Bayes’ theorem describes the probability of occurrence of an event related to any condition. For example: if we have to calculate the probability of taking a blue ball from the second bag out of three different bags of balls, where each bag contains three different color balls viz. red, blue, black. Such case where probability of occurrence of an event is calculated depending on other conditions is known as conditional probability.

Derivation of Bayes Theorem:

Statement:Let \(E_1, E_2,…,E_n\)

\(P(E_i│A)~=~\frac{P(E_i)P(E_i│A)}{\sum\limits_{k=0}^{n}P(E_k)P(A| E_k)}\)

Proof:According to conditional probability formula,

\(P(E_i│A)~=~\frac{P(E_i ∩ A)}{P(A)}\)

Using multiplication rule of probability,

\(P(E_i ∩ A)~= ~P(E_i)P(E_i│A)\)

Using total probability theorem,

\(P(A)~=~\sum\limits_{k=0}^{n}~P(E_k)P(A| E_k)\)

Putting the values from equations (2) and (3) in equation 1, we get

\(P(E_i│A)~=~\frac{P(E_i)P(E_i│A)}{\sum\limits_{k=0}^n~P(E_k)P(A| E_k)}\)

Some illustrations will improve the understanding of the concept.

Illustration 1:Bag I contains 4 white and 6 black balls while another Bag II contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag I.

Solution:Let \(E_1\)

Then,\(P(E_1)~ = ~P(E_2)~ =~\frac{1}{2}\)

Also,\(P(A|E_1) ~= ~P\)

\(P(A|E_2) ~=~ P\)

By using Bayes’ theorem, the probability of drawing a black ball from bag I out of two bags,

\(P(E_1 |A)~ =~\frac{P(E_1)P(A|E_1)}{P(E_1 )P(A│E_1 )+ P(E_2)P(A|E_2)}\)

=\(\large\frac{\frac{1}{2}~\times~\frac{3}{5}}{\frac{1}{2}~\times~\frac{3}{7}~+~\frac{1}{2}~ ×~\frac{3}{5}}\)

Illustration 2:A man is known to speak truth 2 out of 3 times. He throws a die andreports that number obtained is a four. Find the probability that the number obtained is actually a four.

Solution:Let \(A\)

Let \(E_1\)

Then, \(P(E_1)\)

\(P(E_2)\)

Also, \(P(A|E_1)\)

\(P(A|E_2)\)

By using Bayes’ theorem, probability that number obtained is actually a four,

\(P(E_1 |A)~ \)

=~\frac{\frac{1}{6} ~ ×~ \frac{2}{3}}{\frac{1}{6} ~×~ \frac{2}{3}~ +~ \frac{5}{6}~ ×~\frac{1}{3}}\)

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