2.10 Bayes’ rule

By the definition of conditional probability

\[ \text{P}(A \mid B)=\frac{\text{P}(A \cap B)}{\text{P}(B)} \]

By the multiplication rule

\[ \text{P}(A \cap B) = \text{P}(A) \cdot \text{P}(B \mid A) \]

Then, we have

\[ \text{P}(A \mid B)=\frac{\text{P}(A) \cdot \text{P}(B \mid A)}{\text{P}(B)} \]

Bayes’ rule

Let \(A_1, A_2, \cdots, A_n\) form a partition of sample space \(\Omega\).

Then, for any event \(B\) such that \(\text{P}(B)>0\), we have

\[ \text{P}(A_i \mid B) = \frac{\text{P}(A_i) \cdot \text{P}(B \mid A_i)}{\text{P}(B)} \]

\[\text{where}\]

\[ \small{\text{P}(B) = \text{P}(A_1) \cdot \text{P}(B \mid A_1) + \cdots + \text{P}(A_n) \cdot \text{P}(B \mid A_n)} \]

The false positive puzzle

  • 1 in 1,000 people is afflicted with a rare disease for which a diagnostic test has been developed.
  • When an individual has the disease, it would test positive (i.e., state disease present) 99% of the time.
  • When an individual does not have the disease, it would test negative 98% of the time.
  • A person just tested positive.
  • What is the chance this person has the disease?

Let’s define the following events

  • \(A\): Someone has the disease.
  • \(A^c\): Someone does not have the disease.
  • \(B\): Someone tested positive.
  • \(B^c\): Someone tested negative.

A person just tested positive.

What is the chance this person has the disease?

Exercise

  • We were given three coins.
    • One is regular.
    • One has heads in both faces.
    • One has tails in both faces.
  • Without looking at it, we chose a coin at random and tossed it. It landed on heads.
  • What is the probability that the opposite is tails?