Question

Each of an urn contains 4 white and 6 black balls,while another urn contains 6 white and 4 black balls,and urns chosen at random from the (n+1) urns and two balls are drawn from it,both being black.The probability that 6 white and 2 black balls remain in the chosen urn is 1/11.Then the value of n is

Answer 1

Given that there are n+1 bags of which first n bags contain 4W+6B balls

Probabily of choosing each of the first n bags=1/n+1

so total probablitl of choosing each of first n bag is =1/n+!

so total probabilty of drwaing 2 black balls =drawingn 2 black balls from frst n bags+drawing 2 black balls from te last bag

=n/(n+1)*6C2/10C2+1/n+1(5C2/10C2)

=probablilty of draing 2 balck balls from the last bag is=1/(n+1)*5C2/10C2

given that

1/(n+1)*5C2/10C2=1/7*[n/(n+!)*6C2/10C2+1/n+1*(5C2/10C2)]

solving this we et n=4