11. Every year, a large number of students participate in the International Mathematical Olympiad. One twelfth of them win a gold medal, another sixth win a silver medal, a further quarter win a bronze medal, and the remaining half receive a handshake. If we encounter a group of six participants chosen at random, what is the probability that it consists of two gold medalists, two silver medalists, and two handshake winners?
(A) About $40 \%$
(E) About $0.004 \%$.
(B) About $4 \%$
(C) About $0.4 \%$
(D) About $0.04 \%$
11. The answer is $\mathbf{( C )}$. First, let's count the number of ways to divide 6 students such that two of them have won the gold medal, two have won the silver medal, and two have not won any medal. The two who have won the gold medal can be chosen in $\left(\begin{array}{l}6 \\ 2\end{array}\right)=15$ ways; once these are chosen, the two who have won the silver medal can be chosen in $\left(\begin{array}{l}4 \\ 2\end{array}\right)=6$ ways; once these are chosen, the two who have not won any medal are determined. Therefore, the number of ways to divide the students is $15 \times 6=90$.

By symmetry, we can count only the probability that the first two students have won the gold medal, the next two have won the silver medal, and the last two have won no medal. The probability that the first student has won the gold medal is $\frac{1}{12}$; given that the total number of students is assumed to be large, the probability that the second student has won the gold medal is $\approx \frac{1}{12}$ (in reality, knowing that the first student has won the gold medal, the probability is slightly lower, and precisely, if $N$ is the total number of students, it is equal to $\left.\left(\frac{1}{12} N-1\right) / N\right)$. Similarly, we can approximate the probabilities of the third and fourth students winning the silver medal with $\frac{1}{6}$ and that of the fifth and sixth students not winning any medal with $\frac{1}{2}$. The desired probability is therefore approximately equal to

$$
90 \cdot\left(\frac{1}{12} \cdot \frac{1}{6} \cdot \frac{1}{2}\right)^{2}=\frac{5}{1152} \approx 0.4 \%
$$