Question

In a survey of 100 persons, it was found that 28 read magazine A, 30 read ,magazine 142 read magazine C, 8 read magazine A and B, 10 read magazine A and C, 5 read magazine B and C and 3 read all the three magazines. Find how many persons read magazine C only .

Answer 1

We have, $n\left(U\right)=100,n\left(A\right)=28,n\left(B\right)=30,n\left(C\right)=42,n(A\cap B)=8,n(A\cap C)=10,n(B\cap C)=5$

and $n(A\cap B\cap C)=3$

Number of persons who read none of the magazine,

i.e., $n(A\cup B\cup C{)}^{\prime }=n(U)-n(A\cup B\cup C)$

$=n\left(U\right)-\left[n\right(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C\left)\right]$

$=100-(28+30+42-8-10-5+3)$

= 100-80=20

Number of persons who read magazine C only.

i.e. $n(C\cap A^{\prime }\cap B^{\prime })=n[C\cap (A\cup B{)}^{\prime }]=n(C\cap (A\cup B))$

$=n\left(C\right)-n[C\cap (A\cup B{)}^{\prime }]=n(C\cap (A\cup B))$

$=n\left(C\right)-\left[n\right(C\cap A)+n(C\cap B\left)\right]-n[C\cap A\cap B]$

$=42-(10+5-3)=42-12=30$