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               38     CHAPTER 2 PROBABILITY


                                 Table 2-3  Parts Classified
                                                                                   Surface Flaws
                                                                      Yes (event F)       No          Total
                                  Defective        Yes (event D)      10                  18            38
                                                   No                 30                  342          362
                                                   Total              40                  360          400




               EXAMPLE 2-16      Table 2-3 provides an example of 400 parts classified by surface flaws and as (functionally)
                                 defective. For this table the conditional probabilities match those discussed previously in this
                                 section. For example, of the parts with surface flaws (40 parts) the number defective is 10.
                                 Therefore,


                                                            P1D ƒ F2   10
40   0.25

                                 and of the parts without surface flaws (360 parts) the number defective is 18. Therefore,


                                                           P1D ƒ F¿2   18
360   0.05

                                    In Example 2-16 conditional probabilities were calculated directly. These probabilities
                                 can also be determined from the formal definition of conditional probability.



                       Definition
                                    The conditional probability of an event B given an event A, denoted as P1B ƒ A2 , is

                                                           P1B ƒ A2   P1A ¨ B2
P1A2                 (2-5)

                                    for P1A2   0 .





                                 This definition can be understood in a special case in which all outcomes of a random exper-
                                 iment are equally likely. If there are n total outcomes,

                                                       P 1A2   1number of outcomes in A2
n

                                 Also,

                                                   P 1A ¨ B2   1number of outcomes in A ¨ B2
n

                                 Consequently,

                                                                 number of outcomes in A ¨ B
                                                  P 1A ¨ B2
P1A2
                                                                   number of outcomes in A
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