SOLUTION: Problem 4 (15%) A study considered risk factors of HIV infection among intravenous drug users. It is found that 40% of these users were HIV positive. 2 a- A group of 10 pers

Question 1182766: Problem 4 (15%)
A study considered risk factors of HIV infection among intravenous drug users. It is found that 40% of these users were HIV positive.
2
a- A group of 10 persons taking more than 100 injections per month is considered and 5 persons were found to be HIV positive. Does this number represent an excess of HIV positive users?
b- What is the probability that exactly 2 out 5 users are HIV positive?
c- If in a given community the probability of HIV positive is 0.000125, what is the probability that at least 3 out of 10000 tested persons are positive?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
a. Not clear what the 100 injections per month has to do with the answer. If we assume 40% HIV +, then 5 out of 10 would have probability 10C5(0.4)^5*0.6^5=0.2007. This would not appear to be an excessive number.
b. for 2 out of 5 it is 5C2*0.4^2*0.6^3=0.3456
c. Find the probability 0, 1, or 2 are negative and subtract that from 1.
Can do this with Poisson approach with parameter=np=1.25
Look at probability of 0 which is e^(-1.25) =0.2865
p(1) which is e^(-1.25)*1.25=0.3581
p(2) which is e^(-1.25)*1.25^2/2=0.2238
They add to 0.8684
The complement is the answer, 0.1316 probability
-
probability of 0 testing positive is 0.999875^100000=0.2865
probability of 1 testing positive is 10000*0.999875^9999*0.000125=0.3581
probability of 2 testing positive is 10000C2*999875^9998*0.000125^2=0.2238
which is the same answer as above, again subtracting the sum from 1.
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