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52 CHAPTER 2 PROBABILITY

EXAMPLE 2-29 We can answer the question posed at the start of this section as follows: The probability
requested can be expressed as P1H 0 F2. Then,

P1F ƒ H2P1H2 0.1010.202
P1H ƒ F2 0.85
P1F2 0.0235

The value of P(F) in the denominator of our solution was found in Example 2-20.

In general, if P(B) in the denominator of Equation 2-11 is written using the Total
Probability Rule in Equation 2-8, we obtain the following general result, which is known as
Bayes’Theorem.

Bayes’
Theorem If E , E , p , E k are k mutually exclusive and exhaustive events and B is any
1
2
event,
P1B ƒ E 1 2P1E 1 2
ƒ B2 (2-12)
P1E 1 p
P1B ƒ E 2P1E 2 P1B ƒ E 2P1E 2 P1B ƒ E 2P1E 2
2
k
1
1
k
2
for P1B2 0
EXAMPLE 2-30 Because a new medical procedure has been shown to be effective in the early detection of an
illness, a medical screening of the population is proposed. The probability that the test cor-
rectly identifies someone with the illness as positive is 0.99, and the probability that the test
correctly identifies someone without the illness as negative is 0.95. The incidence of the
illness in the general population is 0.0001. You take the test, and the result is positive. What is
the probability that you have the illness?
Let D denote the event that you have the illness, and let S denote the event that the test
signals positive. The probability requested can be denoted as P1D ƒ S2 . The probability that the
test correctly signals someone without the illness as negative is 0.95. Consequently, the prob-
ability of a positive test without the illness is

P1S ƒ D¿2 0.05

From Bayes’Theorem,

P1D ƒ S2 P1S ƒ D2P1D2
3P1S ƒ D2P1D2 P1S ƒ D¿2P1D¿24
0.9910.00012
30.9910.00012 0.0511 0.000124
1
506 0.002

Surprisingly, even though the test is effective, in the sense that P1S 0 D2 is high and
P1S ƒ D¿2 is low, because the incidence of the illness in the general population is low, the
chances are quite small that you actually have the disease even if the test is positive.