Question

Consider a quick test for diabetes. It correctly detects diabetes with probability 0.7 if the patient has diabetes, and it correctly detects absence of diabetes if the patient does not have diabetes with probability 0.9.

Assuming that a random person is positively tested for diabetes and assuming that 5% of the population have diabetes, then the probability that the person indeed has diabetes is

Answer 1

We need to assign variables to events.
Let \(A\) be the event that the have diabetes.
Let \(B\) be the probability that they tested positive for diabetes.
The question is asking for \(Pr(A|B)\).

Answer 2

Remember that: \[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}
\]

Answer 3

We don't know that much about \(\Pr(B)\) yet though... We need another event:
Let \(C\) be the event that the test is correct.

Answer 4

so wat is 0.7 and 0.9 isnt it P(A|B) abd P(-A|-B)?

Answer 5

\[
\Pr(B) = \Pr(B|C)+\Pr(B|\overline{C})
\]

Answer 6

It correctly detects diabetes with probability 0.7
This is saying \[
\Pr(B|A) = 0.7
\]

it correctly detects absence of diabetes if the patient does not have diabetes with probability 0.9.
This is saying: \[
\Pr(\overline{B}|\overline{A})=0.9
\]

Answer 7

But what we want is \[
\Pr(A|B)
\]Get it?

Answer 8

this part i get it. thanks:) wat about C? how do i find it?

Answer 9

Oh, I'm not completely sure you need it... I was just throwing it out there for the moment.

Answer 10

Since \[
\Pr(B|A) = \frac{\Pr(B\cap A)}{\Pr(A)} \implies\Pr(B\cap A)=\Pr(B|A) \Pr(A) \]We plug it in to get:\[
\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)} =\frac{\Pr(B|A)\Pr(A)}{\Pr(B)}
\]

Answer 11

Of course now all we need is that nasty \(\Pr(B)\).

Answer 12

\[
\Pr(B) = \Pr(B|A)+\Pr(B|\overline{A})
\]

Answer 13

So we just need to find out when it tests positively but they don't actually have diabetes.

Answer 14

ok, i will try now. thanks :)

Answer 15

\[
\Pr(B|\overline{A})\Pr(\overline{A}) = \Pr(B\cap\overline{A} )\]\[
\Pr(\overline{B}|\overline{A})\Pr(\overline{A}) = \Pr(\overline{B}\cap\overline{A} )\]\[
\Pr(\overline{A})=\Pr(B\cap\overline{A} )+\Pr(\overline{B}\cap\overline{A} )
\]

Answer 16

P(dia|+)={P(dia)*P(+|dia)} / {P(dia)*P(+|dia)+P(-dia)*(P(+|-dia)}

Answer 17

is this way correct?

Answer 18

\[
\Pr(B|\overline{A})\Pr(\overline{A})=\Pr(\overline{A})-\Pr(\overline{B}|\overline{A})\Pr(\overline{A})\]\[
\Pr(B|\overline{A})=1-\Pr(\overline{B}|\overline{A})
\]

Answer 19

So I'm getting: \[
\Pr(A|B) = \frac{\Pr(B|A)\Pr(A)}{\Pr(B|A)+(1-\Pr(\overline{B}|\overline{A}))}
\]

Answer 20

Is there a way to check your answer?

Answer 21

i'll get back to you when i get back my assignment. thanks so much:)