# Bayes Theorem

## Bayes Theorem

In Probability, Bayes theorem describes the probability of an event, based on the prior knowledge of conditions that might be related to the event.

If A₁, A₂, A₃, … An are n mutually exclusive events from the Sample Space S, B is any other event from S and if probability of occurrence of Ai’s and probability of occurrence of B given that Ai, i = 1, 2, 3, … n has occurred are known, then probabilities of occurrence of Ai’s given that B has occurred (or) If A₁, A₂, A₃, … An are mutually exclusive and exhaustive event in a sample space S such that P(Ai) > 0 for i = 1, 2, 3, … n and E is any event with P(E) > 0 then $$P\left( \frac{{{A}_{k}}}{E} \right)=\frac{P({{A}_{k}})P\left( \frac{E}{{{A}_{k}}} \right)}{\sum\limits_{i=1}^{n}{P({{A}_{i}})P\left( \frac{E}{{{A}_{i}}} \right)}}$$ for k = 1, 2, … n

Proof: From condition probability $$P(E)=\sum\limits_{i=1}^{n}{P({{A}_{i}})P\left( \frac{E}{{{A}_{i}}} \right)}$$,

For k = 1, 2, … n : $$P(E\cap {{A}_{k}})=P(E)P\left( \frac{{{A}_{k}}}{E} \right)$$,

$$P\left( \frac{{{A}_{K}}}{E} \right)=\frac{P(E\cap {{A}_{k}})}{P(E)}=\frac{P({{A}_{k}})P\left( \frac{E}{{{A}_{k}}} \right)}{\sum\limits_{i=1}^{n}{P({{A}_{i}})P\left( \frac{E}{{{A}_{i}}} \right)}}$$.

Example: In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is ⅓ and the probability that he copies the answer is ⅙. The probability that his answer is correct given that he copied it is ⅛. Find the probability that he knew the answer to the questions given that he correctly answered it.

Solution: let us use the following symbols for denoting the various options

A for guesses

B for copies

C denotes the possibility that the examinee knows

R if the answer is right

So, P (A) = ⅓

P (B) = ⅙

P(C) = 1 – (⅓ + ⅙) = ½

Now R = (R∩A) ∪ (R∩B) ∪ (R∩C)

P(R) = P (A) P(R/A) + P (B) P(R/B) + P(C) P(R/C) … (1)

Now, P(R/A) = ¼

P(R/B) = ⅛

P(R/C) = 1

Putting this in equation (1), we obtain

P(R) = ⅓.¼ + ⅙.⅛ + 3/6. 1

= 1/12 +1/48 + 3/6

= 29/48

Hence, the required probability = P (C∩R)/ P(R)

= 24/29.