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Statistics



                      Notes                         2  5  3  4  22
                                                                      ´  +  ´      0.49
                                                    5  9  5  9  45
                                    Hence, A and B are likely to contradict each other in 49% of the cases.

                                                                                                         2
                                           Example 45: The probability that a student A solves a mathematics problem is    and the
                                                                                                         5
                                                                     2
                                    probability that a student B solves it is   . What is the probability that (a) the problem is not
                                                                     3
                                    solved, (b) the problem is solved, (c) Both A and B, working independently of each other, solve
                                    the problem?
                                    Solution.

                                    Let A and B be the respective events that students A and B solve the problem. We note that A and
                                    B are independent events.

                                                                      3  1  1
                                                ( ) ( A   B )   P A  .P B  ´  
                                                 a
                                                    P
                                                             ( ) ( ) 
                                                                      5  3  5
                                                                          1  4
                                                ( )  b P (A   B ) 1 P A   (  B ) 1   
                                                                          5  5
                                                                     2  2   4
                                                ( )  c P  (A   ) B   P A P B  ´  
                                                             ( ) ( ) 
                                                                     5  3  15

                                           Example 46: A bag contains 8 red and 5 white balls. Two successive drawings of 3 balls
                                    each are made such that
                                    (i) balls are replaced before the second trial, (ii) balls are not replaced before the second trial.
                                    Find the probability that the first drawing will give 3 white and the second 3 red balls.
                                    Solution.
                                    Let A be the event that all the 3 balls obtained at the first draw are white and B be the event that
                                    all the 3 balls obtained at the second draw are red.
                                    (a)  When balls are replaced before the second draw, we have

                                                 5 C  5            8 C   28
                                            ( ) 
                                          P A     3       and  P B  3  
                                                               ( ) 
                                                13                 13
                                                  C   143           C   143
                                                   3                  3
                                         The required probability is given by  P A B b  g , where A and B are independent. Thus, we
                                         have
                                                              5   28   140
                                          P (A   ) B   P A  .P B  ´  
                                                    ( ) ( ) 
                                                             143  143  20449
                                    (b)  When the balls are not replaced before the second draw
                                                         8
                                                      )
                                         W e have  P ( /B A   C 3    7  . Thus, we have
                                                         10
                                                          C 3  15
                                                                5   7   7
                                         P (A   ) B   P ( ) ( /A  .P B A )  ´  
                                                               143  15  429



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