SOLUTION: A computer company advertises its computers in PC World, in PC Magazine, and on television. A survey of 780 customers finds that the numbers of customers who are familiar with the

We have one "universal" set C of 780 surveyed Customers and many its subsets:


    subset W           (PC World, of 305 customers)

    subset M           (PC Magazine, of 280 customers)

    subset T           (television, of 310 customers)

    subset (WM)        (intersection of the sets W and M, of 140 customers)

    subset (MT)        (intersection of the sets W and M, of 132 customers)

    subset (WT)        (intersection of the sets W and T, of 120 customers)

    subset (WMT)       (intersection of the sets W, M and T, of 83 customers)

     How many know from exactly one source W ?

     Take subset W, subtract intersections WM and WT from it, and then add intersection WMT;

     so the number Wo (W only) is equal to  305 - 140 - 120 + 83 = 128.



     How many know from exactly one source M ?

     Take subset M, subtract intersections WM and MT from it, and then add intersection WMT;

     so the number Mo (M only) is equal to  280 - 140 - 132 + 83 = 91.



     How many know from exactly one source T ?

     Take subset T, subtract intersections MT and WT from it, and then add intersection WMT;

     so the number To (T only) is equal to  310 - 132 - 120 + 83 = 141.



     Therefore, the answer to the question  (a)  is the sum

         Wo + Mo + To = 128 + 91 + 141 = 360.     ANSWER to question (a)

     To get the number, add elements in WM, MT and WT; then subtract 3 times the number of elements in WMT:

     140 + 132 + 120 - 3*83 = 143.      ANSWER to question (b)

     I just answered this question above, but will repeat it again for your convenience.


     How many know from exactly one source W ?

     Take subset W, subtract intersections WM and WT from it, and then add intersection WMT;

     so the number Wo (W only) is equal to  305 - 140 - 120 + 83 = 128.    ANSWER to question (c)