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Chapter 7: Probability
Figure 7-3:
Tree diagram for the selection of three items
Example 7-6:
For
Example 7-5
, what is the probability of observing at
least two defective items?
Solution:
Let A be the event of at least two defectives. Then
A
= {
DDD
,
DDN
,
DND
,
NDD
}. Event
A
is made up of 4 simple events, and there are 8
simple events in the sample space. Thus,
P
(
A
) = 4/8 = 0.5.
Example 7-7:
A game is played in which you select three chips from a bag
which has both red and blue chips. You win if all three chips are of the
same color. Let
R
represent the event of a red chip, and let
B
represent the
event of a blue chip. List the possible outcomes for the sample space.
Note:
This experiment is equivalent to the three-child family.