Answer:
D) 7
Step-by-step explanation:
Let y = number of yellow marbles in the bag
For the first marble
P(yellow) = number of yellow marbles over total
=y/12
For the second draw, there are y-1 yellow marbles in the bag, and only 11 marbles left since we keep the first marble
P(2nd marble yellow) = (y-1)/11
The probability that the first and second marble are yellow when we keep the first marble is
y/12 * (y-1)/11 = 5/33
y * (y-1) 5
---------- = ----------
121 33
We can use cross products to solve
33 y(y-1) = 660
Divide each side by 33
y(y-1) = 660/33
y(y-1) = 20
Distribute
y^2 -y =20
Subtract 20
y^2 -y -20 =0
Factor
(y-5)(y+4) =0
Using the zero product property
y = 5 or y = -4
Since we cant have negative marbles
We have 5 yellow marbles
Since there are 12 marbles
12-5 =7
There are 7 green marbles