Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black ball

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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

Options:

A . 3255

B . 2155

C . 1955

D . None of these

Answer: Option D
:
D
Let the events are

R_{1}

= A red ball is drawn from urn A and placed in B

B_{1}

= A black ball is drawn from urn A and placed in B

R_{2}

= A red ball is drawn from urn A and placed in A

B_{2}

= 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 (R_{1} R_{2} R) + (R_{1} B_{2} R) + P (B_{1} R_{2} R) + P (B_{1} B_{2} R)

= P (R_{1}) P (R_{2}) P (R) + P (R_{1}) P (B_{2}) P (R) + P (B_{1}) P (R_{2}) P (R) + P (B_{1}) P (B_{2}) P (R)

= \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10} \times \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10} \times \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10} + \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}

= \frac{32}{55}

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