mathsurgently needed(probability)
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.

mathsurgently needed(probability) 
Reiny
I will assume the first ball drawn is NOT returned.
prob(2green) = (n/(9+n) * ((n1)/(8+n)) = 1/7
7n^2  7n = n^2 + 17n + 72
6n^2  24n  72 = 0
n^2  4n  12 = 0
(n6)(n+2) = 0
n = 6 or n = a negative, which is no good
there were 6 green balls
check:
prob(2greens) = (6/15)(5/14) = 1/7
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Somebody please solve this A bag contains 5 red balls 4 blue balls and m green balls.Two balls are drawn at random from the bag.If probability of both being green is 1/7,then find m.