Question
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
Answer: Option D
:
D
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 A and placed in A
B2= A black ball is drawn from urn A and placed in B
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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:
D
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 A and placed in A
B2= A black ball is drawn from urn A and placed in B
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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