union and intersection of events

Question

So basically I evaluate the following probability in two different ways and I get different answers

$P(A \cap B \cup C) = P((A \cap B) \cup C) = P(A \cap B) + P(C) - P(A \cap B \cap C)$

$P(A \cap B \cup C) = P(A \cap (B \cup C)) = P(A) + P(B \cup C) - P(A \cup B \cup C)$

But the terms on the RHS are not equal. Can someone please tell me what is wrong?

Answer 1

It's wrong because in general $(A \cap B) \cup C\neq A \cap (B \cup C)$.

Take for example $(\{0\} \cap \{1\}) \cup \{2\}=\{2\}\neq \varnothing=\{0\} \cap (\{1\} \cup \{2\})$.

Therefore it's also better not to write $A \cap B \cup C$ without brackets...

Answer 2

To rephrase @Phenotype’s answer: this is not one probability, these are two different probabilities:

• $P((A\cap B)\cup C)$ is the probability that either both $A$ and $B$ happened, or $C$ happened.
• $P(A\cap(B\cup C))$ is the probability that $A$ happened and one of $B$ or $C$ happened.

They are not the same e.g. if $C$ happened but $A$ didn’t. If in doubt, draw some Venn diagrams.

Those two probabilities happen to be the same if $C\subseteq A$ ( i.e. if $C$ implies $A$), but in general they are different. More precisely, $A\cap(B\cup C)\subseteq(A\cap B)\cup C$, i.e. the second event above implies the first.