Red balls = 6
Blue balls = 8
White balls = 5
Total balls = 19
(I) When two balls are drawn at random, what is the probability that the two balls are of the same color.
$$\frac{\left(6\cdot5\right)}{\left(19\cdot8\right)}+\frac{\left(8\cdot7\right)}{19\cdot18}+\frac{\left(5\cdot4\right)}{19\cdot18}$$ $$=\frac{\left(6\cdot5\right)+\left(8\cdot7\right)+\left(5\cdot4\right)}{19\cdot18}$$
$$\frac{30+56+20}{342}$$
$$=\frac{106}{342}$$
$$=\frac{53}{171}$$

(II)When 2 balls are drawn at random, what is the probability of picking two random balls of same color
$$=\frac{53}{171}$$
$$different\ colors=1-\frac{53}{171}=\frac{\left(171-53\right)}{171}=\frac{118}{171}$$
Or we find the probability of of picking either Red and Blue ball, Red and White ball and Blue White ball and then find the total sum of the 3.
$$Red\ and\ blue=2\cdot\frac{6}{19}\cdot\frac{8}{18}=\frac{96}{342}=\frac{48}{171}$$
$$Red\ and\ White=2\cdot\frac{6}{19}\cdot\frac{5}{18}=\frac{60}{342}=\frac{30}{171}$$
$$Blue\ and\ White=2\cdot\frac{8}{19}\cdot\frac{5}{18}=\frac{80}{342}=\frac{40}{171}$$
$$Blue\ and\ White=\frac{48}{171}+\frac{30}{171}+\frac{40}{171}=\frac{80}{342}=\frac{118}{171}$$