Question:
There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.
Solution:
Given, we have 3 urns:
Urn 1 = 2 white and 3 black balls
Urn 2 = 3 white and 2 black balls
Urn 3 = 4 white and 1 black balls
Now, the probabilities of choosing either of the urns are
P(U1) = P(U2) = P(U3) = 1/3
Let H be the event of drawing white ball from the chosen urn.
So,
P(H/U1) = 2/5, P(H/U2) = 3/5 and P(H/U3) = 4/5
By using Baye’s Theorem, we have
Therefore, the required probability is 1/3.