Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red, is
Let the events are
R1 = A red ball is drawn from urn A and placed in B
B1 = A black ball is drawn from urn A and placed in B
R2 = A red ball is drawn from urn B and placed in A
B2 = A black ball is drawn from urn B and placed in A
R = A red ball is drawn in the second attempt from A
Then the required probability
= P(R1R2R)+(R1B2R)+P(B1R2R)+P(B1B2R)
= P(R1)P(R2)P(R)+P(R1)P(B2)P(R)+P(B1)P(R2)P(R)+P(B1)P(B2)P(R)
= 610×511×610+610×611×510+410×411×710+410×711×610
= 3255
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