Question 1182766
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 
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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.