Question 1132870
.
<pre>
We have one "universal" set C of 780 surveyed <U>C</U>ustomers 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)
</pre>

(a)  &nbsp;&nbsp;How many of the surveyed customers know about the computers because of exactly one of these forms of advertising?


<pre>
     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.     <U>ANSWER to question (a)</U>
</pre>


(b)  &nbsp;&nbsp;How many of the surveyed customers know about the computers because of exactly two of these forms of advertising?


<pre>
     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.      <U>ANSWER to question (b)</U>
</pre>


(c) &nbsp;&nbsp;How many of the surveyed customers know about the computers because of PC World and neither of the other two forms of advertising?


<pre>
     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.    <U>ANSWER to question (c)</U>
</pre>


Everything is answered; &nbsp;the problem is solved and completed.


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Regarding similar problems, &nbsp;see the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Counting elements in sub-sets of a given finite set</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Advanced-probs-counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Advanced problems on counting elements in sub-sets of a given finite set</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Challenging-problems-on-counting-elements-in-subsets-of-a-given-finite-set.lesson>Challenging problems on counting elements in subsets of a given finite set</A> 

in this site.


Look also into the links


https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html

https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Rate-of-work-word-problems.faq.question.1126097.html

https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1126099.html


to similar solved problems in the archive of this forum.