Probability that someone has the disease The percentage of people that have a disease A is $0,01$. 
  We apply twice a test for that disease, each of which give the correct answer with probability $0,95$. 
  What is the probability that someone has that disease if at least one test is positive and what is the probability if both tests are positive? 


I have done the following: 

We have that $$P(\text{at least one positive})=1-P(\text{no positive})=1-P(NN)$$ where N: "negative". 

Let D:"has the disease". 

Then we have that $$P(NN)=P(NN\cap D)+P(NN\cap D^C)=P(NN\mid D)\cdot P(D)+P(NN\mid D^C)\cdot P(D^C)$$ We have that $P(D)=0,01$ and $P(D^C)=0,99$. 

Does it hold that $P(NN\mid D)=P(N\mid D)^2$ and $P(NN\mid D^C)=P(N\mid D^C)^2$ ? 

If yes, does it hold that $P(N\mid D)=0,05$ and $P(N\mid D^C)=0,95$ ?
Suppose a population of 100000.  0,01 of them, 1000, have the disease and 99000 do not.  Of the 1000 who have the disease 95%, 950, test positive and 50 test negative.

Testing the positives again, 95% of 950, 902.5 test positive again and 5%, 47.5 test negative. Testing the negatives again 95% of 50, 47.5, test negative again and 5% of 50, 2.5, test positive.

That is, 902.5 test positive twice, 50 test negative twice, and 47.5+ 2.5= 50, test positive once, negative once.

Of the 99000 who do not have the disease 5%, 4950, test positive and 95%, 94050, test negative.

Testing the positives again, 5% of 4950, 247.5, test positive again and 95% of 4950, 4702.5, test negative on the second test. Testing the negatives again, 5% of 94050, 4702.5, test positive and 95% of 94050, 89347.5, test negative again.

That is, of the 99000 who do not have the disease, 247.5 test positive twice,89347.5 test negative twice, and 4702.5+ 4702.5= 9405 test positive once, negative once.

Of the 902.5+ 247.5+ 9405+ 50= 10605 who tested positive [I]at least once, 902.5+ 50= 952.5 actually have the disease.  If a person test positive at least once, the probability they have the disease is 952.5/10605= 0.085 or 8.5%.

Of the 4950+ 247.5= 5197.5 who tested positive twice, 902.5 actually have the disease.  If a person tested positive twice, the probability they actually have the disease is 902.5/5197.5= 0.173 or 17.3%.
Suppose a population of 100000.  0,01 of them, 1000, have the disease and 99000 do not.  Of the 1000 who have the disease 95%, 950, test positive and 50 test negative.

Testing the positives again, 95% of 950, 902.5 test positive again and 5%, 47.5 test negative. Testing the negatives again 95% of 50, 47.5, test negative again and 5% of 50, 2.5, test positive.

That is, 902.5 test positive twice, 50 test negative twice, and 47.5+ 2.5= 50, test positive once, negative once.

Of the 99000 who do not have the disease 5%, 4950, test positive and 95%, 94050, test negative.

Testing the positives again, 5% of 4950, 247.5, test positive again and 95% of 4950, 4702.5, test negative on the second test. Testing the negatives again, 5% of 94050, 4702.5, test positive and 95% of 94050, 89347.5, test negative again.

That is, of the 99000 who do not have the disease, 247.5 test positive twice,89347.5 test negative twice, and 4702.5+ 4702.5= 9405 test positive once, negative once.

Of the 902.5+ 247.5+ 9405+ 50= 10605 who tested positive [I]at least once, 902.5+ 50= 952.5 actually have the disease.  If a person test positive at least once, the probability they have the disease is 952.5/10605= 0.085 or 8.5%.

Of the 4950+ 247.5= 5197.5 who tested positive twice, 902.5 actually have the disease.  If a person tested positive twice, the probability they actually have the disease is 902.5/5197.5= 0.173 or 17.3%.