# Probability of tourists taking tours

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.

• 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.
– mlc
Oct 1, 2016 at 17:51