Answer: 29
Step-by-step explanation:
Let S denotes the total number of people surveyed, A denotes the event of having dog , B denotes the event of having cats and C denotes the event of having fish.
Given : n(S)=50 ;n(A)=11 ; n(B) =13 and n(C)=6
Also, n(A∩B)=5 ; n(A∩C) = 2 and n(B∩C)=3 and n(A∩B∩C)=1
We know that,
[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)-n(A \cap B\cap C)\\\\=11+13+6-5-2-3+1=21[/tex]
Now, the number of people owned none of these pets :-
[tex]n(S)-n(A\cup B\cup C)\\\\=50-21=29[/tex]
Hence, the number of people owned none of these pets =29