SOLUTION: A bag contains five red marbles, three green marbles, and four blue marbles. How many green marbles must be added to the bag so that the probability of drawing a green marble is 3

Question 240249: A bag contains five red marbles, three green marbles, and four blue marbles. How many green marbles must be added to the bag so that the probability of drawing a green marble is 3/4?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
T = 5+3+4 = 12
This means the total number of marbles is equal to 5 red plus 3 green plus 4 blue which equals a total of 12.

G = 3

This means the number of green marbles you are starting out with is 3.

P(G) is now 3/12 = 1/4

P(G) means the probability that you will draw a green marble is equal to .25 right now.

We want:

P(G) = 3/4

We need to add X number of green marbles.

P(G) = 3/4 = (X+3) / (12 + X)

This means that we need to add x number of green marbles to the total. This will change both the numerator of the equation (number of green marbles) and the denominator of the equation (number of total marbles).

We need to solve for X.

(X + 3) / (X + 12) = 3/4

Multiply both sides by (X+12) to get:

(X+3 = 3/4 * (X + 12)

Simplify to get:

X + 3 = (3/4)*X + (3/4)*12

Simplify further to get:

X + 3 = (3/4)*X + 9

Subtract 3 from both sides of this equation to get:

X = (3/4)*X + 9 - 3

Subtract (3/4)*X from both sides of this equation to get:

X - (3/4)*X = 9 - 3

Combine like terms to get:

(1/4)*X = 6

Multiply both sides of the equation by 4 to get:

X = 6 * 4 = 24

Number of green marbles should be 24.

Add 24 green marbles to 3 and you get 27.

Add 24 to 12 and you get 36

27/36 = .75 = 3/4

the probability of getting a green marble now is .75 because 75% of the marbles out of the total number of marbles are green.

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