At least one color missing when drawing three balls

Question

An urn holds 9 balls: 3 red, 3 green, and 3 blue. Three balls are drawn at random without replacement. What is the probability that at least one color is completely missing from the draw?

Accepted Answer

Let Ac be the event that color c is absent, and apply inclusion-exclusion over the three colors. Counts are over \binom{9}{3} = 84 equally likely draws. - One color absent: the three balls come from the other 6, so \binom{6}{3} = 20, with \binom{3}{1} = 3 colors. - Two colors absent: all three balls are one color, so \binom{3}{3} = 1, with \binom{3}{2} = 3 pairs. - All three colors absent is impossible. Alternating signs: 3(20) - 3(1) = 60 - 3 = 57. So P(	ext{some color missing}) = \frac{57}{84} = \frac{19}{28} \approx 0.6786. As a check, the complement (all three colors present) means one o…

At least one color missing when drawing three balls

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