I need to prove that
$$P(A \backslash B) = P(A) - P(A\cap B)$$
and
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
using the axioms of probability but can't see where to start. Note: this is homework, so a hint only would be nice :)
I need to prove that
$$P(A \backslash B) = P(A) - P(A\cap B)$$
and
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
using the axioms of probability but can't see where to start. Note: this is homework, so a hint only would be nice :)
Proof The events A\B, B \A and A∩B are pairwise disjoint and their union is A∪B. By Axiom 3 P(A∪B) = P(A\B) +P(B\A) +P(A∩B) (1) Axiom 3 also implies that P(A) = P(A\B) +P(A∩B) so P(A\B) = P(A)−P(A∩B), (2) and P(B) = P(B\A) +P(A∩B) so P(B) = P(B\A) +P(A∩B). (3) We can now substitute equations (2) and (3) into (1) to obtain P(A∪B) = P(A) +P(B)−P(A∩B).
Well,
Let $B \subset A$, then $A = B \cup (A \setminus B)$ which is a disjoint union so...
Now, $$A \cup B = (A \setminus (A \cap B)) \cup (A \cap B) \cup (B \setminus (A \cap B))$$ which is again a disjoint union. So...
Hint: Think "countable additivity". What can you say about $A\setminus B$ and $A\cap B$ if you are hoping to use this axiom.
The second one as written is false. I'm sure you meant
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Jonas' answer is a good hint on how to apply the first part to prove the second part.