Page 73 - Probability Demystified
P. 73

62                            CHAPTER 4 The Multiplication Rules

                                                                        1
                                 probability of getting a specific card is  52 , but the probability of getting a
                                 specific card on the second draw is  1  since 51 cards remain.
                                                                   51


                                 EXAMPLE: Two cards are selected from a deck and the first card is not
                                 replaced. Find the probability of getting two kings.


                                 SOLUTION:
                                                                                    4
                                 The probability of getting a king on the first draw is  52  and the probability
                                 of getting a king on the second draw is  3  , since there are 3 kings left and
                                                                        51
                                 51 cards left. Hence the probability of getting 2 kings when the first card is
                                 not replaced is  4     3  ¼  12  ¼  1  .
                                                52  51  2652  221

                                   When the two events A and B are dependent, the probability that the
                                 second event B occurs after the first event A has already occurred is written as
                                 P(B | A). This does not mean that B is divided by A; rather, it means and is
                                 read as ‘‘the probability that event B occurs given that event A has already
                                 occurred.’’ P(B | A) also means the conditional probability that event B occurs
                                 given event A has occurred. The second multiplication rule follows.


                                 Multiplication Rule II: When two events are dependent, the probability of
                                 both events occurring is PðA and BÞ¼ PðAÞ  PðB j AÞ



                                 EXAMPLE: A box contains 24 toasters, 3 of which are defective. If
                                 two toasters are selected and tested, find the probability that both are
                                 defective.


                                 SOLUTION:
                                 Since there are 3 defective toasters out of 24, the probability that the first
                                                           1
                                 toaster is defective is  3  ¼ . Since the second toaster is selected from the
                                                      24   8
                                 remaining 23 and there are two defective toasters left, the probability that
                                 it is defective is  2  . Hence, the probability that both toasters are defective is
                                                23
                                                                          1    1
                                                                         3    2    1
                                      PðD and D Þ¼ PðD Þ  PðD jD Þ¼        4     ¼
                                                         1
                                                 2
                                          1
                                                                    1
                                                                2
                                                                        24 8  23   92
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