False positive in probability and stats 
A cancer test is 90 percent positive when cancer is present. It gives a false positive in 10 percent of the tests when the cancer is not present. If 2 percent of the population has this cancer what is the probability that someone has cancer given that the test is positive?

I multiplied the 90 by 10 divided by 90 times 10 plus 2.

Do you know Bayes Theorem?
If not you can get a feeling for it using an "expected average tree diagram".
Imagine 1000 patients.
2% have cancer so you expect to have a split:


*

*980 cancer free

*20 with cancer


Look at the 980 cancer free. We have 10% false positives (test indicates cancer when there is none); so you expect this splits:


*

*98 test positive cancer but are cancer free

*882 test negative for cancer but are cancer free


Look at the 20 with cancer. We have 90% true positives (test indicates cancer when there is cancer); so you expect this splits:


*

*18 test positive cancer and have cancer

*2 test negative for cancer and have cancer


Therefore 98+18=116 test positive for cancer and of these only 18 have cancer. 
So the probability of having cancer, given a positive test, is small: $18/116=9/58\approx 15.52\%$.
Bayes Theorem works as follows:
$$\mathbb{P}[\text{cancer | positive}]=\frac{\mathbb{P}[\text{cancer and positive}]}{\mathbb{P}[\text{positive}]}.$$
Note 
$$
\begin{align}
\mathbb{P}[\text{positive}]&=\mathbb{P}[\text{(positive | cancer) or (positive | no cancer)}]\\&=\mathbb{P}[\text{positive | cancer}]+\mathbb{P}[\text{positive | no cancer}].
\end{align},$$
and so
$$\mathbb{P}[\text{cancer | positive}]=\frac{0.02(0.9)}{0.02(0.9)+(0.98)(0.1)}=\frac{9}{58}.$$
