Question 263800: A box contains some green marbles and exactly four red marbles. The probability of selecting a red marble is x%. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x – 15)%. How many green marbles are in the box before the number of green marbles is doubled?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A box contains some green marbles and exactly four red marbles. The probability of selecting a red marble is x%. If the number of green marbles is doubled, the probability of selecting one of the four red marbles from the box is (x – 15)%. How many green marbles are in the box before the number of green marbles is doubled?
----
Original setup
P(1 red) = 4/(g+4) = x/100
Then g + 4 = 4/x/100
g = (4/(x/100))-4
g = (400/x) -4
----------------------
New setup
P(1 red) = 4/(2g+4) = (x-15)/100
(2g+4) = 400/(x-15)
g+2 = 200/(x-15)
g = 200/(x-15) -2
----------------------------
Solve for "g":
(400/x) -4 = (200/(x-15) -2
Divide thru by 200 to get:
(2/x) - 2/100 = (1/(x-15) - 1/100
(2/x) - (1/(x-15) = 1/100
200(x-15) - 100x = x(x-15)
200x - 3000 - 100x = x^2-15x
x^2 -115x +3000 = 0
(x-75)(x-40) = 0
----------------------------
x is 75 or x = 40
Need to solve for "g"
g = 200/(x-15) -2
================================
If x = 75,
g = 200/(50) -2 = 2
If x = 40,
g = 200/25 - 2 = 6
---------------------------
Looks like original # of green is 2 or 6.
Hummmm!!!
---------------------------
Cheers,
Stan H.
----------------------------------
|
|
|
