There are two urns containing 4 red, 5 black and 5 red balls, 6 black balls.One ball is drawn at random from the first and transferred to the second and the one ball is drawn from the second and transferred to the first. After this mutual transfer one ball is drawn at random from the first urn. Let P be the probability that it will be red ball $\frac{1}{16}$ then the least inter greater than $(\frac{243}{13}) P$ is

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answer is 9.

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Detailed Solution

First urn contains 4 Red, 5 black balls second balls is transferred from 1^sturn to 2^nd urn. Then one is transferred from 2^nd to 1^st urn. Now on ball is from 1st urn.
P(getting red) + P(RRR) + P(RBR) + P(BRR) + P(BBR)
$= \frac{433}{12 \times 81} = P$ (given)
Now, $(\frac{243}{13}) P = \frac{243}{13} \times (\frac{433}{12 \times 81}) = \frac{433}{52} \neq integer$
$[\frac{433}{52}] = 8$ or least integer that is greater than
$[\frac{433}{52}]$ is 9

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