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Using ${\mathbf{1}}_{A^c} = 1 - {\mathbf{1}}_A$ and expanding ${\mathbf{1}}_{\cup_i E_i} = 1 - \prod_{i=1}^n (1-{\mathbf{1}}_{E_i})$, then taking expectations yields
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$$
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P\Big( \bigcup_{i=1}^n\ E_i \Big)
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P\Big( \bigcup_{i=1}^n\ {\E}_i \Big)
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= \sum_{k=1}^n (-1)^{k-1}
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\sum_{1 \leq i_1 < \cdots < i_k \leq n}
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P(\E_{i_1} \cap \cdots \cap \E_{i_k}).
@@ -61,7 +74,8 @@ If $X\in\{0,1\}$ takes value $1$ with probability $p$, we write $X\sim\operatorn
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### Binomial
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Let $X$ be the total number of successes in $n$ independent Bernoulli$(p)$ trials. Then $X\sim \operatorname{Bin}(n,p)$ and can be written $X=X_1+\cdots+X_n$ with $X_i\sim\operatorname{Ber}(p)$. The pmf is
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Let $X$ be the total number of successes in $n$ independent Bernoulli$(p)$ trials. Then $X\sim \operatorname{Bin}(n,p)$ and can be written $X=X_1+\cdots+X_n$ with $X_i\sim\operatorname{Ber}(p)$. The pmf is
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