Independence and application: the medical test

Defining independence

Two events A and B are independent if the occurrence of one changes nothing about the probability of the other, that is if P(A given B) = P(A). In this case, the multiplication rule simplifies: P(A intersection B) = P(A) x P(B).

Application: a medical test

We test for a disease D whose prevalence in the population is P(D) = 1/100 (1 percent). The test has a sensitivity of 99 percent (P(test positive given ill) = 0.99) and a false-positive rate of 5 percent (P(test positive given not ill) = 0.05).

Probability tree of the medical test:

                       0.99             (test +  given D)
                 /----------- +
                /
     D (0.01) --
                \
                 \----------- -        0.01              (test -  given D)

                       0.05             (test +  given not-D)
                 /----------- +
                /
 not-D (0.99) --
                \
                 \----------- -        0.95              (test -  given not-D)

(D = ill, not-D = not ill; + = test positive, - = test negative)
(each pair of branches sums to 1: 0.99+0.01=1 and 0.05+0.95=1)

Computing the probability of being ill given a positive test

We first compute P(test positive) with the law of total probability:

P(+) = P(D) x P(+|D) + P(not-D) x P(+|not-D) = 0.01 x 0.99 + 0.99 x 0.05 = 0.0099 + 0.0495 = 0.0594

Then we apply the definition of conditional probability:

P(D given +) = P(D intersection +) / P(+) = 0.0099 / 0.0594 = 1/6, i.e. about 16.7 percent.

Common pitfall

Even with a very sensitive test (99 percent), the probability of actually being ill given a positive test stays low (only about 16.7 percent) when the disease is rare: this is because the false positives, applied to a huge population of non-ill people, end up outnumbering the true positives. Confusing P(+ given D), which is 99 percent, with P(D given +), which is only 16.7 percent, is the central pitfall of this lesson.