You choose one of two identical looking bags at random. One bag has three black marbles and one white marble. The other has three white marbles and one black marble.

After choosing a bag you draw one marble out at random. You notice it is black. You then put it back and draw another marble out of the same bag at random.

What is the probability that the second marble drawn is black?

Let A be the event that the first marble drawn is black, B denote the event that the second marble drawn is black. Also let C denote the event that the bag chosen for the drawing is the one with 3 black and 1 white marbles and D denote the event that the bag chosen is the other one with 1 black and 3 white marbles.

We need to find the conditional probability of B given A, denoted P(B|A) which is P(A and B)/P(A).

Now P(A and B)= P(A and B and C) + P(A and B and D) since C and D are exclusive and exhaustive.

P(A and B and C) = P(A and B | C).P(C) = (3/4).(3/4).(1/2) = 9/32.

P(A and B and D) = P(A and B | D).P(D) = (1/4).(1/4).(1/2) = 1/32

So P(A and B) = (9/32)+(1/32)=(10/32)=(5/16)

Similiarly, P(A) = P(A and C) + P(A and D)
P(A and C) = P(A | C).P(C)=(3/4).(1/2)=(3/8)
P(A and D) = P(A | D).P(D)=(1/4).(1/2)=(1/8)

SO P(A)=(3/8)+(1/8)=(4/8)=(1/2).

Thus, the required probability

P(B|A) =(5/16)/(1/2)=(5/16).(2)=(5/8)

Posted by Prab on 2003-11-05 09:07:55