A bag contains ten b***s, some of which are red and the rest of which are yellow. When two b***s are drawn at random at the same time, the probability that

Let the number of yellow b***s = Y

Then the number of red b***s =  10 - Y

And we know that   the chance of drawing a red ball on the first draw = [10 - Y] / 10

And the chance of drawing a red ball on the second draw [without rplacement]  = [9 -Y] / 9

So.....the chance of drawing a red ball on both draws is given as :

[ 10 - Y] / 10 * [9 - Y] / 9  = 1/15     simplify

[90 - 19Y + Y^2] / 90  = 1/15     multiply both sides by 90

Y^2 - 19Y + 90 = 90/15

Y^2 - 19Y + 90  = 6     subtract 6 from both sides

Y^2 - 19Y + 84 = 0    factor

(Y - 12)(Y - 7)  = 0     ans setting each factor to 0, we have that  Y = 12  [impossible]   or Y  = 7

So.....the number of red ones = 10 - Y   = 10 - 7   = 3

Proof :    (3/10) (2/9)  =  6/90   = 1/15