Find the probability that two balls drawn at random without replacement of the same color

The probability of drawing the first white ball is $\frac{6}{15}$.
After drawing one white ball, there are 5 white balls left and a total of 14 balls remaining.
The probability of drawing a second white ball (given the first was white) is $\frac{5}{14}$.
The probability of drawing two white balls is:

$$P(\text{2 White}) = \frac{6}{15} \times \frac{5}{14} = \frac{30}{210} = \frac{1}{7}$$

The probability of drawing the first blue ball is $\frac{9}{15}$.
After drawing one blue ball, there are 8 blue balls left and a total of 14 balls remaining.
The probability of drawing a second blue ball (given the first was blue) is $\frac{8}{14}$.
The probability of drawing two blue balls is:

$$P(\text{2 Blue}) = \frac{9}{15} \times \frac{8}{14} = \frac{72}{210} = \frac{12}{35}$$

The probability of drawing two balls of the same color is the sum of the probabilities of drawing two white balls and drawing two blue balls.

$$P(\text{Same Color}) = P(\text{2 White}) + P(\text{2 Blue})$$ $$P(\text{Same Color}) = \frac{1}{7} + \frac{12}{35}$$

To add these fractions, find a common denominator, which is 35.

$$P(\text{Same Color}) = \frac{1 \times 5}{7 \times 5} + \frac{12}{35} = \frac{5}{35} + \frac{12}{35}$$ $$P(\text{Same Color}) = \frac{5 + 12}{35} = \frac{17}{35}$$

The probability that two balls drawn at random without replacement are of the same color is $\boxed{\frac{17}{35}}$.