Applied Statistics and Probability for Engineers

52 CHAPTER 2 PROBABILITY

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 Ddenote the event that you have the illness, and let Sdenote the event that the test

signals positive. The probability requested can be denoted as. 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

From Bayes’Theorem,

Surprisingly, even though the test is effective, in the sense that is high and

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.

P 1 SƒD¿ 2

P 1 S^0 D 2

 1
506 0.002

0.99 1 0.0001 2
3 0.99 1 0.0001 2 0.05 11 0.0001 24

P 1 DƒS 2 P 1 SƒD 2 P 1 D 2
3 P 1 SƒD 2 P 1 D 2 P 1 SƒD¿ 2 P 1 D¿ 24

P 1 SƒD¿ 2 0.05

P 1 DƒS 2

EXAMPLE 2-29 We can answer the question posed at the start of this section as follows: The probability

requested can be expressed as Then,

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.

P 1 HƒF 2 

P 1 FƒH 2 P 1 H 2

P 1 F 2



0.10 1 0.20 2

0.0235

0.85

P 1 H^0 F 2.

If are k mutually exclusive and exhaustive events and Bis any

event,

(2-12)

for P 1 B 2  0

P 1 E 1 ƒB 2 

P 1 BƒE 12 P 1 E 12

P 1 BƒE 12 P 1 E 12 P 1 BƒE 22 P 1 E 22 pP 1 BƒEk 2 P 1 Ek 2

E 1 , E 2 , p, Ek

Bayes’

Theorem

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