Page 46 - Applied Statistics And Probability For Engineers

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2-2 INTERPRETATIONS OF PROBABILITY 29

E
Diodes
Figure 2-10
Probability of the
S
event E is the sum of
the probabilities of the
outcomes in E. P(E) = 30(0.01) = 0.30

It is frequently necessary to assign probabilities to events that are composed of several
outcomes from the sample space. This is straightforward for a discrete sample space.

EXAMPLE 2-9 Assume that 30% of the laser diodes in a batch of 100 meet the minimum power requirements
of a specific customer. If a laser diode is selected randomly, that is, each laser diode is equally
likely to be selected, our intuitive feeling is that the probability of meeting the customer’s
requirements is 0.30.
Let E denote the subset of 30 diodes that meet the customer’s requirements. Because
E contains 30 outcomes and each outcome has probability 0.01, we conclude that the prob-
ability of E is 0.3. The conclusion matches our intuition. Figure 2-10 illustrates this
example.

For a discrete sample space, the probability of an event can be defined by the reasoning
used in the example above.

Definition
For a discrete sample space, the probability of an event E, denoted as P(E), equals the
sum of the probabilities of the outcomes in E.

EXAMPLE 2-10 A random experiment can result in one of the outcomes {a, b, c, d} with probabilities 0.1, 0.3,
0.5, and 0.1, respectively. Let A denote the event {a, b}, B the event {b, c, d}, and C the event
{d}.Then,

P1A2 0.1 0.3 0.4
P1B2 0.3 0.5 0.1 0.9
P1C2 0.1

Also, P1A¿2 0.6, P1B¿2 0.1 , and P1C¿2 0.9 . Furthermore, because A ¨ B 5b6,
P1A ¨ B2 0.3 . Because A ´ B 5a, b, c, d6, P1A ´ B2 0.1 0.3 0.5 0.1 1.
Because A ¨ C is the null set, P1A ¨ C2 0 .