What are Belgium's chances on getting a medal at the 400m finals, given that they have 2 of the 8 athletes?
http://en.wikipedia.org/wiki/Combination
There are $\binom82$ possibilites for the ranks those athletes finish in.
In $\binom52$ of those, neither of them gets a medal,
so in $\binom82-\binom52$ of those, at least one of the athletes gets a medal.
Therefore the chance is $\frac{\binom82-\binom52}{\binom82} = \frac9{14}$ .
(A simple way to convince yourself that 6/8 should be wrong is that when
you apply that reasoning to 3 athletes, you get a 'probability' greater than 1.)
It is easier to compute the probability of them not winning a medal. There are ${5\choose 2}=10$ different ways for the two Borlées to be placed in position fourth to eighth (5 positions, two brothers). Altogether there are ${8\choose 2}=28$ ways to put the two athletes in the 8 slots (1st to 8th). Assuming that positions are random (obviously a stretch of facts, but we ignore that), the chances for them coming out empty is thus $$ \frac{10}{28}=\frac{5}{14}. $$ So the chances for a Belgian medal are $$ 1-\frac{5}{14}=\frac{9}{14}\approx0{,}643 $$ or about 64%.
I'm rooting for them anyway :-)
This has been answered very well, but I'll answer for people bad at math.
The first brother has a 3/8 chance of winning, the questioner was correct.
The tricky bit is why the second brother doesn't add another 3/8, and that's because the question was asking what was the chance that at least one brother got a medal. Therefore, if the first brother got a medal, we just don't care what the second brother does.
So the only way the second brother can help is if both the first brother doesn't get a medal, and the second brother does.
To find the chance of 2 independent things both happening, we multiply. The chance the first brother doesn't get a medal is 5/8, and the chance the second brother does is 3/7. (O.K., 3 out of 7 is another tricky bit. We have assumed the first brother lost, so with him "out of the running" there are only 7 people who have a chance of being in the 3 winning spots). Multiplying gives 15/56, which is the chance that only the second brother wins a medal.
So adding the 3/8 from the first brother to the 15/56 of help from the second brother, we get 36/56 (the same as everyone else, whew!).
The calculations other people used are based on generalizations of this simple case to any number of successes out of any number of tries, and on the realization that the easier way to think about this in general is to count the number of possible successes out of the number of possible outcomes.
So, there are 3 medals. You have 8 finalists. You are interested in only the winning chances of two of them. You can compute it as follows: count how many ways you can pick 3 winners out of 8 such that none of them is Belgian. (So, for the gold medal, you have 6 choices, for the silver 5 and for the bronze 4.) Divide this by the total amount of possible combinations of 3 winners, regardless of nationality. (Now you have 8 choices for gold, 7 for silver and 6 for bronze.) You then get the probability that no Belgian wins a medal. The complement of that probability is the one you are looking after.
The probability of no Belgian getting a medal is:
$$\frac{6 \times 5 \times 4}{8 \times 7 \times 6} = 0.3571 \; , $$
the complement thus being
$$\text{Probability of a Belgian getting a medal} = 1- 0.3571 = 0.6429 \; .$$
This is making the quite unrealistic assumption though that all athletes are equally fit, that conditions are equal for them, etc...
There are $\binom83$ ways to pick three winners out of eight. If the Belgians do not win, there are $\binom63$ ways to pick three winners from the non-Belgians. Hence the probability that some Belgian would get a medel is $1-\binom63/\binom83 = 1-\frac{6!5!3!}{3!3!8!} = 1-\frac{5}{14}=\frac{9}{14}.$