2-4 CONDITIONAL PROBABILITY 39
Therefore, can be interpreted as the relative frequency of event Bamong the trials that
produce an outcome in event A.
EXAMPLE 2-17 Again consider the 400 parts in Table 2-3. From this table
Note that in this example all four of the following probabilities are different:
Here, P(D) and are probabilities of the same event, but they are computed under two
different states of knowledge. Similarly, P(F) and are computed under two different
states of knowledge.
The tree diagram in Fig. 2-13 can also be used to display conditional probabilities. The
first branch is on surface flaw. Of the 40 parts with surface flaws, 10 are functionally defec-
tive and 30 are not. Therefore,
Of the 360 parts without surface flaws, 18 are functionally defective and 342 are not. Therefore,
Random Samples from a Batch
Recall that to select one item randomly from a batch implies that each item is equally likely.
If more than one item is selected, randomlyimplies that each element of the sample space is
equally likely. For example, when sample spaces were presented earlier in this chapter, sam-
pling with and without replacement were defined and illustrated for the simple case of a batch
with three items {a,b,c}. If two items are selected randomly from this batch without replace-
ment, each of the six outcomes in the ordered sample space
has probability. If the unordered sample space is used, each of the three outcomes in
{{a,b}, {a, c}, {b, c}} has probability 1
3.
1
6
Swithout 5 ab, ac, ba, bc, ca, cb 6
P 1 DƒF¿ 2 342
360 and P 1 D¿ƒF¿ 2 18
360
P 1 DƒF 2 10
40 and P 1 D¿ƒF 2 30
40
P 1 FƒD 2
P 1 DƒF 2
P 1 D 2 28
400 P 1 DƒF 2 10
40
P 1 F 2 40
400 P 1 FƒD 2 10
28
P 1 DƒF 2 P 1 D ̈F 2
P 1 F 2
10
400 ^
40
400
10
40
P 1 BƒA 2
Surface flaw
No Yes
No Yes No Yes
Defective
360
400
40
400
342
360
18
360
30
40
10
Figure 2-13 Tree 40
diagram for parts
classified
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