A5 - law of total probability

for two sets:

E = (E \cap F) \cup (E \cap \bar{F})

$(E \cap F)$ and $(E \cap \bar{F})$ are mutually exclusive, so:

P (E) = (E \cap F) + (E \cap \bar{F}) = P (E | F) P (F) + P (E | \bar{F}) P (\bar{F})

law of total probability

let $F_{1}, F_{2}, \dots, F_{n}$ partition the state space $Ω$ , then:

P (E) = \sum_{i = 1}^{n} P (E \cap F_{i}) = \sum_{i = 1}^{n} P (E | F_{i}) P (F_{i})

the above equation also holds if $F_{n}$ partition $E$
it also extends to the case where in there are countably infinite partitions
proof:

E = \cup_{i = 1}^{n} (E \cap F_{i})

since $E \cap F_{i}$ are mutually exclusive:

P (E) = \sum_{i = 1}^{n} P (E \cap F_{i}) = \sum_{i = 1}^{n} P (E | F_{i}) P (F_{i})

note that events conditioned on must be:
- mutually exclusive
- sum to the whole space

law of total probability for conditional probability

let $F_{1}, F_{2}, \dots F_{n}$ partition the $Ω$ , then:

P (A | B) = \sum_{i = 1}^{n} P (A | B \cap F_{i}) P (F_{i} | B)

proof:

\begin{aligned} P (A | B) & = \frac{P (A \cap B)}{P (B)} \\ = \frac{\sum_{i} P (A \cap B \cap F_{i})}{P (B)} \\ = \frac{\sum_{i} P (B) P (F_{i} | B) P (A | B \cap F_{i})}{P (B)} \\ = \sum_{i} P (F_{1} | B) P (A | B \cap F_{i}) \end{aligned}