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fenix001 [56]
3 years ago
8

The local pet store surveyed 50 people about pets. Eleven of these people owned dogs, 13 owned cats, and 6 owned fish. One perso

n owned all three types of pets, 2 people owned only fish and dogs, 3 people only fish and cats, and 5 people owned only cats and dogs. How many people owned none of these pets?
Mathematics
1 answer:
siniylev [52]3 years ago
7 0

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,

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

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

n(S)-n(A\cup B\cup C)\\\\=50-21=29

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

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