Bayes’s Theorem for Conditional Probability
We strongly recommend to refer below post as a pre-requisite for this.
Below is Bayes’s formula.
The formula provides the relationship between P(A|B) and P(B|A). It is mainly derived from conditional probability formula discussed in the previous post.
Consider the below formulas for conditional probabilities P(A|B) and P(B|A)
Since P(B ∩ A) = P(A ∩ B), we can replace P(A ∩ B) in the first formula with P(B|A)P(A)
After replacing, we get the given formula.
Product rule states that
So the joint probability that both X and Y will occur is equal to the product of two terms:
From the product rule :
implies P(X|Y) = P(X)/P(Y)
implies P(X|Y) = 1
When the above product rule is generalized we lead to chain rule. Let there are n events . Then, the joint probability is given by
From the product rule, and . As and are same .
Example : Box P has 2 red balls and 3 blue balls and box Q has 3 red balls and 1 blue ball. A ball is selected as follows:
(i) Select a box (ii) Choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are (1/3) and (2/3), respectively.
Given that a ball selected in the above process is a red ball, the probability that it came from the box P is (GATE CS 2005)
R --> Event that red ball is selected B --> Event that blue ball is selected P --> Event that box P is selected Q --> Event that box Q is selected We need to calculate P(P|R)? P(R|P) = A red ball selected from box P = 2/5 P(P) = 1/3 P(R) = P(P)*P(R|P) + P(Q)*P(R|Q) = (1/3)*(2/5) + (2/3)*(3/4) = 2/15 + 1/2 = 19/30 Putting above values in the Bayes's Formula P(P|R) = (2/5)*(1/3) / (19/30) = 4/19
Exercise A company buys 70% of its computers from company X and 30% from company Y. Company X produces 1 faulty computer per 5 computers and company Y produces 1 faulty computer per 20 computers. A computer is found faulty what is the probability that it was bought from company X?
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