Question

There are 10 marbles in a bag, and the marbles are either red or blue. Eric will randomly choose two marbles from the bag, without replacing the first one. If the probability of both marbles' being red is
2/15, how many BLUE marbles are in the bag?

Answer 1

Let a be the number of red balls in the bag, so that there are 10-a blue balls in the bag.

Chance of both balls being red is (a/10)*(a-1)/9=(a^2-a)/90
Equating this to 2/15, we see that
a^2-a=90*2/15=12
or a^2-a-12=0
Solve the quadratic (in fact easy to factor it) (a-4)(a+3)=0
a=4 or -3; But a=-3 is inadmissible (negative number of balls!)
So a=4
This means there are 6 blue balls.