So, this is a problem that we'll utilize Bayes Theorem that we've already given.

The probability that the event B will occur,

given that the even A has already occurred when they're both dependent,

is given by this equation which we've already had.

So, to use that, let's suppose that A1 is

the event that the individual does have the disease,

and A2 is that the individual does not have

the disease, and B is a positive test result.

So, then we have the probability of A1, in other words,

the probability that the person does have the disease

is 1 in 1000 In other words, .001.

On the other hand, the probability of A2,

in other words the probability that the individual

does not have the disease, is one minus that, or

point 0.999 and the probability of B occurring

when A1 has occurred is 99%, this one here.

In other words, the probability of B when A1 has occurred is .99,

probability of a positive result when the person does have the disease.

Conversely, the probability of B occurring

when A2 occurs is .02, 2% of the time.

In other words, the probability of a positive

result when the person does not have the disease is .02.

So now we can substitute these values into our basic equation for

Bayes Theorem which then looks like this.

Probability of A1 is .001 P B A2, is 0.99,

etc, .99, substituting in the numbers,

the answer is 0.0472, and the answer is C.