Page 68 - Applied statistics and probability for engineers

46 Chapter 2/Probability

Sometimes the probability of an event is given under each of several conditions. With enough
of these conditional probabilities, the probability of the event can be recovered. For example,
suppose that in semiconductor manufacturing, the probability is 0.10 that a chip subjected to
high levels of contamination during manufacturing causes a product failure. The probability
is 0.005 that a chip not subjected to high contamination levels during manufacturing causes a
product failure. In a particular production run, 20% of the chips are subject to high levels of
contamination. What is the probability that a product using one of these chips fails?
Clearly, the requested probability depends on whether or not the chip was exposed to high
levels of contamination. For any event B, we can write B as the union of the part of B in A and
the part of B in A′. That is,
B = ( A∩ B)∪( A ∩ B)
′
This result is shown in the Venn diagram in Fig. 2-15. Because A and A′ are mutually exclu-
sive, A∩ B and A′ ′ B are mutually exclusive. Therefore, from the probability of the union
of mutually exclusive events in Equation 2-6 and the multiplication rule in Equation 2-10,
the following total probability rule is obtained.
Total Probability Rule
(Two Events)
For any events A and B,
)
P B ∩
P B A P A ( ) ′|
P B ∩ ) = ( |
P B ( ) = ( A) + ( A′ P B A P A ( ) + ( ) ′ (2-11)
Example 2-27 Semiconductor Contamination Consider the contamination discussion at the start of this sec-
tion. The information is summarized here.

Probability of Level of Probability
Failure Contamination of Level
0.1 High 0.2
0.005 Not high 0.8
Let F denote the event that the product fails, and let H denote the event that the chip is exposed to high levels of con-
tamination. The requested probability is P F( ), and the information provided can be represented as
( ( ′
P F H| ) = . 0 10 and P F H ) = .005
|
0
(
P H ( ) = . 0 20 and P H ) = .0 80′
From Equation 2-11,
0 10
20
0
P F ( ) = . (0 . ) + .005 . (0 80 ) = .0 024
which can be interpreted as just the weighted average of the two probabilities of failure.
The reasoning used to develop Equation 2-11 can be applied more generally. Because
(
′
A∪ A′ = S, we know A∩ B)∪( A ∩ B) equals B, and because A∩ A′ = φ, we know A∩ B
and A′ ′ B are mutually exclusive. In general, a collection of sets E ,E , ,E k1 2 … such that
E 1 ∪ E 2 ∪…∪ E k = S is said to be exhaustive. A graphical display of partitioning an event B
among a collection of mutually exclusive and exhaustive events is shown in Fig. 2-16.
Total Probability
Rule (Multiple …
Events) Assume E ,E , ,E k1 2 are k mutually exclusive and exhaustive sets. Then
(
P B ∩ ) + (
P B ( ) = ( E 1 P B ∩ ) + … + P B ∩ )
E 2
E k
+ (
k)
)
P B E P E ( ) + ( |
|
= ( | 1 ) 1 P B E P E ( ) + … + P B E P E k ( ) (2-12)
2
2