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Statistics



                      Notes         From the above table, we can write

                                              d    500   1
                                    (i)   P A Bi       
                                                   1000   2

                                          P A  d   250   1
                                    (ii)       Bi      
                                                   1000   4
                                               d
                                          P A B  i d       500  + 250  3
                                    (iii)           A   Bi        
                                                              1000    4
                                    (iv) b        150  +500 250    9
                                                           +
                                          P A Bg 
                                                       1000       10
                                    This can, alternatively, be written as  P A B b  g  1  P A  d  Bi    100    9  .
                                                                                        1
                                                                                           1000  10
                                           Example 24:  What is the probability of drawing a black card or a king from a well-
                                    shuffled pack of playing cards?
                                    Solution.
                                    There are 52 cards in a pack,    n(S) = 52.

                                    Let A be the event that the drawn card is black and B be the event that it is a king. We have to find
                                         b
                                     P A Bg .
                                    Since there are 26 black cards, 4 kings and two black kings in a pack, we have n(A) = 26, n(B) = 4

                                                                    +
                                                                       
                                    and  (  B = 2n A  )   Thus,  (A   ) B   26 4 2    7
                                                         P
                                                                    52     13
                                    Alternative Method
                                    The given information can be written in the form of the following table:
                                                                           B B Total
                                                                    A    2 24 26
                                                                    A    2 24 26
                                                                  Total  4 48 52

                                    From the above, we can write
                                         b      P A  d     24   7
                                     P A Bg         Bi      
                                              1
                                                          1
                                                             52   13
                                           Example 25: A pair of unbiased dice is thrown. Find the probability that (i) the sum of
                                    spots is either 5 or 10, (ii) either there is a doublet or a sum less than 6.
                                    Solution.
                                    Since the first die can be thrown in 6 ways and the second also in 6 ways, therefore, both can be
                                    thrown in 36 ways (fundamental principle of counting). Since both the  dice are given to be
                                    unbiased, 36 elementary outcomes are equally likely.







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