# Bayes’ Theorem

### Partition

**A partition of Ω **is a sequence of** disjoint sets A _{1}, A_{2}, …** such that

## The Law of Total Probability

Let A_{1}…A_{k}be a partition of Ω. Then, for any event B,

**Example**: There probability of accident of an accident-prone person is 0.4 and that of a not accident-prone person is 0.2. If there are 30% accident-prone person in society, what is the probability of accident of a random person?**Solution**: Note that **accident-prone** and **not-accident prone** is the partition of types of people.

P(A): (probability of a person to be accident-prone): 0.3

P(A^{c}): 0.7

P(X|A): (probability of accident if the person is accident prone): 0.4

P(X|A^{c}): 0.2

Therefore, P(X) = P(X|A)P(A) + P(X|A^{c})P(A^{c}) = 0.4×0.3 + 0.2×0.7 = 0.26

## Bayes’ Theorem

Let A_{1}, …, A_{k}be a partition of Ω such that P(A_{i})>0 for each i.If P(B)>0 then, for each i=1,…, k,

We call **P(A _{i})** the

**prior probability**of A and

**P(A**the

_{i}|B)**posterior probability**of A. The Bayes’ theorem is commonly seen in this form,

Notice that this is nothing but a consolidated form of the law of total probability in the denominator. Try deriving both the forms of Bayes’ theorem. Hint: Use the conditional probability formula.

**Example:** I divide my email into three categories: A1 = “spam,” A2 = “low priority” and A3 = “high priority.” From previous experience, I find that

P(A1) = 0.7

P(A2) = 0.2

P(A3) = 0.1

Of course, 0.7 + 0.2 + 0.1 = 1

Let B be the event that the email contains the word “free.” From previous experience, P(B|A1) = 0.9, P(B|A2) = 0.01, P(B|A1) = 0.01. (Note: .9 + .01 + .01 (= 1.) I receive an email with the word “free.” What is the probability that it is spam? Bayes’ theorem yields,