Page 77 - Applied statistics and probability for engineers

Section 2-7/Bayes’ Theorem 55

Example 2-36 Reconsider Example 2-27. The conditional probability that a high level of contamination was present
when a failure occurred is to be determined. The information from Example 2-27 is summarized here.
Probability of Failure Level of Contamination Probability of Level
0.1 High 0.2
0.005 Not high 0.8
(
|
The probability of P H F) is determined from . )
. (0 20
(
|
)
(
P H F) = P F H P H) = 0 10 = . 0 83
(
|
P F)
(
. 0 024
The value of P F( ) in the denominator of our solution was found from P F( ) = ( P F H P H ( ) + ( | ′) (
)
P F H P H′).
P F) = ( |
In general, if P B( ) in the denominator of Equation 2-15 is written using the total probability
rule in Equation 2-12, we obtain the following general result, which is known as Bayes’ theorem.
Bayes’ Theorem
…
If E ,E , ,E k1 2 are k mutually exclusive and exhaustive events and B is any event,
P B E P E 1 ( )
| (
1 (
P E B) = 1) (2-16)
|
k)
(
| (
| (
1)
P B E P E 2 ( ) + … +
P B E P E 1 ( ) + ( | 2) P B E P E k ( )
for P B ( ) > 0
Notice that the numerator always equals one of the terms in the sum in the denominator.
Example 2-37 Medical Diagnostic 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 correctly identiies someone with the illness as positive is 0.99, and the probability that the test correctly identi-
ies 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 P D S| ). The probability that the test correctly signals someone without the illness as nega-
tive is 0.95. Consequently, the probability of a positive test without the illness is
| (
P S D′) = .05
0
From Bayes’ theorem,
(
(
(
(
P D S) = P S D P D) /[ P S D) ( D) + P S D P D′)]α
′
(
(
)
)
|
|
|
P
|
.
.
= . 0 99 ( .0001 [ .99 0 0001) + . 0 05(1− 0 0001)]
(
) /
0
0
[
= / 1 506 = 0 002
.
Practical Interpretation: The probability of your having the illness given a positive result from the test is only 0.002. Surpris-
| (
(
ingly, even though the test is effective, in the sense that P S D) is high and P S D| ′) 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.
Example 2-38 Bayesian Network Bayesian networks are used on the Web sites of high-technology manufac-
turers to allow customers to quickly diagnose problems with products. An oversimplii ed example
is presented here.
A printer manufacturer obtained the following probabilities from a database of test results. Printer failures are asso-
ciated with three types of problems: hardware, software, and other (such as connectors) with probabilities of 0.1, 0.6,
and 0.3, respectively. The probability of a printer failure given a hardware problem is 0.9, given a software problem is
0.2, and given any other type of problem is 0.5. If a customer enters the manufacturer’s Web site to diagnose a printer
failure, what is the most likely cause of the problem?