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Does the union of two events exclude that those events may occur together? In other words, is it the same as an exclusive or?

$p(E_1 \cup E_2) = p(E_1) + p(E_2) - p(E_1 \cap E_2)$

The reason I ask is because the above equation explicitly subtracts the probability of the events occurring together.

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No. The intersection event $E_1\cap E_2$ occurs in both $E_1$ and $E_2$, so if you don't subtract one out it will be counted twice. Another way to look at it is to look at the following equality: $p(E_1 \cup E_2) = p(E_1\setminus E_2) + P( E_1\cap E_2) + p(E_2\setminus E_1)$. –  copper.hat Oct 29 '12 at 23:28
    
Informally, you can think of $\Pr(A)$ as the "weight" of $A$. So if you add the weights of $E_1$ and of $E_2$, and $E_1\cap E_2)$ has non-zero weight, you get more than the weight of $E_1\cup E_2)$. How much more? By the weight of $E_1\cap E_2$. –  André Nicolas Oct 29 '12 at 23:36

2 Answers 2

up vote 5 down vote accepted

We know that if $A$, $B$ and $C$ are mutually disjoint events i.e. if $A \cap B = B \cap C = C \cap A = \emptyset$, then $$\mathbb{P}(A \cup B \cup C) = P(A) + P(B) + P(C)$$

Now consider two events $E_1$ and $E_2$ that are not mutually disjoint events i.e. if $E_1 \cap E_2 \neq \emptyset$ and we want to evaluate $P(E_1 \cup E_2)$.

From the figure, it is apparent that the intersection $E_1 \cap E_2$ is counted twice: once as part of $E_1$ and once as part of $E_2$. Hence, it needs to be subtracted off. For a more rigorous derivation, see below the figure.

enter image description here

The idea is to split $E_1 \cup E_2$ into three disjoint sets as follows. $$E_1 \cup E_2 = (E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1)$$

Note that $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ and $(E_2 \backslash E_1)$ are mutually disjoint sets. Hence, we have that $$P(E_1 \cup E_2) = P((E_1 \backslash E_2) \cup (E_1 \cap E_2) \cup (E_2 \backslash E_1))\\ = P(E_1 \backslash E_2) + P(E_1 \cap E_2) + P(E_2 \backslash E_1)$$ Call the above equation $\star$.

Now note that $$(E_1 \backslash E_2) \cup (E_1 \cap E_2) = E_1$$ Since $(E_1 \backslash E_2)$, $(E_1 \cap E_2)$ are mutually disjoint sets, we have that $$P(E_1 \backslash E_2) + P(E_1 \cap E_2) = P(E_1)$$ Hence, $$P(E_1 \backslash E_2) = P(E_1) - P(E_1 \cap E_2)$$ Similarly, since $$(E_1 \cap E_2) \cup (E_2 \backslash E_1) = E_2$$ are mutually disjoint sets, we have that $$ P(E_1 \cap E_2) + P(E_2 \backslash E_1) = P(E_2)$$ Hence, $$ P(E_2 \backslash E_1) = P(E_2) - P(E_1 \cap E_2)$$ Now plug in for $P(E_1 \backslash E_2)$ and $P(E_2 \backslash E_1)$ in $\star$, to get what you want.

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What does the notation with \ mean? –  user1038665 Oct 29 '12 at 23:45
    
@user1038665 The notation $A \backslash B$ means the set of elements in the set $A$ but not in the set $B$. Also, as Austin Mohr has pointed out, note that the intersection $E_1 \cap E_2$ is counted twice: once as part of $E_1$ and once as part of $E_2$. Hence, it needs to be subtracted off. –  user17762 Oct 29 '12 at 23:49
    
@marvis which program was used for the graphics? –  yiyi Oct 30 '12 at 1:05
    
@MaoYiyi I drew it using TikZ in LaTeX and cut copy pasted it. –  user17762 Oct 30 '12 at 1:08

The formula you give is the correct way to determine $\Pr(E_1 \cup E_2)$ (often referred to as the inclusion-exclusion principle). The union of the two events, however, does include outcomes occurring in both events. Formally, $$ E_1 \cup E_2 = \{\omega \mid \omega \in E_1 \text{ (inclusive) or } \omega \in E_2 \}. $$ The reason we subtract $\Pr(E_1 \cap E_2)$ in the formula you give is because outcomes occurring in the intersection would otherwise be counted twice. To see this, it is easier to just think of sets. Is it true that $$ |A \cup B| = |A| + |B|? $$ No. For a simple counterexample to this claim, let $A = B = \{1\}$. We know $A \cup B = \{1\}$, so $|A \cup B| = 1$, while $|A| + |B| = 2$. The reason is because the lefthand side counts the element $1$ a single time, while it is counted twice on the righthand side (once because it belongs to $A$ and once because it belongs to $B$). An analogous argument applies to probabilities.

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Why would the intersection be counted twice? –  user1038665 Oct 29 '12 at 23:46

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