I'm working through my probability textbook and came across a question that I don't fully understand.
Given the sample space $S = A \bigcup B \bigcup C$
$$P(A) = \frac{1}{2}$$ $$P(B) = \frac{2}{3}$$ $$P(A \bigcup B) = \frac{5}{6}$$ $$P(C \bigcap (A \bigcup B)) = \frac{5}{12}$$
What is $P(C)?$
My solution is:
$$P(C \bigcap (A \bigcup B)) = P(C) * P(A \bigcup B)$$
$$P(C) = \frac{P(C \bigcap (A \bigcup B))}{P(A \bigcup B)}$$
Which yields $$P(C) = \frac{1}{2}$$
The answer in the textbook says $P(C) = \frac{7}{12}$
I'm obviously missing something and was hoping someone could explain where my logic breaks down.
Your assertion is only valid if $C$ and $A \cup B$ are independent.
In this problem, $P(A)$ and $P(B)$ are irrelevant. Think of the entire sample space $S$ as $D \cup C$, where $D = A \cup B$. We then have
$$ P(D) = \frac56 $$
$$ P(C \cap D) = \frac{5}{12} $$
The key is to observe that $P(C \cup D) = 1$, because $S = C \cup D$ is the entire sample space; it has to have probability one. We can now approach this in one of (at least) two ways. One is to use the identity
$$ P(C) + P(D) = P(C \cup D) + P(C \cap D) $$
which yields
$$ P(C) + \frac56 = 1 + \frac{5}{12} $$
whereupon
$$ P(C) = \frac{7}{12} $$
Another way is to recognize that whenever $D$ doesn't hold—which is $\frac16$ of the time—$C$ must hold. Therefore, $P(C)$ must be at least $1-\frac56 = \frac16$. Whatever $P(C)$ is in excess of that quantity must be the probability that $C$ and $D$ both hold at the same time. We are told this probability is $\frac{5}{12}$. Therefore,
$$ P(C) = \frac16 + \frac{5}{12} = \frac{7}{12} $$
In essence, these two approaches are equivalent, but from a pedagogical perspective, sometimes one is easier to see than the other.