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Unit 7: Modern Approach to Probability



            Then, we can write                                                                    Notes

                             1        2           6          3
                                               )
                                                          )
                        P A    ,  ( ) B   ,  ( /P D A   ,  ( /P D B 
                                P
                         ( ) 
                             3        3          10          8
                          1  6  2  3   9
                    P D
            Further,   ( )   ´  +  ´    .
                          3  10  3  8  20
                       P (A   D )  1  6  20  4
                     )
              P ( /A D         ´   ´   
                          ( )
                         P D     3  10  9  9
                   Example 57: In a certain recruitment test there are multiple-choice questions. There are
            4  possible answers to each question out of which only one  is correct.  An intelligent student
            knows 90% of the answers while a weak student knows only 20% of the answers.

            (i)  An intelligent student gets the correct answer, what is the probability that he was guessing?
            (ii)  A weak student gets the correct answer, what is the probability that he was guessing?
            Solution.
            Let A be the event that an intelligent student knows the answer, B be the event that the weak
            student knows the answer and C be the event that the student gets a correct answer.
                               d
            (i)  We have to find P A /Ci. We can write

                          P ( A C  )    P ( ) ( /A P C A )
                        )
                  P ( /A C                                                             .... (1)
                             ( )
                            P C    P ( ) ( /A P C A ) P+  ( ) ( /A P C A )
                 It is given that P(A) = 0.90,   1    and  ( /P C A ) 1.0
                                            )
                                       P ( /C A     0.25
                                               4
                 From the above, we can also write  P A d i = 0.10
                 Substituting these values, we get

                                  ´
                        )
                  P ( /A C    0.10 0.25     0.025    0.027
                          0.10 0.25 0.90 1.0  0.925
                                   +
                                       ´
                              ´
                               d
            (ii)  We have to find P B /Ci. Replacing A  by B , in equation (1), we can get this probability.
                                        d                a
                 It is given that P(B) = 0.20, P C/Bi =0.25 and P C/Bf =1.0
                 From the above, we can also write  P B d i =0.80
                                         0.80 0.25     0.20
                                            ´
                                  )
                 Thus, we get   ( /P B C                   0.50
                                        ´
                                    0.80 0.25 0.20 1.0  0.40
                                                 ´
                                             +
                   Example 58:  An electronic manufacturer has  two lines A and B assembling identical
            electronic units. 5% of the units assembled on line A and 10%of those assembled on line B are
            defective. All defective units must be reworked at a significant increase in cost. During the last
            eight-hour shift, line A produced 200 units while the line B produced 300 units. One unit is
            selected at random from the 500 units produced and is found to be defective. What is the probability
            that it was assembled (i) on line A, (ii) on line B?
            Answer the above questions if the selected unit was found to be non-defective.



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