Bayes' Theorem in Clincal Settings: Testing For Cancer a Second Time

In the problem from before the probability that a patient who tested positive once on this cancer test actually having cancer is $8 /103$ or $7.8$ %.

Bayes theorem is $P(A\mid B) = \frac{P(B \mid A) \times P(A)} {P(B) } = \frac{ P(B\mid A) \times P(A)} {P(B\mid A) \times P(A) + P(B\mid -A) \times P(-A)}$

Let's use similar notation to before $P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid -ca) \times P(-ca)}$
Where: $P(ca\mid pos)$ is the probability that the patient has cancer given a second positive result on this second test.
$P(pos \mid ca)$ is the probability that if a positive result appears on the test, then patient has cancer.
$P(ca)$ is the probability that any patient has cancer given that the patient already tested positive once. This is where the previous problem comes in. It is no longer a 1% change of having cancer, but a $7.8\%$ chance.
$P(pos\mid -ca)$ is the probability that if a a positive result appears on the test, then patient does not have cancer.
$P(-ca)$ is the probability that any patient does not have cancer given that they tested positive one the first test. Which is now no longer $99$ % but $100$ % $- 7.8$ % $= 92.2$ %

Now we can plug in the given variables:
$P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid -ca) \times P(-ca)} = \frac{(0.80) \times (0.078)} {(0.80 \times 0.078) + (0.096 \times 0.922)}$ = .41... = 41%

Which means, that after two positive results, the odds of the patient having cancer increase from 7.8% to 41%.