Question

State and prove Bayes’ theorem.

Answer 1

Hint: Bayes’ theorem describes the probability of occurrence of an event related to any condition. To prove the Bayes’ theorem, use the concept of conditional probability formula, which is $P({E_i}|A) = \dfrac{{P({E_i} \cap A)}}{{P(A)}}$.

Complete step-by-step answer:
Bayes’ Theorem-
Bayes’ Theorem describes the probability of occurrence of an event related to any condition. It is also considered for the case of conditional probability.
Statement: Let $E_1, E_2,…...,E_n$ be a set of events associated with a sample space S, where all the events $E_1, E_2,…...,E_n$ have non zero probability of occurrence and they form a partition of S. Let A be any event associated with S, then according to Bayes’ theorem,
$P({E_i}|A) = \dfrac{{P({E_i})P(A|{E_i})}}{{\sum\limits_{k = 1}^n {P({E_k})P(A|{E_k})} }},k = 1,2,3,....,n$
Proof: According to conditional probability formula,
$P({E_i}|A) = \dfrac{{P({E_i} \cap A)}}{{P(A)}} - (1)$
Using multiplication rule of probability,
$P({E_i} \cap A) = P({E_i})P(A|{E_i}) - (2)$
Using total probability theorem,
$P(A) = \sum\limits_{k = 1}^n {P({E_k})P(A|{E_k}) - (3)} $
Putting the values from equation (2) and (3) in equation (1), we get-
$P({E_i}|A) = \dfrac{{P({E_i})P(A|{E_i})}}{{\sum\limits_{k = 1}^n {P({E_k})P(A|{E_k})} }},k = 1,2,3,....,n$ [Hence proved]

Note: Whenever it is asked to state a theorem and prove it, write the statement of the theorem first and then prove it step by step. As mentioned in the solution, after stating the Bayes’ theorem, use the standard formula of conditional probability to prove the Bayes’ theorem.