In: Statistics and Probability
Of 30 personal computers (PCs) owned by faculty members in a certain university department, 20 run Windows, eight have 21-inch monitors, 25 have CD-ROM drives, 20 have at least two of these features, and six have all three.
a) How many PCs have at least one of these features?
b) How many have none of these features?
c) How many have exactly one feature?
Given that set of PCs running under Windows (W) = 20, set of PCs with 21-inch monitors (M) = 8 and set of PCs with CD-ROM drives (C) =25. Also, 20 have at least two of these features, and six have all three. So,
(1)
and
(2)
We expand the equation (1) using the Principle of Inclusion-Exclusion
(a) The number of PCs with at least one feature is
(b) Since there are 27 PCs with at least one feature, then there are 30 – 7 = 3 PCs that have none of these features.
(c) We know that the number of PCs with at least one feature is 27. We know that the number of PCs with at least two features is 20. Thus, the number of PCs with exactly one feature is simply 27 – 20 = 7.