The shipment for the computers in the retail outlet has the 3 defectives. If the random purchase had been done for the 2 of the computers, determine the probability that the number of defectives can be chosen.

Solution:

Let the variable Y whose values are the probable values numbers of the defectives computers that can be purchased. This means that the y can be used for the numbers starting from 0, 1, 2.

Now,

f(0) = P(y = 0) = (30) (52) = 10

                               (82)        28

f(1) = p(x=1)= (31) (51) = 15

                            (82)        28

f(2) = p(x=0) = (32)(50) = 3

                             (82)      28

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