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.
Best Answer
You are off to a good start: Let $X$ be the number of different colors drawn.
$$P(X = 3) = \frac{{3\choose 1}{2\choose 1}{1\choose 1}}{{6\choose 3}}.$$
From that you can answer your question.
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?)