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 :) |
||||
|
|
Hint: Think "countable additivity". What can you say about $A\setminus B$ and $A\cap B$ if you are hoping to use this axiom. |
|||
|
|
|
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... |
|||
|
|
|
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. |
|||
|
|