Probability of testing positive

if we reframe the question

consider that in a population of 100000 only 500 people have the disease. Hence the probablity of a person having a disease is still 0.005

Therefore the people having the disease is 500 and people not having the disease is 99500

now since the test is, 95% of the time positive for people having the disease, it means out of 500 people 475 will be tested positive.

also since the test is positive 1% of the time for people not having the disease, it means 995 people out of 99500 people will be tested positive.

therefore total number of people tested positive 995+475 = 1470. But the number of people actually having the disease is just 475.

Therefore the probablity of having it if tested positive is 475/140 = 95/294

Let P(B) the probability that really has the disease. Then P(B) = 0.005.

Let P(B') the probability that really has not the disease. Then P(B') = 0.995.

Let P(A) the probability that when tested gives positive. Unknown.

Let P( A | B ) the probability that a person gives positive for the disease given that the person really has the disease. That is, P(A | B) = 0.95.

Let P( A | B' ) the probability that the person gives positive for the disease given that the person doesn't really has the disease.

Let P( B | A ) the probability that a person really has the disease given that when tested gives positive by the device. To find this probability is our goal.

Using the definition of conditional probability:

$P(B|A)=\frac{ P(B\cap A) }{ P(A) }$

Let's find P(A) using

$P(A) = P((A\cap B) \cup (A \cap B')) = P(A\cap B) + P (A \cap B') - P((A\cap B) \cup (A \cap B'))$

where the last term is equal to zero by using Venn diagrams.

Substituting with all the extra information we have already, we get:

$P(B|A) = \frac{P(A | B) P(B)}{P(B) P(A|B) + P(B')P(A|B')}$

which gives 95/294.

using probability distribution..................... p(E) = ( 0.005 * (95/100) ) / ( 0.005 * (95/100) + 0.995 * (1/100) ) = (95/294)