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Unit 3: Conditional Probability and Independence Baye’s Theorem



            Notice that the sum of the probabilities of the eight points in  is                  Notes

            which is as it should be.
            Summarising the discussion so far, consider two random experiments    and   with sample
                                                                       l     2
            spaces S  and S , respectively. Let u , u  . . . be the points of S  and let v , v  . . ., be the points of S .
                  1     2               1  2               1       1  2                2
            Suppose p , p , . . ,a nd q , q ,. . . are the associated probabilities, i.e., P(u ) = p  and P(v) = q, with
                    1  2       l  2                                  i   i      j  j
            p , q   0    p   1,  q   1.  We say that   and   are independent experiments if the events “first
             i  j     i      j              1     2
                    i      j
            outcome is u ” and the event “second outcome is v ”, are independent,
                      i                             j
            i.e., if the assignment of probabilities on the product space S1 x S2 is such that
                 P{(u, v)} = P(u ) P(v) = p q
                     j  j    i   j   i j
            This assignment is a valid assignment because P((u , v))  0 and
                                                     i  j
                P(u ,v )  =   p q  i
                                 i
                       j
                     i
               i  j         i  j
                         =   p i q  1.
                                  j
                            i   j
            where the sums are taken over all values of i and j.
            Can we extend these concepts to the case of n (n > 2) random experiments?
            Let us denote the n random experiments by  ,  , . . . ,  . Let S . S , . . . , S  be the corresponding
                                                1  2    n     1  2    n
            sample spaces.  Let  P(x )  denote the  probability assigned  to the outcome  x   of the  random
                               j                                           j
            experiment . We say that  . . . ,   are independent experiments, if the assignment of probabilities
                      j          1    n
            on the product space S  × S  × . . . × S  is such that
                              1  2       n
                 P{(x , x , . . ., x )} = P(x ) P(x ) . . . P(x ).
                     1  2    n     1   2      n
            The random experiments  . . . ,   are said to be repeated independent trials of an experiment  if
                                 1    n
            the sample space of  ,. . .   are all identical and so are the assignment of probabilities, it is in this
                            1    n
            sense that the experiment discussed in Example 26 corresponds to 3 independent repetitions of
            the experiment of inspecting a unit, where the probability of a unit being defective is P.
            Before we conclude ow discussion of product spaces and repeated aials, let us revert to the case
            of two independent experiments   and   with sample spaces S  and S
                                       1    2                 1     2
            Suppose
            S  = (u , u , . . . ), P(u ) = p , i  1
             1   1  2       j   i
            S  = (v , v  ,... ), P(v) = q, j  1
             2   1  2      j   j
            Let A  =  (u ,u ,....)  and A  =  (v ,v ,...)  be two events in S  and S . Then A  × A  is event in
                1    1 i  2 i     2    1 j  2 j              1     2       1   2
            S  × S  and
             1  2
                 A  × A  = {(u , v ) | r, s = 1, 2, . . .}.
                   1   2    r i  s j
            Under the assumption that E~ and E~ are independent, we can write

               P(A1 × A2) =   P{(u , v )}
                                      s j
                                  r i
                           r  s
                        =   p q  j
                              i





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