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Three tours, A, B, and C, are offered to a group of 100 tourists. It turns out that 28 tourists sign for A, 26 for B, 16 for C, 12 for both A and B, 4 for both A and C, 6 for both B and C, and 2 for all three tours.

(a) What is the probability that a randomly chosen tourist is taking none of these tours?

(b) What is the probability that a randomly chosen tourist is taking exactly one of these tours?

(c) What is the probability that two randomly chosen tourists are both taking at least one of these tours?


Let $A,B,C$ be the event that a tourist takes tour $A,B$ and $C$, respectively.

I see that for part a) we compute $P((A\cup B \cup C)^c)=1-P(A\cup B\cup C)$, which yields $P((A\cup B \cup C)^c)=1/2$ from principle of inclusion exclusion.

However, for part b) the given solution is
$$P(\text{a tourist takes exactly one tour})=P(A)+P(B)+P(C)-2[P(A\cap B)+P(A\cap C)+P(B\cap C)]+3P(A\cap B \cap C).$$ I'm not seeing why we must multiply the probability of the intersection of events by $2$ and $3$, respectively. I was under the impression that applying P.I.E. was enough to account for overcounting the intersection of events.

Also, the solution for part c was given to be $$\frac{\binom{50}{2}}{\binom{100}{2}}.$$ The denominator makes sense, since we choose two tourists at random w/o replacement. I tried combining different groups of tourists to get $\binom{50}{2}$, but I don't see where this is coming from either.

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    $\begingroup$ For part b), consider only $A$ and $B$ for simplicity. You know that $P(A \cup B) = P(A) +P(B) - P(A \cap B)$. If you are interested in $P ( (A \cup B) \setminus (A \cap B))$ or "exactly in one of $A$ and $B$ (but not both)" you need to subtract $P(A \cap B)$ again, leading to $P ( (A \cup B) \setminus (A \cap B)) = P(A) +P(B) - 2P(A \cap B)$. The same reasoning extends to three sets as in your question. $\endgroup$
    – mlc
    Oct 1, 2016 at 17:51

2 Answers 2

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For the second part of your question:

The probability space can be visualized using a venn diagram for your problem

The reason behind the multiplication of intersection P(A∩B) by 2 is while subtracting the intersection you need to subtract it both from space of P(A) and P(B) and similarly for others.

While subtracting this intersection we also subtract the intersection of P(A∩B∩C) three times which explains the multiplication by 3 while adding the same.

Refer to the image.

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You know by point a) that there are 50 people which signed for at least one trip, hence the answer at point c) means: pick 2 out of 50 people who signed at at least one trip and divide by the number of all possible pairs.

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