4) In an urn, there are 9 balls, 3 white, 3 red, and 3 blue. Tullio draws 3 balls simultaneously. What is the probability that he draws one white, one red, and one blue ball?
(A) $\frac{1}{27}$
(B) $\frac{1}{9}$
(C) $\frac{9}{28}$
(D) $\frac{6}{27}$
(E) $\frac{3}{14}$.
4) The answer is $(\mathbf{C})$. Tullio hopes that the three balls he draws are of three different colors. The first ball that Tullio observes, among the three drawn, can be of any color. The probability that the second is of a different color from the first is $\frac{6}{8}$; the probability that the third is of a different color from the first two is $\frac{3}{7}$. The probability that the three balls are of different colors is therefore $\frac{6}{8} \cdot \frac{3}{7}=\frac{9}{28}$. SECOND SOLUTION.

The desired probability is the ratio between the number of "winning" sets and the number of subsets with 3 elements of a set that has 9 elements; therefore

$$
\frac{3 \cdot 3 \cdot 3}{\left(\begin{array}{l}
9 \\
3
\end{array}\right)}=\frac{27 \cdot 3 !}{9 \cdot 8 \cdot 7}=\frac{9}{28}
$$