Bayes's theorem

Bayes's theorem, named after Thomas Bayes, gives the probability of a cause given the occurrence of its effect.

Formally,
given events $A, B$ with $P (B) \neq 0$ ,

P (A ∣ B) = \frac{P (B ∣ A) \cdot P (A)}{P (B)}

And more generally,
let ${A_{1}, A_{2}, \dots}$ be a set of mutually exclusive and collectively exhaustive events. Then

P (A_{j} ∣ B) = \frac{P (B ∣ A_{j}) \cdot P (A_{j})}{\sum_{i} P (B ∣ A_{i}) \cdot P (A_{i})}

Let $A, B$ be events with $P (B) \neq 0$ .

\begin{aligned} P (A ∣ B) & = \frac{P (A \cap B)}{P (B)} \\ P (B ∣ A) & = \frac{P (B \cap A)}{P (A)} \end{aligned}

Rearranging both equations,

\begin{aligned} P (A \cap B) & = P (A ∣ B) \cdot P (B) \\ P (A \cap B) & = P (B ∣ A) \cdot P (A) \end{aligned}

setting both sides equal to each other,

P (A ∣ B) \cdot P (B) = P (B ∣ A) \cdot P (A)

dividing both sides by $P (B)$ ,

P (A ∣ B) = \frac{P (B ∣ A) \cdot P (A)}{P (B)}

Now let ${A_{1}, A_{2}, \dots}$ be a set of mutually exclusive and collectively exhaustive events.

By the law of total probability, $P (B) = \sum_{i} P (B ∣ A_{i}) \cdot P (A_{i})$

Therefore,

P (A_{j} ∣ B) = \frac{P (B ∣ A_{j}) \cdot P (A_{j})}{\sum_{i} P (B ∣ A_{i}) \cdot P (A_{i})}

QED.