# Probability that at least one color is not drawn

Question: A bucket contains $3$ red balls, $2$ green balls, and $1$ yellow balls.
Three balls are chosen randomly and without replacement. What is the probability that at least one color is not drawn?

Hint: You will be using binomial coefficients. It will be easier to calculate $1-P$(all $3$ colors drawn)

By using the hint, the $P$(all $3$ colors drawn) is red=$\binom{6}{3}$, green=$\binom{6}{2}$,and yellow=$\binom{6}{1}$
If I am following the right concept, how do I use the hint to get to get at least one color not drawn. If I am not following the concept of the hint, I need some guidance.

• You will be drawing $3$ balls, no matter what. The only thing that will change is the set of balls you get to pick from. So the coefficients will all be of the form $\binom{x}{3}$. Make sure you understand the concept of combinations and counting them with binomial coefficients; your approach looks a bit confused. – Akay Mar 13 '17 at 7:00
• so $\binom{6}{3}$ then. okay – behold Mar 13 '17 at 7:01
• No, that counts all possible combinations. Correction in previous comment: of the form $\binom{x}{1}$ as BruceET said. – Akay Mar 13 '17 at 7:07

You are off to a good start: Let $X$ be the number of different colors drawn.
$$P(X = 3) = \frac{{3\choose 1}{2\choose 1}{1\choose 1}}{{6\choose 3}}.$$
The distribution of $X$ is $P(X=1) = 1/20,\; P(X = 2) = 13/20,$ and $P(X = 3) = 3/10.$ (Maybe you'd want to extend the problem to find all three probabilities. Why is there only one 'way' to get only one color?)
• I am not seeing this clearly. Doing the $P(X=3)$, I get $\frac{3}{10}$ from your equation. I suppose take $1-P(X=3)$? – behold Mar 13 '17 at 18:36