Here are 12 marbles in a bag, and the marbles are either yellow or green. two marbles will be randomly picked from the bag, without replacing the first one picked. the probability that both marbles will be yellow is 5 33 . how many green marbles are in the bag? a) 4 b) 5 c) 6 d) 7

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

a

b

d

step-by-step explanation: