A new test has been developed to determine whether a given student is overstressed. This test is $95\%$ accurate if the student is not overstressed, but only $85\%$ accurate if the student is in fact overstressed. It is known that $99.5\%$ of all students are overstressed. Given that a particular student tests negative for stress, what is the probability that the test results are correct, and that this student is not overstressed?
I figure that $A=${accurate test result}, $S=${student is stressed}, and $N=${Negative test result}, so $P(A|S^c)=.95$, $P(A|S)=.85$, and $P(S)=.995$.
I'm not sure exactly what I'm supposed to solve for for. I think I'm trying to solve for $P(A|N)$, and the fact that this student is not overstressed just tells me how accurate the results are. The thing I don't know how to get is any of the probabilities surrounding a negative test result.
I would greatly appreciate some help.
Hint: Accuracy is a compound of the events of test result and stress state. It is better not to consider it a seperate event. Rather think about what an accurate results says is about the student when given the result, and vice versa. When the result is accurate and negative, that says that the student is unstressed. When the result is accurate and the student is stressed that says that the result is positive.
$A=(S\cap N^\complement)\cup(S^\complement\cap N)\\A\cap N=S^\complement\cap N\\A\cap S=S\cap N^\complement\\\ddots\text{et cetera}$
$$\begin{align}\mathsf P(A\mid N)~&=~\mathsf P(S^\complement\mid N) \\[2ex]&=~\dfrac{\mathsf P(S^\complement)~\mathsf P(N\mid S^\complement)}{\mathsf P(S^\complement)~\mathsf P(N\mid S^\complement)+\mathsf P(S)~\mathsf P(N\mid S)}\\[2ex]&=~\dfrac{(1-0.995)~0.95}{(1-0.995)~0.95+0.995~(1-0.85)}\end{align}$$