bayes theorem

Question 1:
An urn B1 contains 2 white and 3 black balls and another urn B2 contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was B1.
Solution:
Step1:
Let E1, E2 denote the events of selecting urns B1 and B2 respectively.
Then P(E1) = P(E2) = 1/2.
Then we have to find P(E1/B).

Step 2:
By hypothesis P(B/E1) = 3/5 and P(B/E2) = 4/7
By Bayes theorem
P(E1/B) = P(E1)P(B/E1) / P(E1)P(B/E1)+P(E2)P(B/E2)
= (1/2?3/5)/ (1/2?3/5+1/2?4/7)
= 21/41
Question 2:
The urns contain 6 green, 4 black; 4 green, 6 black and 5 green, 5 black balls respectively. Randomly selected an urn and a ball is drawn from it. If the ball drawn is Green, find the probability that it is drawn from the first urn.


Solution:
Let E1, E2, E3 and A be the events defined as follows:
E1 = urn first is chosen, E2 = urn second is chosen,
E3 = urn third is chosen, and A = ball drawn is Green.
Step 1:
Since there are three urns and one of the three urns is chosen at random, therefore
P(E1) = P(E2) = P(E3) = 1/3
If E1 is already occurred, then urn first has been chosen which contains 6 Green and 4 Black balls. The probability of drawing a green ball from it is 6/10.
So, P(A/E1) = 6/10
Similarly, we have P(A/E2) = 4/10 and P(A/E3) = 5/10
Step 2:
We are required to find P(E1/A) , i.e. given that the ball drawn is Green , what is the probability that it is drawn from the first urn. By Bayes’ theorem, we have
P(E1/A) = P(E1)P(A/E1)/ P(E1)P(A/E1)+P(E2)P(A/E2)+P(E3)P(A/E3)
= (1/3?6/10)/ (1/3?6/10)+(1/3?4/10) +(1/3?5/10)
= 6/15
= 2/5
=> P(E1/A)= 2/5.