# Law of total probability

## Statement

Let [ilmath](S,\Omega,\mathbb{P})[/ilmath] be a probability space, let [ilmath](U_i)_{i\eq 1}^n\subseteq\Omega[/ilmath] be a finite collection of [ilmath]\mathbb{P} [/ilmath]-measurable sets such that:

• [ilmath]S\subseteq\bigcup_{i\eq 1}^n U_i[/ilmath][Note 1], and,
• the [ilmath](U_i)_{i\eq 1}^n[/ilmath] are pairwise disjoint, [ilmath]\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big][/ilmath][Note 2]

then we claim:

• $\forall A\in\Omega\left[\mathbb{P}[A]\eq\sum_{i\eq 1}^n\mathbb{P}[A\cap B_i]\right]$
• There are a few alternate forms:
1. [ilmath]\forall A\in\Omega\Big[\mathbb{P}[A]\eq\sum_{i\eq 1}^n\mathbb{P}[A\vert B_i]\mathbb{P}[B_i]\Big][/ilmath], or even
2. [ilmath]\forall A\in\Omega\Big[\mathbb{P}[A]\eq\sum_{i\eq 1}^n\mathbb{P}[B_i\vert A]\mathbb{P}[A]\Big][/ilmath]

## Notes

1. As:
• [ilmath]\forall A\in\Omega[A\subseteq S][/ilmath], specifically each [ilmath]U_i\subseteq S[/ilmath], and,
• as a union of subsets is a subset of the union (in this case: [ilmath]\bigcup_{i\eq 1}^n U_i \subseteq \bigcup_{A\in\{S\} } A[/ilmath] specifically)
we automatically have:
• [ilmath]\bigcup_{i\eq 1}^n U_i\subseteq S[/ilmath]
Combine this with the requirement that
• [ilmath]S\subseteq\bigcup_{i\eq 1}^n U_i[/ilmath]
We see:
• [ilmath]S\eq\bigcup_{i\eq 1}^n U_i[/ilmath]
Thus:
• [ilmath]\big[S\eq\bigcup_{i\eq 1}^n U_i\big]\iff\big[S\subseteq\bigcup_{i\eq 1}^n U_i\big][/ilmath]
So it doesn't matter if we use [ilmath]\eq[/ilmath] or [ilmath]\subseteq[/ilmath]
2. The property of pairwise disjointedness is often stated using the contrapositive, that is:
• [ilmath]\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big]\iff\big[(U_i\cap U_j\neq\emptyset)\implies(i\eq j)\big][/ilmath]
giving:
• [ilmath]\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(U_i\cap U_j\neq\emptyset)\implies(i\eq j)\big][/ilmath] as opposed to
• [ilmath]\forall i,j\in\{1,\ldots,n\}\subseteq\mathbb{N}\big[(i\neq j)\implies(U_i\cap U_j\eq\emptyset)\big][/ilmath] which we gave above