Problem (Combinatorics):
In a short-track speed skating event, there are five finalists including two Canadians. The first three skaters to finish the race win a medal. If all finalists have the same chance of finishing in any position, what is the probability that neither Canadian wins a medal?

Solution:
We label the 5 skaters A, B, C, D, and E, where D and E are the two Canadians.

There are then $5 !=5 \times 4 \times 3 \times 2 \times 1=120$ ways of arranging these skaters in their order of finish (for example, $\mathrm{ADBCE}$ indicates that A finished first, $\mathrm{D}$ second, etc.), because there are 5 choices for the winner, 4 choices for the second place finisher, 3 choices for the third place finisher, etc.



If the two Canadians finish without winning medals, then they must finish fourth and fifth. So the $\mathrm{D}$ and $\mathrm{E}$ are in the final two positions, and $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ in the first three. There are $3 !=6$ ways of arranging the $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$, and $2 !=2$ ways to arrange the $\mathrm{D}$ and E. Thus, there are $6 \times 2=12$ ways or arranging the skaters so that neither Canadian wins a medal.

Therefore, the probability that neither Canadian wins a medal is

$$
\frac{\# \text { of ways where Canadians don't win medals }}{\text { Total } \# \text { of arrangements }}=\frac{12}{120}=\frac{1}{10}
$$

We label the 5 skaters as A, B, C, D, and E, where D and E are the two Canadians. In any race, two of the skaters finish fourth and fifth. Also, any pair of skaters are equally as likely to finish fourth and fifth, since the probability of every skater is equally likely to finish in a given position.

How many pairs of 2 skaters can we form from the 5 skaters? There are ten such pairs:

$$
\{A, B\},\{A, C\},\{A, D\},\{A, E\},\{B, C\},\{B, D\},\{B, E\},\{C, D\},\{C, E\},\{D, E\}
$$

Only one of these ten pairs is made up of the two Canadians. Therefore, the probability is $\frac{1}{10}$, since one out of ten choices gives the desired result.

Final answer: $\frac{1}{10}$