Class 12 Maths Probability | Bayes Theorem |

**Bayes' Theorem**:

If E1, E2,..., En are mutually exclusive and exhaustive events associated

with a sample space, and A is any event of non zero probability, then

** Example**: In a survey in a classroom it was found that chance of a person becoming a topper is 10% . if a person is a topper, there is a 99% chance that he watches examfear videos. One who is not a topper there is 1% chance of watching Examfear videos. Then, what is the Probability for one to be a Topper , given if he watches Examfear videos?

* Solution*:

Let “A person to be a Topper” is Event A and “To watch Examfear videos” is Event B

Given P(A)=0.1 = probability for a person to be a Topper.

So P(A')= 1-P(A)=1-0.1=0.9=probability for a person to be not a Topper

Given P(B|A)=0.99= probability that a person watches examfear videos, if he is a topper,

so P(B'|A)=1-P(B|A)=1-0.99=0.01=probability that a person not watches

examfear videos, if he is a topper

Given P(B|A')=0.01=probability that a person watches Examfear videos , if he

is not a topper

so P(B'|A')=1-0.01=0.99=probability that a person not watches Examfear

videos , if he is not a topper

Therefore according to Bayes' Theorem,Probability for one to be a Topper , if he watches Examfear videos is

= (0.1x0.99)/ {( (0.1x0.99)+(0.01x0.9))}

= 0.9909

∴if one watches Examfear videos , for him there is 99.09% chance to become topper.

* Example*: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

* Solution*:

Let E1= the event of selecting first bag.

E2=the event of selecting second bag.

A = The event of getting red ball.

Since there is equal chance of selecting first bag or selecting second bag,

P(E1)=P(E2)= 1/2

now P(A|E1)=P(Drawing a red ball from first bag )= 4/8

and P(A|E2)=P(Drawing a red ball from second bag )= 2/8 =1/4

probability that ball is drawn from the first bag given that the ball drawn is red is P(E1|A)

* Example*:Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

**Solution:**

Let E1= event that the first group wins the competition

E2=event that the second group wins the competition

A = event of introducing a new product.

Then,

P(E1) = Probability that the first group wins the competition = 0.6

P(E2) = Probability that the second group wins the competition = 0.4

P(A|E1)=Probability of introducing new product if the first group wins =0.7

P(A|E2)=Probability of introducing new product if the second group wins =0.3

The probability that the new product is introduced by the second group is given by

.