A bag contains 6 red balls, 8 blue balls and 5 white balls. When two balls are drawn at random from this bag, what is the probability that the two balls are
(i) of the same colour
(ii) are of different colours
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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.
$$Red\ and\ blue=2\cdot\frac{6}{19}\cdot\frac{8}{18}=\frac{96}{342}=\frac{48}{171}$$
$$Red\ and\ White=2\cdot\frac{6}{19}\cdot\frac{5}{18}=\frac{60}{342}=\frac{30}{171}$$
$$Blue\ and\ White=2\cdot\frac{8}{19}\cdot\frac{5}{18}=\frac{80}{342}=\frac{40}{171}$$
$$Blue\ and\ White=\frac{48}{171}+\frac{30}{171}+\frac{40}{171}=\frac{80}{342}=\frac{118}{171}$$
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.
$$Red\ and\ blue=2\cdot\frac{6}{19}\cdot\frac{8}{18}=\frac{96}{342}=\frac{48}{171}$$
$$Red\ and\ White=2\cdot\frac{6}{19}\cdot\frac{5}{18}=\frac{60}{342}=\frac{30}{171}$$
$$Blue\ and\ White=2\cdot\frac{8}{19}\cdot\frac{5}{18}=\frac{80}{342}=\frac{40}{171}$$
$$Blue\ and\ White=\frac{48}{171}+\frac{30}{171}+\frac{40}{171}=\frac{80}{342}=\frac{118}{171}$$