Probability problem on a cancer test. There is a test to find out if a person has cancer.
1.) 95% of people who will take this test AND have cancer will get a positive ID that they have cancer.
2.) 2% of people who take this test who DON'T have cancer will also get a positive ID that they have cancer.
3.) The chances that a person who is taking this test has cancer is 8%.
What is the probability for a person to have cancer given that he will take the test and get a positive result.

I don't know how to tackle this problem... I assumed that since we don't know if person actually has cancer or not, that he would fall under both 95% of the first point and under 2% of people who don't have cancer but still have positive ID. 
Therefore,
95% * 2% = 1.9%
But I don't see how the third point ties it all up...
 A: Point $1$ describes the true positive rate. Point $2$ describes the false positive rate. Point $3$ describes the overall prevalence. 
The simplest way to solve this problem is to build a $2\times 2$ table. For instance, let us assume that we are screening a population of $2500$ subjects (this number can be taken arbitrarily, since we are interested in percentages). Since the prevalence is $8\%$, we have $200$ cancer subjects. Because the true positive rate is $95\%$, we have that the test is positive in $190$ of them, and negative in the remaining $10$. On the other hand, because the false positive rate is $2\%$, we have that, among the $2300$ healthy subjects, $46$ of them have a positive test. So the total number of subjects with positive test is $190+46=236$, and the probability for a subject to have cancer given a positive result (that is to say, the positive predictive value) is $\displaystyle \frac{190}{236}\approx 80.5\%$.
A: This is an application to Baye's Theorem stating that $P(A|B)=\frac{P(A)\cdot P(B|A)}{P(B)}$.
In your example, let $A$ represent the event that the person has cancer and let $B$ represent the event that they test positive for cancer.
We know that $P(B|A) = P$(They tested positive if they have cancer) = $.95$ and that
$P(A) = P($They have cancer) = $.08$.
The only piece of information we are missing is $P(B)$ which can be calculated by doing $P(B) = P(B\cap A) + P(B\cap A^c) = P(A)\cdot P(B|A) + P(A^c)\cdot P(B|A^c) = .08\cdot .95 + .92\cdot .02 = .0944$
Plugging these values in, we get $P(A|B) = \frac{P(A)\cdot P(B|A)}{P(B)} = \frac{.08\cdot .95}{.0944}\approx .8051$
