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                                                                              2-4 CONDITIONAL PROBABILITY  39


                                                        Surface flaw
                                               360                     40
                                                  No               Yes
                                               400                    400
                                     Defective
                 Figure 2-13  Tree  342  No       Yes  18        30  No        Yes  10
                                   360                360        40               40
                 diagram for parts
                 classified




                                   Therefore, P1B ƒ A2  can be interpreted as the relative frequency of event B among the trials that
                                   produce an outcome in event A.

                 EXAMPLE 2-17      Again consider the 400 parts in Table 2-3. From this table

                                                                               10   40   10
                                                      P1D ƒ F2   P1D ¨ F2
P1F2    ^
                                                                              400 400    40
                                   Note that in this example all four of the following probabilities are different:

                                                         P1F2   40
400   P1F ƒ D2   10
28
                                                         P1D2   28
400   P1D ƒ F2   10
40

                                   Here, P(D) and P1D ƒ F2  are probabilities of the same event, but they are computed under two
                                   different states of knowledge. Similarly, P(F) and P1F ƒ D2  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,
                                                     P1D ƒ F2   10
40   and   P1D¿ƒ F2   30
40


                                   Of the 360 parts without surface flaws, 18 are functionally defective and 342 are not. Therefore,
                                                   P1D ƒ F¿2   342
360   and   P1D¿ƒ F¿2   18
360


                                   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, randomly implies 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


                                                            S without    5ab, ac, ba, bc, ca, cb6

                                   has probability 1
6 . If the unordered sample space is used, each of the three outcomes in
                                   {{a, b}, {a, c}, {b, c}} has probability 1
3 .
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