Answer: 29
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,

Now, the number of people owned none of these pets :-

Hence, the number of people owned none of these pets =29