Conditional probability problem - how did they arrive at this number?

[Rigamarole]

I am reading an essay about common statistical errors, and they give this example to illustrate confusion over conditional probabilities:

As an example, Eddy (1982) presented this scenario to 100
physicians:

One of your patients has a lump in her breast. You are almost certain
that it is benign, in fact you would say there is only a 1% chance that
it is malignant. But just to be sure, you have the patient undergo a
mammogram, a breast X-ray designed to detect cancer.

You know from themedical literature that mammograms are 80% accurate
for malignant lumps and 90% accurate for benign lumps. In other
words, if the lump is truly malignant, the test results will say that it is
malignant 80% of the time and will falsely say it is benign 20% of the
time. If the lump is truly benign, the test results will say so 90% of the
time and will falsely declare that it is malignant only 10% of the time.
Sadly, the mammogram for your patient is returned with the news that
the lump is malignant. What are the chances that it is truly malignant?

Most of the physicians responded with an answer close to
75%. But in fact, given the probabilities presented, the correct
answer is only about 7.5%!

Now I understand the concept of the problem and how people made an error in assessing the probability. The stupidity of the average physician aside, how did they arrive at the correct probability being “about 7.5%?” I’m thinking it should be almost 10%. Is this just a case of extreme rounding or am I missing something?

If we have a 0.1 probability of a benign tumor returning a result of malignant, and a 0.8 probability of a malignant tumor returning a result of malignant, and a 0.01 chance that the tumor is actually malignant, then shouldn’t the probability that it is actually malignant in light of the test result be (0.1 * 0.99) - (0.8 * 0.01) = 0.091?

[Rigamarole]

OK, I figured it out. The correct formula seems to in fact be (0.8 * 0.01) / (0.1 * 0.99 + 0.8 * 0.1) = 0.0748

[ultrafilter]

This is a standard illustration of Bayes’ theorem.

[Indistinguishable]

Right, the conditional probability of A given B [aka, P(A | B)] is the same as the probability of (A and B) divided by the probability of B. This tells us that P(actually malignant | test says malignant) = P(actually malignant and test says malignant)/P(test says malignant) = P(actually malignant and test says malignant)/(P(actually malignant and test says malignant) + P(not actually malignant but test says malignant)). Then expanding out these probabilities using the provided data gives you the answer you wrote.

[Indistinguishable]

Put another way, which is perhaps more intuitively obvious though of course entirely equivalent, the probability of two events occurring is the probability of the first one * the conditional probability of the second one given the first one

[HardlySanguine]

Since the question’s been answered … When I read the problem, I was assuming the 1% was just a matter of experience (most lumps are benign) and the mammogram was a separate test.

So, I think this is probably a bad example question to test physicians’ knowledge.