Question

Suppose that 6% of the people in the world have a particular genetic defect, and that a screening test is 89% accurate for people who have it and 76% accurate for people who do not.
If 1,810 people are screened for the defect, which is the best prediction for the number of people with the defect who are identified as having it?

Answer 1

Let \(D\) denote the event that a person has the defect, and \(D'\) the event that a person does not, so \(P(D)=0.06\) and \(P(D')=1-P(D)=0.94\).

Let \(A\) denote the event that the test is positive, and \(A'\) that it's negative. The test is 89% accurate for those that have the defect, meaning \(P(A|D)=0.89\); the test is 76% accurate for people without the defect, meaning \(P(A|D')=0.76\).

You want to find the probability \(P(D|A)\). Let me just look this over to make sure I'm understanding the question properly...

Answer 2

this confused me more than i already was lol

Answer 3

Have you learned Bayes' formula? I'm pretty sure you'll need it for this question.

Answer 4

And in case it wasn't clear, \(P(A|D)\) is the probability of \(A\) given \(D\), i.e. the probability of a person getting a positive result on the test knowing that he/she has the defect.

Answer 5

no i have to take online classes and i dont have anyone to teach me any of it

Answer 6

This video might help wrap your mind around the problem at hand: http://www.youtube.com/watch?v=JGeTcRfKgBo

I haven't watched the entire thing, so I don't know if they do any examples.

Answer 7

thank you

Answer 8

In any case, Bayes' formula states that
\[P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A')P(A')}\]
Then
\[\begin{align*}P(D|A)&=\frac{P(A|D)P(D)}{P(A|D)P(D)+P(A|D')P(D')}\\\\
&=\frac{0.89\times0.06}{0.89\times0.06+0.76\times0.94}\\\\
&\approx0.0695
\end{align*}\]