. Treat $P(\mathrm{Death})$ as a base rate, like $P(\mathrm{HIV})$, and use the same technique you used in question one.

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