Bayes’s Theorem for Conditional Probability



We strongly recommend to refer below post as a per-requisite for this.

Conditional Probability



Bayes’s formula
Below is Bayes’s formula.
P(A|B) = \frac{P(B|A)P(A)}{P(B)}

The formula provides relationship between P(A|B) and P(B|A). It is mainly derived form conditional probability formula discussed in the previous post.
Consider the below forrmulas for conditional probabilities P(A|B) and P(B|A)
P(A|B) = \frac{P(A \cap B)}{P(B)} —-(1)

P(B|A) = \frac{P(B \cap A)}{P(A)} —-(2)



Since P(B ∩ A) = P(A ∩ B), we can replace P(A ∩ B) in first formula with P(B|A)P(A)

After replacing, we get the given formula.

Product Rule

Product rule states that

(1)                   \begin{equation*}  P(X \cap Y) = P(X|Y)*P(Y)   \end{equation*}

So the joint probability that both X and Y will occur is equal to the product of two terms:

From the product rule :
X \subseteq Y implies P(X|Y) = P(X)/P(Y)
Y \subseteq X implies P(X|Y) = 1

Chain rule

When the above product rule is generalized we lead to chain rule . Let there are E_{1}, E_{2},....E_{n} n events . Then , the joint probability is given by

(2)   \begin{equation*} P(\bigcap_{i=1,..,n}E_{i}) = P(E_{n}|\bigcap_{i=1,..,n-1}E_{i})*P(\bigcap_{i=1,..,n-1}E_{i}) \end{equation*}



Bayes’ Theorem

From the product rule,  P(X \cap Y) = P(X|Y)P(Y) and  P(Y \cap X) = P(Y|X)P(X) . As  P(X \cap Y) and  P(Y \cap X) are same .

(3)    \begin{equation*}  P(Y|X) = \frac{P(X|Y)*P(Y)}{P(X)}   \end{equation*}

where  P(X) = P(X \cap Y) + P(X \cap Y^{c}) .



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)
(A) 4/19
(B) 5/19
(C) 2/9
(D) 19/30

Solution:

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(P|R) = \frac{P(R|P)P(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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