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Chapter 7: Probability
273
Solution:
Recall that the sample space was
S
= {
BB
,
BG
,
GB
,
GG
}. From
this sample space, the event of one boy occurs twice, and there are 4 simple
events in the sample space. Thus,
P
(one boy) =
P
(
BG
or
GB
) = 2/4 = 0.5.
Example 7-5:
Given that three items are selected from a manufacturing
process to check whether an item may be classified as defective or non-
defective. What is the probability of observing two defective items?
Let
D
represent the event of a defective item, and let
N
represent the event of
a nondefective item. List the possible outcomes for the sample space.
Solution:
The possible points in the sample space are given in the set
S
.
S
= {
DDD
,
DDN
,
DND
,
DNN
,
NDD
,
NDN
,
NND
,
NNN
}.
The sample space can be obtained from the tree diagram given in
Figure 7-3
. There are two possible outcomes when the first item is selected:
a defective (
D
) or a non-defective (
N
). If a defective is selected, then on the
second selection, there are again two possibilities,
D
or
N
. If a nondefective
was selected on the first selection, then on the second selection, a defective
or a nondefective can be selected. Continuing in this manner, you can
display the outcomes by the tree diagram given in
Figure 7-3
.