**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, … A_{n} are n mutually exclusive events from the Sample Space S, B is any other event from S and if probability of occurrence of A_{i}’s and probability of occurrence of B given that A_{i}, i = 1, 2, 3, … n has occurred are known, then probabilities of occurrence of A_{i}’s given that B has occurred (or) If A₁, A₂, A₃, … A_{n} are mutually exclusive and exhaustive event in a sample space S such that P(A_{i}) > 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.