Each of the $n$ urns contains $4$ white and $6$ black balls. The $(n+1)th$ urn contains $5$ white and $5$ black balls. Out of the $(n+1)$ urns is chosen an urn 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 ball is $\displaystyle \frac { 1 }{ 6 } $, then the value of $n$ is

Using Bayes' theorem, the probability that the balls were drawn from the (n+1)th urn is calculated as the probability of drawing two black balls from that urn divided by the total probability of drawing two black balls from any urn. The probability of drawing two black balls from the (n+1)th urn is 5C2 out of 10C2, which is 10/45 or 2/9, while the probability from the other n urns is 6C2 out of 10C2, which is 15/45 or 1/3. Setting up the equation for a total probability of 1/6 gives (n multiplied by 1/3 plus 2/9) divided by n plus 1 equal to 1/6, and solving for n yields n equals 10.

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