jmc

The probability that both marbles are red is given by: $$P(\text{both red})=P(\text{1st red})\times P(\text{2nd red after 1st red is drawn}).$$The probability that the first marble is red is $\frac{4}{10}$. After drawing a red marble, there are 3 red marbles and 9 marbles total left in the bag, so the probability that the second marble is also red is $\frac{3}{9}$. Therefore $$P(\text{both red})=\frac{4}{10}\times \frac{3}{9}=\frac{2}{15}.$$Similarly, the probability that both marbles are blue is given by: $$P(\text{both blue})=P(\text{1st blue})\times P(\text{2nd blue after 1st blue drawn}).$$The probability that that the first marble is blue is $\frac{6}{10}$. After drawing a blue marble, there are 5 blue marbles and 9 marbles total left in the bag, so the probability that the second marble is also blue is $\frac{5}{9}$. Therefore $$P(\text{both blue})=\frac{6}{10}\times \frac{5}{9}=\frac{1}{3}.$$Since drawing two red marbles and drawing two blue marbles are exclusive events, we add the individual probabilities to get the probability of one or the other occurring. Therefore: $$\begin{array}{rl}P(\text{both same color})& =P(\text{both red})+P(\text{both blue})\\ & =\frac{2}{15}+\frac{1}{3}=\boxed{\frac{7}{15}}.\end{array}$$