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` .

Answer

10

Explanation

Let `E_(1)` denote the event that one of the first n urns in chosen and `E_(2)` denote the event that (n+1)th urn is selected. A denote the event that two balls drawn are black. Then
`p(E_(1))=(n)/(n+1),p(E_(2))=(1)/(n+1)`
`p((A)/(E_(1)))=(.^(6)C_(2))/(.^(10)C_(2))=(1)/(3) ` and `p((A)/(E_(2)))=(.^(5)C_(2))/(.^(10)C_(2))=(2)/(9)`
Using Bayes's theorem, required probability `=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))))`
`rArr = (1)/(16)=(((1)/(n+1))(2)/(9))/(((n)/(n+1))((1)/(3))+((1)/(n+1))((2)/(9)))=(2)/(3n+2)rArr n=10`