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?

Guest May 29, 2020

#1**0 **

At first:

Number of red marbles: R

Number of green marbles: 4

Probability of drawing a red marble: R / (R + 4) = x

After doubling the number of green marbles:

Number of red marbles: R

Number of greeen marbles: 8

Probability of drawing a red marble: R / (R + 8) = x - 0.15

--- solving this equation for x: R / (R + 8) + 0.15 = x

Setting the two equations equal to each other:

R / (R + 4) = R / (R + 8) + 0.15

Multiplying all terms by (R + 4) · (R + 8)

R · (R + 8) = R · (R + 4) + (0.15) · (R + 8) · (R + 4)

Multipling out:

R^{2} + 8R = R^{2} + 4R + 0.15(R^{2} + 12R + 32)

R^{2} + 8R = R^{2} + 4R + 0.15R^{2} + 1.8R + 4.8

Simplifying:

0 = 0.15R^{2} - 2.2R + 4.8

Multiplying by 100:

0 = 15R^{2} - 220R + 480

Dividing by 5:

0 = 3R^{2} - 44R + 96

Factoring:

0 = (3R + 8)(R - 12)

Answers: either -8/3 (impossible) or **12**

geno3141 May 29, 2020