# H0w many children were surveyed

I am studying for the p exam and realized I really need to brush up on my basic set theory, and am having trouble with this question.

In a survey on Popsicle flavor preferences of kids aged 3-5, it was found that:

• 22 like strawberry.

• 25 like blueberry.

• 39 like grape.

• 9 like blueberry and strawberry.

• 17 like strawberry and grape.

• 20 like blueberry and grape.

• 6 like all flavors.

• 4 like none.

Apparently the answer is 50, but I can't seem to figure out how to arrive at this

Note that

$$P(A\cup B\cup C) = P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).$$

Now let $A$ be "kid likes strawberry", $B$ be "kid likes blueberry", $C$ be "kid likes grape", and $n$ be the number of kids.

$P(A)=22/n$, $P(B)=25/n$, $P(C)=39/n$, with intersections $9/n$, $17/n$, $20/n$, $6/n$.

Then the probability of liking any flavor at all is $$P(A\cup B\cup C) = (22+25+39-9-17-20+6)/n=46/n.$$

Also, the probability of not liking any flavor at all is $4/n$, but also $$4/n=P(A^c \cap B^c \cap C^c)=1-P(A\cup B\cup C)=1-46/n.$$

Therefore, $$\frac{4}{n}=1-\frac{46}{n}=\frac{n-46}{n},$$

so $4=n-46$, and so $n=50$.

• Thank you for the clarification, I actually found a similar example in the text I have been using that got me the answer right after posting, but I didn't know how to factor in the 4 properly other than just adding it on in the end. – Heavenly96 Feb 21 '17 at 2:30