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 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

Let `E_(1)` denoter rthe evetn that one of the first n urns is chosen and `E_(2)` donate the evetn that (n + 1)th urn is selected. A denotes 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_(1))=""^(6)C_(2)//^(10)C_(2)=2//9.`
Usign Bayes's theorem, we have
`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)))`
`or 1/16=(((1)/(n+1))2/9)/(((n)/(n+1))((1)/(3))+((1)/(n+1))((2)/(9)))=(2)/(3n+2)or n=10`