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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