Two red and two blue from three colors

Question

A jar holds 4 red balls, 4 blue balls, and 4 green balls, 12 in total. You draw 4 balls at random. What is the probability of drawing exactly 2 red and 2 blue (and therefore no green)?

Accepted Answer

"No green" means you must choose 0 from the green pile, which still fits the recipe: multiply each color's chosen count, divide by all ways to draw 4. Here \binom{n}{k} counts ways to choose k from n, and \binom{4}{0}=1. Total ways to draw 4 from 12: \binom{12}{4}=495. Favorable ways: \binom{4}{2}	imes\binom{4}{2}	imes\binom{4}{0}=6	imes 6	imes 1=36. Divide: \frac{36}{495}=\frac{4}{55}\approx 0.073. Takeaway. A color you want none of contributes a factor of \binom{k}{0}=1, so it quietly drops out of the product while still forcing zero of that color. The recipe handles "none of this color…

Two red and two blue from three colors

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