An even more convenient example:

1% of women at age forty who participate in routine screening have breast cancer. 80% of women with breast cancer will get positive mammographies. 9.6% of women without breast cancer will also get positive mammographies. A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

inf(C) = -log2(0.01) = 6.64
iev(+,C) = inf(+) + inf(C) - inf(+ & C)
= - log2(0.01*0.8+0.99*0.096) - log2(0.01) + log2(0.01*0.8)
= 3.28 + 6.64 - 6.97 = 2.95
inf(C|+) = inf(C) - iev(+,C) = 6.64 - 2.95 = 3.69
P(C|+) = 2^(-3.69) = 7.7%

It works! That was fun.