Probability Made Easy

Another great Steven Strogartz math column in the NY Times, this one about probability:

The probability that one of these women has breast cancer is 0.8 percent. If a woman has breast cancer, the probability is 90 percent that she will have a positive mammogram. If a woman does not have breast cancer, the probability is 7 percent that she will still have a positive mammogram. Imagine a woman who has a positive mammogram. What is the probability that she actually has breast cancer?. . .

The right answer is 9 percent.

How can it be so low? . . . the analysis becomes almost transparent if we translate the original information from percentages and probabilities into natural frequencies:

Eight out of every 1,000 women have breast cancer. Of these 8 women with breast cancer, 7 will have a positive mammogram. Of the remaining 992 women who don’t have breast cancer, some 70 will still have a positive mammogram. Imagine a sample of women who have positive mammograms in screening. How many of these women actually have breast cancer?

Since a total of 7 + 70 = 77 women have positive mammograms, and only 7 of them truly have breast cancer, the probability of having breast cancer given a positive mammogram is 7 out of 77, which is 1 in 11, or about 9 percent.

As this example shows, many probability problems become much simpler when you start by imagining a complete set of possible outcomes, in this case, for 1000 women who have had mammograms. To figure out the probability of rolling an 8 on two dice -- regular six-sided dice -- you can start by listing all the possible outcomes (there are 36, 6x6) and just counting how many add up to 8.