The following are data regarding a mammorgram:
In the absence of any special information, the probability that a woman (of the age and health status of this patient) has breast cancer is 1%.
If the patient has breast cancer, the probability that the radiologist will correctly diagnose it is 80%.
If the patient has a benign lesion (no breast cancer), the probability that the radiologist will incorrectly diagnose it as cancer is 10%.
What is the probability that a patient with a positive mammogram actually has breast cancer?
Let $B$ indicate breast cancer and $D$ indicate a correct diagnosis. Then $P(B) = .01, \ P(\neg B) = .99, \ P(D|B) = .8, \ P(\neg D | B) = .2, \ P(\neg D | \neg B) = .1, \ P(D | \neg B) = .9$
From Baye's Rule:
$\begin{equation} \begin{split} P(B|D) &= P(D|B)\times P(B) / \big [ P(D|B)\times P(B) + P(D|\neg B) \times P(\neg B)\big] \\ &= .8 \times .01 / [.8 \times .01 + .9 \times .99] \\ &= .039 \end{split} \end{equation}$
But my textbook (Rice, Mathematical Statistics and Data Analysis 3e, p. 23) says the answer is .075. What am I doing wrong?
You define $D$ to be a correct diagnosis. That is not the same as a positive diagnosis.
So, let's use $P$ for Positive diagnosis.
Now you want to find $P(B|P)$.
You 'still' have:
$P(P|B)=.8$ and $P(\neg P|B)=.2$
But the crucial difference is:
$P(P|\neg B)=.1$ and $P(\neg P|\neg B)=.9$
OK, so try that.