Let $A,B,C$ be events in a probability space. Suppose that $\mathbb P(A) = 1/4\\ \mathbb P(B^c) = 2/3\\ \mathbb P(C) = 1/2\\ \mathbb P(A^c\cap B) = 1/4\\ \mathbb P(A\cap C) = 0\\ \mathbb P(B^c\cup C^c) = 5/6. $
I want to calculate the value of $\mathbb P(A\cup B\cup C)$.
My work:
I calculated that $\mathbb P(B)= 1/3, \mathbb P(A \cap B) = 1/12$ and $\mathbb P(B \cap C) =1/6 $.
And I know that $\Bbb P (A \cup B \cup C) = \Bbb P (A) + \Bbb P (B) + \Bbb P (C) - \Bbb P (A\cap B) - \Bbb P (B\cap C) -\Bbb P (A\cap C) + \Bbb P (A \cap B \cap C)\\ = 5/6 + \Bbb P (A \cap B \cap C) $.
My question is: Can I use the fact that $0 \leq \Bbb P (A\cap B \cap C ) \leq \Bbb P (A\cap C ) = 0 \Rightarrow \Bbb P (A\cap B \cap C)=0$ to say that $\Bbb P (A \cup B \cup C) =5/6$?
Yes. Since $A \cap B \cap C \subseteq A \cap C$, $0 \leq \Pr(A \cap B \cap C) \leq \Pr(A \cap C) = 0 \implies \Pr(A \cap B \cap C) = 0$.