SOLUTION: In a survey of 400 households regarding ownership of HDTVs and DVD players, the following data were obtained. 315 households own one or more HDTVs. 180 households own one or mor

Let s(HDTV) be the set of households that own one or more HDTVs       and n(HDTV) be the number of such households.

Let s(DVD)  be the set of households that own one or more DVD players and n(DVD) be the number of such households.


(1)  We are given that n(HDTV) = 315.


(2)  The households that own one or more HDTVs AND one or more DVD players form the intersection of sets s(HDTV) 
     and s(DVD) inside the universal set of 400 households under the survey.

     We are given that this intersection has 180 households.


(3)  The households that own one or more HDTVs OR one or more DVD players form the union of sets s(HDTV) and s(DVD) 
     inside the universal set of 400 households under the survey.

      The number of households in this union is 400 - 19 = 381.


(4)  Using the formula for the number of elements in two subsets of a universal set, 
     their union and their intersection, we can write this equation

          381 = 315 + n(DVD) - 180.


     From this equation, we can find the number of households n(DVD) that have one or more DVD players

          n(DVD) = 381 - 315 + 180 = 246.


(5)  Now to get the number of households that only own one or more DVD players, we should subtract 180,
     the number of elements in the intersection, from 246, the number of elements in s(DVD).

     Doing it this way, we get

           n(DVD only) = 246 - 180 = 66.


At this point, the problem is solved completely.


ANSWER.  The number of households that only own one or more DVD players is 66.