The Bayesian problem that 95 out of 100 physicians get wrong

Bayes theorem is $P(A\mid B) = \frac{P(B \mid A) \times P(A)} {P(B) } = \frac{ P(B\mid A) \times P(A)} {P(B\mid A) \times P(A) + P(B\mid not A) \times P(not A)}$

In this case, lets borrow James Eddy's own notation to state this as: $P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid benign) \times P(benign)}$
Where: $P(ca\mid pos)$ is the probability that the patient has cancer given a positive result on the test (the posterior possibility).
$P(pos \mid ca)$ 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 (the prior probability).
$P(pos\mid benign)$ is the probability that if a a positive result appears on the test, then patient does not have cancer, that any cancer is benign.
$P(benign)$ is the probability that any patient does not have cancer, that any tumors they have are benign (the prior probability of not having cancer).

Now we can plug in the given variables:
$P(ca\mid pos) = \frac{ P(pos \mid ca) \times P(ca)} {P(pos\mid ca) \times P(ca) + P(pos\mid benign) \times P(benign)} = \frac{(0.80) \times (0.01)} {(0.80 \times 0.01) + (0.096 \times 0.99)}$ = .078 = 7.8%

Which means, that even if a patient gets a positive result back on this particular test, in this particular scenario where 1% of the population has cancer, then they still only have a 7.8% cancer of actually having cancer.

In his informal study, 95 out of 100 physicians listed the probability as between 70% and 80%, similar to the prior probability, but one order of magnitude off.