Suppose that you are diagnosed with microscopic hematuria (blood in the urine that is only visible under a microscope). This symptom occurs in 10 percent of all people and 100 percent of people with kidney cancer. You would like to know the probability that you have kidney cancer, which occurs in 0.0002 percent of all people.
Now, when I solve it with Bayes' Theorem, the answer is as follows;
[1.0][0.000002]/[0.1] = .00002
But the thing is, what if the chance of the symptom occuring increased to 20%? Then the answer would be .00001, but shouldn't the answer also increase? What am I missing?
Going through your calculations, we have
Probability of hematuria $= P(H) = 0.10$,
Probability of kidney cancer $= P(C)=0.000002$,
Probability of hematuria given kidney cancer = $P(H\mid C) = 1$,
So, you working (which appears to be correct) shows that the probability of kidney cancer given hematuria is
$$ P(C\mid H)=\frac{P(H\mid C)p(C)}{p(H)}=\frac{1\times 0.000002}{0.1}=0.00002$$
If, however, $P(H)=0.20$, then $P(C\mid H)$ is halved to $0.00001$ which is expected.
This is because the probability of kidney cancer is the same, but you have twice as many people exhibiting the symptom of hematuria. In other words you will have more people diagnosed with hematuria but who do not have kidney cancer.
So the probability of someone having kidney cancer given that they exhibit hematuria will reduce - it will be halved, as you have twice the number diagnosed with hematuria.