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 BLUE marbles are in the bag?
Assuming the probability of both marbles being red is 2/15, the solution can be found as follows: Let r be the number of red marbles. The probability of the first marble chosen being red is r/10. If the first marble chose is red, the probability of the second marble chosen being red is (r - 1)/9. The probability of both marbles being red can be expressed as: [tex]\frac{r}{10}\times\frac{r-1}{9}=\frac{r^{2}-r}{90}=\frac{2}{15}[/tex] This can be rearranged to give the quadratic equation: [tex]r^{2}-r-12=0[/tex] which has the positive solution r = 4. The answer is: there are 6 blue marbles in the bag.
