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Mathematics Semester 3  STPM  Chapter 2 Probability

                    Example 33


                 Two events C and D are such that P(C) =   2  and P(D) =   1 . If
                                                    5            3
                 (a)  P(C  D) =   11 ,                 (b)  P(C  D) =   9  ,
                                15                                     15
                 determine if C and D are mutually exclusive and find also if C and D are independent.

                  Solution:          (a)  By using P(C  D)  = P(C) + P(D) – P(C  D)
                                                       11   =   2   +   1  – P(C  D)
                                                       15   5    3
           2                                     P(C  D)  = 0
                                         \ C and D are mutually exclusive.
                                         P(C) × P(D) =   2   ×   1    =   2
                                                      5    3    15
                                         Since P(C  D) ≠ P(C) × P(D), C and D are not independent.

                                      (b)  By using P(C  D) = P(C) + P(D) – P(C  D)
                                                       9   =   2   +   1  – P(C  D)
                                                       15   5    3
                                                 P(C  D) =   2
                                                            15
                                         Since P(C  D) ≠ 0, C and D are not mutually exclusive.

                                         As P(C  D) = P(C) × P(D), C and D are independent.
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                    Example 34

                 The probability that a new released model of printer will develop a fault within a year is 0.2. If two new
                 printers are selected at random from a store, determine the probability that only one printer will develop
                 a fault.
                  Solution:          We start with a tree diagram showing all the possible combined outcomes of the
                                     two experiments, the happening of the first and second printers.
                                     The first set of branches of the tree shows what could happen to the first printer
                                     and the second set of branches indicates what could happen to the second printer.
                                     Since  a  printer  has  two  possible  outcomes,  fault  or  no  fault  and  each  of  these
                                     may lead to two other possible outcomes of the second printer, we have a total
                                     of 2 × 2 = 4 possible outcomes.
                                                               First printer  Second printer
                                                                               fault
                                                                        0.2
                                                                 fault
                                                        0.2
                                                                        0.8    no fault
                                                                        0.2    fault
                                                        0.8
                                                                no fault
                                                                        0.8
                                                                               no fault


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         02 STPM Math(T) T3.indd   98                                                                 28/10/2021   10:21 AM
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