posted by Anonymous .
A bag contains 5 red, 4 blue and an unknown number of m green balls.
Two balls are drawn. If probability of both being green is 1/7 find m.
I will assume the first ball drawn is NOT returned.
prob(2green) = (n/(9+n) * ((n-1)/(8+n)) = 1/7
7n^2 - 7n = n^2 + 17n + 72
6n^2 - 24n - 72 = 0
n^2 - 4n - 12 = 0
(n-6)(n+2) = 0
n = 6 or n = a negative, which is no good
there were 6 green balls
prob(2greens) = (6/15)(5/14) = 1/7