Let $A,B,C$ be events. The event "$A$ and $B$ occur but $C$ does not" may be expressed as $A \cap B \cap C^c$.
(a) Find an expression for the event "at least one of B and C occur, but A does not"
(b) Show that the probability of event in (a) is equal to
$\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(B\cap C)-\mathbb{P}(A\cap C)+\mathbb{P}(A\cap B \cap C)$
I claim that the answer to (a) is $(B \cup C)\cap A^c$. Can someone confirm or deny?
I have no idea how to proceed from here. From previous work, I have proven the following results:
$\mathbb{P}(A\setminus B)=\mathbb{P}(A)-\mathbb{P}(A \cap B)$
$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)$
My main concern about (a) is what exactly they mean by "at least one" and the use of the operator "and".
Best Answer
"At least one of" a series of events means you have an inclusive union. $A\cup B$ means the event of $A$ happening, or $B$ happening, or both happening.
"and" means the intersection. $A\cap B$ means the event of $A$ and $B$ both happening.
So your answer is confirmed, $(B\cup C)\cap A^c$ and as a hint this is equal to $(B\cup C)\setminus A$. Now apply your rules.