Suppose we have three urns, namely, A B and C. A has 3 black balls and 7 white balls. B has 7 black balls and 13 white balls. C has 12 black balls and 8 white balls. We first choose one urn from A, B and C. Then we randomly pick up two balls from that urn without replacement. Let Ai, i 1,2,3 denote the event that the urn we choose is A, B and C respectively. Suppose P(A1): P(A2): P(A3) =1:2:2. Compute :
(a) The probability that the first ball is black.
(b) The probability that the first ball is black given that the second ball is white.

Three urns A, B and C contains ( 3 black balls 7 white balls), (7 black balls and 13 white balls) and (12 black balls and 8 white balls) respectively.

Now,

Since events of choosing urn A, B and C are denoted by Ai , i=1, 2, 3

Then , P(A1 + P(A2) +P(A3) =1 ....(1)

And P(A1):P(A2):P(A3) = 1: 2: 2 (given) ....(2)

Let P(A1) = x, then using equation (2)

P(A2) = 2x and P(A3) = 2x

(from the ratio given in the question)

Substituting these values in equation (1), we get

x+ 2x + 2x =1

Or 5x =1

Or x =1/5

So, P(A1) =x =1/5 , ....(3)

P(A2) = 2x= 2/5 and ....(4)

P(A3) = 2x= 2/5 ...(5)

Also urns A, B and C has total balls = 10, 20 , 20 respectively.

Now, if we choose one urn and then pick up 2 balls randomly then;

(a) Probability that the first ball is black

=P(A1)×P(Back ball from urn A) +P(A2)×P(Black ball from urn B) + P(A3)×P(Black ball from urn C)

= (1/5)×(3/10) + (2/5)×(7/20) + (2/5)×(12/20)

= (3/50) + (7/50) + (12/50)

=22/50

=11/25

(b) The Probability that the first ball is black given that the second ball is white is same as the probability that first ball is black (11/25). This is because the event of picking of first ball is independent of the event of picking of second ball.

Although the event picking of the second ball is dependent on the event of picking the first ball.

Hence, probability that the first ball is black given that the second ball is white is 11/25