**Bayesâ€™ theorem** describes the probability of occurrence of an event related to any condition. It is also considered for the case of conditional probability. 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 colour balls viz. red, blue, black. Such a case where the probability of occurrence of an event is calculated depending on other conditions is known as conditional probability.

## Bayes Theorem Proof

Statement:Let E_{1}, E_{2},…,E_{n}Â be a set of events associated with a sample space S, where all the events E_{1}, E_{2},…,E_{n}Â have nonzero probability of occurrence and they form a partition of S. Let AÂ be any event associated with S, then according to Bayes theorem,

\(P(E_iâ”‚A)~=~\frac{P(E_i)P(Aâ”‚E_i)}{\sum\limits_{k=1}^{n}P(E_k)P(A| E_k)}\)

for any k = 1, 2, 3, …., n

**Proof:**According to conditional probability formula,

\(P(E_iâ”‚A)~=~\frac{P(E_i âˆ© A)}{P(A)}\) â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯(1)

Using multiplication rule of probability,

\(P(E_i âˆ© A)~= ~P(E_i)P(A â”‚E_i)\)â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯(2)

Using total probability theorem,

\(P(A)~=~\sum\limits_{k=1}^{n}~P(E_k)P(A| E_k)\)â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯â‹¯(3)

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

\(P(E_iâ”‚A)~=~\frac{P(E_i)P(Aâ”‚E_i)}{\sum\limits_{k=1}^n~P(E_k)P(A| E_k)}\)

### Bayes Theorem Formula

If A and B are two events, then the formula for Bayes theorem is given by:

P(A|B) = N(Aâˆ©B)/N(B);Â N(B)â‰ 0 |

Where P(A|B) is the probability of condition when event A is occurring while event B has already occurred.

N(A âˆ© B) is the number of factors common to both A and B.

N(B) is the number of factors in B

### Examples and Solutions

Some illustrations will improve the understanding of the concept.

**Example 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 \(E_1\) be the event of choosing the bag I, \(E_2\) the event of choosing the bag II and A be the event of drawing a black ball.

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

Also,\(P(A|E_1) ~= ~P\)(drawing a black ball from Bag I) = \(\frac{6}{10}~ = ~\frac{3}{5}\)

\(P(A|E_2) ~=~ P\)(drawing a black ball from Bag II) = \(\frac{3}{7}\)

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

\(P(E_1 |A)~ =~\frac{P(E_1)P(A|E_1)}{P(E_1 )P(Aâ”‚E_1 )+ P(E_2)P(A|E_2)}\)

=\(\large\frac{\frac{1}{2}~\times~\frac{3}{5}}{\frac{1}{2}~\times~\frac{3}{7}~+~\frac{1}{2}~ Ã—~\frac{3}{5}}\) = \(\frac{7}{12}\)

**Example 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 \(A\) be the event that the man reports that number four is obtained.

Let \(E_1\) be the event that four is obtained and \(E_2\) be its complementary event.

Then, \(P(E_1)\) = Probability that four occurs = \(\frac{1}{6}\)

\(P(E_2)\) = Probability that four does not occurs = \(1 ~â€“~ P(E_1) ~=~ 1~-\frac{1}{6}~ =~\frac{5}{6}\)

Also, \(P(A|E_1)\) = Probability that man reports four and it is actually a four = \(\frac{2}{3}\)

\(P(A|E_2)\) = Probability that man reports four and it is not a four = \(\frac{1}{3}\)

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

\(P(E_1 |A)~ \) \(= \large \frac{P(E_1)P(A|E_1)}{P(E_1 )P(Aâ”‚E_1 )~+~ P(E_2)P(A|E_2)}~

=~\frac{\frac{1}{6} ~ Ã—~ \frac{2}{3}}{\frac{1}{6} ~Ã—~ \frac{2}{3}~ +~ \frac{5}{6}~ Ã—~\frac{1}{3}}\) = \(\frac{2}{7}\)

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Outstanding explanation