A box contains 2 black, 4 yellow and 6 white balls. Three balls are drawn in succession with replacement. What is the probability that <strong>ALL</strong> three are of the same colour?

Explanation

The box contains 12 balls in total. Because the drawing is with replacement, the probability for each draw remains constant. P(B) = \frac{2}{12} = \frac{1}{6}, P(Y) = \frac{4}{12} = \frac{1}{3}, and P(W) = \frac{6}{12} = \frac{1}{2}. The probability that all three are of the same color is P(BBB) + P(YYY) + P(WWW) = (\frac{1}{6})^3 + (\frac{1}{3})^3 + (\frac{1}{2})^3 = \frac{1}{216} + \frac{1}{27} + \frac{1}{8} = \frac{1 + 8 + 27}{216} = \frac{36}{216} = \frac{1}{6}.

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