Each of the `n` urns contains 4 white and 6 black balls. The `(n+1)` th urn contains 5 white and 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 1/16, then find the value of `n` .

Explanation

`E_1` : one of the n urn is chosen
`E_2 : (n+1)^(th)` urn is chosen
A : both the balls drawn are black
`P(E_1)=(n)/(n+1)`
`P(E_2)=1/(n+1)`
`P(A|E_1)=(""^6C_2)/(""^10C_2)=1/3`
`P(A|E_2)=(""^5C_2)/(""^10C_2)=2/9`
By Bayes' rule
`P(E_2|A)=(P(E_2)P(A|E_2))/(P(E_1)P(A|E_1)+P(E_2)P(A|E_2))`
`P(E_2|A)=2/9 implies n=10`