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I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets: $$P(A \cup B) = P(A) + P(B) - P(A \cap B) $$

Note that the function $P$ is the probability of an event happening. From the third axiom, we are given that for any two disjoint sets, that is $ A \cap B = \emptyset$:

$$P(A \cup B) = P(A) + P(B)$$

My thought process follows from this third axiom, but I run into a problem near the end:

Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so

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

From this, it appears that we are essentially taking the symmetric difference of A and B. This, to me, seems to be equivalent to:

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

Hence, by my logic:

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

Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them. If the sets are disjoint, then the intersection is merely null. If they intersect, then we are removing the "extra pieces".

I feel like this is inadequate proof despite my ability to convince myself that it is true.. Where, if anywhere, did I go wrong?

Thanks!

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4 Answers 4

Here's how I'd do it:

Note that $$ (*)\quad P(A\bigcup B) = P(A \bigcup A^{c}B) = P(A)+P(A^{c}B)$$

Then since $B = AB \bigcup A^{c}B$ we get that

$$P(B) = P(AB) + P(A^{c}B)$$

or equivalently $$P(A^{c}B) = P(B) - P(AB)$$ Plugging this into $*$ gives the desired result and thus completing the proof. Hope that helps!

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For any sets $A$ and $B$, we have the disjoint union

$$A \cup B = (A- B) \cup (A \cap B) \cup (B-A). $$

Then, by the axiom,

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

Since $ P (A-B)= P(A) - P(A \cap B)$ and $ P (B-A)= P(B)- P(A \cap B) $, the result follows.

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Use these three facts to get started. If $C\subset D$ are events, $P(D - C) = P(D) - P(C).$ If $C$ and $D$ are any events $C\cup D = C \cup (D - C)$ is a disjoint union, and $D - C = D - (C\cap D)$.
Fool with this and you should be able to get it.

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One way to prove this is to define indicator random variables $I_A, I_B, I_{A \cap B}$ and $I_{A \cup B}$, where $I_E = 1$ if the event $E$ occurs and $I_E = 0$ if $E$ does not occur.

Now suppose that $I_{A \cup B} = 1$, which means $A$ or $B$ has happened, and consider the three ways in which this can occur. If $A$ occurs but not $B$, then $I_A = 1$, $I_B = 0$ and $I_{A \cap B} = 0$. If $B$ occurs but not $A$ then similarly $I_A = 0$, $I_B = 1$ and $I_{A \cap B} = 0$. Finally if both $A$ and $B$ occur, then $I_A = 1$, $I_B = 1$ and $I_{A \cap B} = 1$.

If on the other hand $I_{A \cup B} = 0$, then neither $A$ nor $B$ has occurred and $I_A = 0$, $I_B = 0$ and $I_{A \cap B} = 0$

Notice we're just shown that $I_{A \cup B} = I_A + I_B - I_{A \cap B}$ and hence $\mathbb{E}(I_{A \cup B}) = \mathbb{E}(I_A) + \mathbb{E}(I_B) - \mathbb{E}(I_{A \cap B})$, or $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$. (The basic idea here is that the inclusion-exclusion formula in probability is nothing but simple counting.)

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