Page 65 - Applied statistics and probability for engineers
P. 65

Section 2-4/Conditional Probability     43



                        Let A and B denote the events that the irst and second part selected are defective, respectively. The probability requested
                     can be expressed as P(B | A). If the irst part is defective, prior to selecting the second part the batch contains 49 parts, of

                     which 2 are defective. Therefore,
                                                               P B A) =  2
                                                                   |
                                                                (
                                                                        49


                     Example 2-25    Continuing Example 2-24, what is the probability that the irst two parts selected are defective
                                     and the third is not defective?
                                        This probability can be described in shorthand notation as P(d d n ), where d and n denote parts
                                                                                         1 2 3
                     that are defective and not defective, respectively. Here
                                   P d d n ) =  P n d d P d d ) =  P n d d P d | d P d ) =  47 2  ⋅  3  =  0 0024
                                                                                        ⋅
                                                                  |
                                                                      )
                                                                                                 .
                                                 |
                                                     )
                                                                              )
                                                   1 2
                                                                             1
                                                       ( 1 2
                                    ( 1 2 3
                                                                                ( 1
                                              ( 3
                                                                        ( 2
                                                                    1 2
                                                               ( 3
                                                                                     48 449 50

                     The probabilities for the irst and second selections are similar to those in the previous example. The P(n |d d ) is
                                                                                                             3  1 2
                     based on the fact that after the irst 2 parts are selected, 1 defective and 47 nondefective parts remain.

                        When the probability is written to account for the order of the selections, it is easy to solve this question from the
                     deinition of conditional probabil ity. There are other ways to express the probability, such as P(d d n ) = P(d |d n )

                                                                                                     1 2 3     2  1 3
                     P(d n ). However, such alternatives do not lead to conditional prob abilities that can be easily calculated.
                        1 3
                     Exercises            FOR SECTION 2-4
                         Problem available in WileyPLUS at instructor’s discretion.
                                 Tutoring problem available in WileyPLUS at instructor’s discretion
                     2-99.     Disks of polycarbonate plastic from a supplier are   2-101.     The analysis of results from a leaf transmutation
                     analyzed for scratch and shock resistance. The results from 100   experiment (turning a leaf into a petal) is summarized by type
                     disks are summarized as follows:                  of transformation completed:
                                                Shock Resistance                                   Total Textural
                                                                                                  Transformation
                                                High       Low
                                                                                                  Yes        No
                      Scratch     High           70          9
                                                                       Total Color      Yes       243         26
                      Resistance  Low            16          5
                                                                       Transformation   No         13         18
                     Let A denote the event that a disk has high shock resistance,
                     and let B denote the event that a disk has high scratch resist-  (a) If a leaf completes the color transformation, what is the
                     ance. Determine the following probabilities:        probability that it will complete the textural transformation?
                     (a)  P A ( )    (b)  P B ( )                      (b) If a leaf does not complete the textural transformation, what
                     (c)  P A B| (  )  (d) P B A| (  )                   is the probability it will complete the color transformation?
                                                                       2-102.     Samples of a cast aluminum part are classiied on the

                     2-100.   Samples of skin experiencing desquamation are

                                                                       basis of surface inish (in microinches) and length measurements.
                     analyzed for both moisture and melanin content. The results
                                                                       The results of 100 parts are summarized as follows:
                     from 100 skin samples are as follows:
                                               Melanin Content                                       Length
                                              High         Low                                 Excellent    Good
                      Moisture  High           13           7          Surface  Excellent        80           2
                      Content  Low             48          32          Finish  Good              10           8
                     Let A denote the event that a sample has low melanin content,   Let  A  denote the event that a sample has excellent surface

                     and let B denote the event that a sample has high moisture con-  inish, and let B denote the event that a sample has excellent
                     tent. Determine the following probabilities:      length. Determine:  (b)  P B ( )
                     (a)  P A ( )    (b)  P B ( )                      (a)  P A ( )
                                                                                           | (
                                                                             | (
                                          | (
                     (c)  P A B)    (d) P B A)                         (c)  P A B)   (d) P B A)
                            | (
   60   61   62   63   64   65   66   67   68   69   70