A bag contains red and blue marbles (and no marbles of any other color). The bag has twice as many blue marbles as red marbles.
Oliver draws two marbles from the bag without replacement. If the probability that Oliver draws two marbles of the same color is 11/21, then how many marbles are in the bag?
Thank you!
Let r be the number of red marbles in the bag. Then the number of blue marbles is 2r.
There are two ways that Oliver can draw two marbles of the same color: he can draw two red marbles, or he can draw two blue marbles.
The probability that he draws two red marbles is r+2rr⋅r+2r−1r−1=(r+2r)(r+2r−1)r(r−1)=5r2−3rr(r−1).
The probability that he draws two blue marbles is r+2r2r⋅r+2r−11=5r2−3r2r.
So, the probability that he draws two marbles of the same color is 5r2−3rr(r−1)+2r=158.
Equating this to 11/21, we get r(r−1)=11. This factors as r=−1 or r=11. We can't have a negative number of marbles, so r=11.
The total number of marbles in the bag is 2r+r=33.
