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                                                                                       2-6 INDEPENDENCE    47


                                     Table 2-4  Parts Classified
                                                                                     Surface Flaws
                                                                         Yes (event F)       No         Total
                                      Defective       Yes (event D)       2                  18          20
                                                      No                 38                 342         380
                                                      Total              40                 360         400


                 EXAMPLE 2-24      The information in Table 2-3 related surface flaws to functionally defective parts. In that case,
                                   we determined that P1D ƒ F2   10
40   0.25  and P1D2   28
400   0.07.  Suppose that the
                                   situation is different and follows Table 2-4. Then,

                                                P1D ƒ F2   2
40   0.05   and   P1D2   20
400   0.05

                                   That is, the probability that the part is defective does not depend on whether it has surface
                                   flaws. Also,

                                                P1F ƒ D2   2
20   0.10   and   P1F2   40
400   0.10

                                   so the probability of a surface  flaw does not depend on whether the part is defective.
                                   Furthermore, the definition of conditional probability implies that

                                                              P1F ¨ D2   P1D ƒ F2P1F2

                                   but in the special case of this problem

                                                                              2   2    1
                                                        P1F ¨ D2   P1D2P1F2
                                                                             40  20   200

                                       The preceding example illustrates the following conclusions. In the special case that
                                   P1B 0 A2   P1B2,  we obtain

                                                         P1A ¨ B2   P1B ƒ A2P1A2   P1B2P1A2

                                   and

                                                                P1A ¨ B2   P1A2P1B2
                                                       P1A ƒ B2                      P1A2
                                                                  P1B2       P1B2

                                   These conclusions lead to an important definition.

                          Definition
                                       Two events are independent if any one of the following equivalent statements is true:
                                          (1)  P1A ƒ B2   P1A2
                                          (2)  P1B ƒ A2   P1B2
                                          (3)  P1A ¨ B2   P1A2P1B2                                    (2-9)
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