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Each of the n urns contains 4 white and black balls. The $(n + 1) th$ urn contains 5 white, 5 black balls. One of the $(n + 1)$ urns 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 value of $n$ is _____

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$^{10} C_{9}$

$^{10} P_{9}$

$^{10} C_{1}$

answer is A, B, D.

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Detailed Solution

$E_{1} -$ One of the first ‘n’ was chosen

$E_{2} - {(n + 1)}^{t h}$ urn is chosen

$A -$ Two blank balls are chosen

$P (\frac{E_{2}}{A}) = \frac{1}{16}$

By Bay’s theorem,

$P (\frac{E_{2}}{A}) = \frac{P (E_{2}) \cdot P (\frac{A}{E_{2}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{E}{E_{2}})}$

$P (E_{1}) = \frac{n}{n + 1}, P (E_{2}) = \frac{1}{n + 1}$

$P (\frac{A}{E_{1}}) = \frac{^{6} C_{2}}{^{10} C_{2}} = \frac{1}{3} P (\frac{A}{E_{2}}) = \frac{^{5} C_{2}}{^{10} C_{2}} = \frac{2}{9}$

$P (\frac{E_{2}}{A}) = \frac{(\frac{1}{n + 1}) \frac{2}{9}}{(\frac{n}{n + 1}) \frac{1}{3} + (\frac{1}{n + 1}) \frac{2}{9}}$

$P (\frac{E_{2}}{A}) = \frac{\frac{2}{9 (n + 1)}}{\frac{3 n + 2}{9 (n + 1)}} = \frac{1}{16}$

$\begin{array}{l} \frac{2}{3 n + 2} = \frac{1}{16} \\ 3 n + 2 = 32 \end{array}$

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