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.