Bayes’ theorem

We again consider the conditional probability statement:

$P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (A \cap B)}{P (B | A_{1}) P (A_{1}) + P (B | A_{2}) P (A_{2}) + \dots + P (B | A_{n}) P (A_{n})}$

in which we have used the theorem of Total Probability to replace $P (B)$ . Now

$P (A \cap B) = P (B \cap A) = P (B | A) \times P (A)$

Substituting this in the expression for $P (A | B)$ we immediately obtain the result

$P (A | B) = \frac{P (B | A) \times P (A)}{P (B | A_{1}) P (A_{1}) + P (B | A_{2}) P (A_{2}) + \dots + P (B | A_{n}) P (A_{n})}$

This is true for any event $A$ and so, replacing $A$ by $A_{i}$ gives the result, known as Bayes’ theorem as

$P (A_{i} | B) = \frac{P (B | A_{i}) \times P (A_{i})}{P (B | A_{1}) P (A_{1}) + P (B | A_{2}) P (A_{2}) + \dots + P (B | A_{n}) P (A_{n})}$