# A cancer test is 90 percent positive when cancer is present. It gives a false positive in 10 percent

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.
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Keely Fernandez
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:
$18/116=9/58\approx 15.52\mathrm{%}$.
Bayes Theorem works as follows:
$\mathbb{P}\left[\text{cancer | positive}\right]=\frac{\mathbb{P}\left[\text{cancer and positive}\right]}{\mathbb{P}\left[\text{positive}\right]}.$
Note
$\begin{array}{rl}\mathbb{P}\left[\text{positive}\right]& =\mathbb{P}\left[\text{(positive | cancer) or (positive | no cancer)}\right]\\ & =\mathbb{P}\left[\text{positive | cancer}\right]+\mathbb{P}\left[\text{positive | no cancer}\right].\end{array},$
and so
$\mathbb{P}\left[\text{cancer | positive}\right]=\frac{0.02\left(0.9\right)}{0.02\left(0.9\right)+\left(0.98\right)\left(0.1\right)}=\frac{9}{58}.$

prirodnogbk
Let A be the event that someone has cancer.
$P\left(A\right)=0.02$
Let B be the event that the cancer test returns positive. There are four cases to consider:
- Test positive, cancer present: $P\left(B|A\right)=0.9$
- Test negative, cancer present: $P\left({B}^{c}|A\right)=0.1$
- Test positive, cancer not present: $P\left(B|{A}^{c}\right)=0.1$
- Test negative, cancer not present: $P\left({B}^{c}|{A}^{c}\right)=0.9$
The two probabilities are the same here, we will need $P\left(B|A\right),P\left(B|{A}^{c}\right)$, which are given in the question.
We need to find $P\left(A|B\right)$.
Bayes' Theorem tells us that:
$P\left(A|B\right)=\frac{P\left(B|A\right)P\left(A\right)}{P\left(B\right)}$
and conditional probabilities tell us:
$\begin{array}{rl}P\left(B\right)& =P\left(B|A\right)P\left(A\right)+P\left(B|{A}^{c}\right)P\left({A}^{c}\right)\\ & =0.9×0.02+0.1×0.98\\ & =0.018+0.098\\ & =0.116\end{array}$
So $P\left(B|A\right)=\frac{0.018}{0.116}\approx 0.15517$