There are three red balls, three white balls and three blue balls in an urn. We will randomly pick three of these balls. What's the probability that we get one ball of each color?
Solution:
Total numbers of ways to pick three balls = $9 \choose 3$
Selecting one red ball = $3 \choose 1$
Selecting one white ball = $3 \choose 1$
Selecting one blue ball = $3 \choose 1$
Number of ways to picking one ball of each color = $\dbinom{3}{1}\dbinom{3}{1}\dbinom{3}{1} $
Probability of picking one ball of each color = $\frac{{3 \choose 1} {3 \choose 1}{3 \choose 1}}{9 \choose 3}$
Did I solve the problem correctly?
Best Answer
Yes. You solved the problem nicely.
If in doubt, expand it out:
$$\dfrac{\binom 3 1 \binom 3 1\binom 3 1}{\binom 9 3} = \dfrac{3!\cdot 3\cdot 3\cdot 3}{\quad\;\; 9\cdot 8\cdot 7}$$