Question

There are 10 marbles in a bag, and the marbles are either red or blue. Eric will randomly choose two marbles from the bag, without replacing the first one. If the probability of both marbles' being red is

2
15
, how many RED marbles are in the bag?

Answer 1

6 on usatestprep

Explanation:

Eric’s probability on the first pick is

x /10

His probability of the second pick is

x-1 /9

This will give a solution of

x(x-1) /90 . 2 /15

is equal to

12 /90 . 4 and 3 are the only consecutive numbers that when multiplied together equal 12. Therefore there are 4 red marbles in the bag and 6 blue marbles.

Answer 2

Let the number of red marbles be x, then
Probability that the first marble is red = x/10
Probability that the second marble is red = (x - 1)/9
Probability that both is red = x/10 * (x - 1)/9 = 2/15
(x^2 - x)/90 = 2/15
15(x^2 - x) = 2(90)
x^2 - x = 12
x^2 - x - 12 = 0
(x - 4)(x + 3) = 0
x - 4 = 0
x = 4

Therefore, there are 4 red marbles in the bag.