Urn A contains 6 red , 4 white balls and urn B contains 4 red and 6 white balls.

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Urn A contains 6 red , 4 white balls and urn B contains 4 red and 6 white balls. One ball is drawn from the urn A and placed in the urn B. Then one ball is drawn at random from urn B and placed in the urn A. if one ball is now drawn from the urn A, then the probability that it is found to be red is

Options:

A . 3255

B . 3040

C . 3250

D . 34

Answer: Option A
:
A
case(i): red ball from A to B, red ball from B to A ,then red ball from A

P_{1} = \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}

Case(ii): red ball from A to B, white ball from B to A, red ball from A

P_{2} = \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}

Case(iii): white ball from A to B, red ball from B to A, red ball from A

P_{3} = \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}

Case(iv): white ball from A to B, white ball from B to A, red ball from A

P_{4} = \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}

Therefore required probability

= P_{1} + P_{2} + P_{3} + P_{4} = \frac{640}{1100} = \frac{32}{55}

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