You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 5 balls. What are the odds that the sample contains 4 balls of one color and 1 of another color?
I thought of it like this:
$$P(\text{4 balls of one color and 1 of another color})=\frac{3 \cdot \binom{6}{4}\cdot 2 \binom{6}{1}}{\binom{18}{5}}\approx 0.0630252$$ since you have three choices for the color of the first four balls and only two (since you've can't use the same color that you did for the first four balls) for the last ball. All of this divided by the total number of ways to pick a group of five from the 18 balls you have in total. Is my approach correct?
