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 both b***s are red is 1/15 . How many b***s in the bag are red?

tertre Sep 10, 2016

#1**0 **

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

CPhill Sep 10, 2016