A medical test has a 95% accuracy of detecting a Condition Z if the person has it. It also has a 97% chance to indicate that the person does not have the condition if they really don't have it. If the incidence rate of this disease is 10 out of every 100:

What is the probability that a person chosen at random will both test positive and actually have the disease (i.e., get a true positive)?

Question

A medical test has a 95% accuracy of detecting a Condition Z if the person has it. It also has a 97% chance to indicate that the person does not have the condition if they really don't have it. If the incidence rate of this disease is 10 out of every 100:

What is the probability that a person chosen at random will both test positive and actually have the disease (i.e., get a true positive)?

anonymous · Answer

Events are going to be
A - tests positive  (false positive or true positive)
B - has desease

anonymous · Answer

And the question asks for $$
\Pr(A\cap B)
$$

anonymous · Answer

"95% accuracy of detecting a Condition Z if the person has it"
This means $$
\Pr(A|B) = 0.95
$$

anonymous · Answer

"incidence rate of this disease is 10 out of every 100"$$
\Pr(B) = \frac{10}{100}
$$

anonymous · Answer

"It also has a 97% chance to indicate that the person does not have the condition if they really don't have it"$$
\Pr(A^C|B^C) = 0.97
$$

anonymous · Answer

Actually, I'm a bit unsure about parts of this, because it seems t hey game too much information.

anonymous · Answer

Well there was another question: What is the probability that a person chosen at random will test positive but not have the disease (i.e., get a false positive)?

But I left it out because I figured once I figured out how to do it I'd be able to figure the 2nd one

anonymous · Answer

Oh, ok.

anonymous · Answer

Do you know this formula: $$
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
$$

anonymous · Answer

We can use it to solve for $\Pr(A\cap B)$.

anonymous · Answer

I only know how to do intersections when there's multiple numbers

anonymous · Answer

Sets I mean

anonymous · Answer

Do you understand the formula I gave?

anonymous · Answer

Yeah, but I'm not sure how to do the intersection when it's only 2 numbers

anonymous · Answer

Well $$
\Pr(A\cap B)
$$means probability of event A and B.

$$
\Pr(A|B)
$$means probability of A if we know B is true

anonymous · Answer

$$
\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}
$$We can multiply by $\Pr(B)$ and then: $$
\Pr(A\cap B) = \Pr(A|B)\times \Pr(B)
$$

anonymous · Answer

But how do I do the upside down U part?

anonymous · Answer

Are you talking about $$
\Pr(A\cup B) = \Pr(A) +\Pr(B) - \Pr(A\cap B)
$$

anonymous · Answer

The Pr(A *Flipped U* B)

anonymous · Answer

Oh, you do `$\cap$` $\cap$

anonymous · Answer

Well, how do I do that part of the equation though

anonymous · Answer

The intersection part

anonymous · Answer

I think it's 9.5%?

anonymous · Answer

Yeah, that looks right.

anonymous · Answer

Thanks :)