Each of the 'n' urns contain 4 white and 6 black balls. The (n + 1)^th urn contains 5 white and 5 black balls. One of the urn is chosen at random and two balls are drawn from it without replacement. Both the balls turn out to be black. If the probability that the (n +1)^th urn was chosen to draw the balls is $\frac{1}{16},$ then the value of n is

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Detailed Solution

Let $E_{1}$ be the event that one of first n urns is chosen, $E_{2}$ be the event that ${(n + 1)}^{th}$ urn is chosen.A be the event that two balls drawn are black.

$P (E_{1}) = \frac{n}{n + 1}; P (E_{2}) = \frac{1}{n + 1} P (\frac{A}{E_{1}}) = \frac{{}^{6}C_{2}}{{}^{10}C_{2}} = \frac{1}{3}; P (\frac{A}{E_{2}}) = \frac{{}^{5}C_{2}}{{}^{10}C_{2}} = \frac{2}{9} P (\frac{E_{2}}{A}) = \frac{P (E_{2}) . P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) . P (\frac{A}{E_{2}})} \Rightarrow \frac{1}{16} = \frac{\frac{1}{n + 1} . \frac{2}{9}}{\frac{n}{n + 1} . \frac{1}{3} + \frac{1}{n + 1} . \frac{2}{9}} \Rightarrow n = 10$