Thread: Assigning swimming tracks!?

Problem: 7 swimmers are participating in a swimming match; 3 Americans, 2 Germans, 1 Australien and 1 Dutch. The swimming tracks 1 till 7 are randomly assigned.

Two questions:
1) What is the chance that 1 of the Germans is swimming in a outer track (so that would mean 1 german in track 1 or track 7)
2) What is the chance that at least 1 of the non-Americans is swimming in a outer track?

Could somebody please tell me how to calculate this?

I think I should divide the number of succesful outcomes by the total number of outcomes.

Total number = 7! = 5040

But how do I find the number of succesful outcomes in this case...

Answers should be
1) 0.476
2) 0.857

Originally Posted by timmeh

Problem: 7 swimmers are participating in a swimming match; 3 Americans, 2 Germans, 1 Australien and 1 Dutch. The swimming tracks 1 till 7 are randomly assigned. Two questions:
1) What is the chance that 1 of the Germans is swimming in a outer track (so that would mean 1 german in track 1 or track 7)
2) What is the chance that at least 1 of the non-Americans is swimming in a outer track?
Answers should be: 1) 0.476 2) 0.857

Are you translating these questions into English?
I ask because it took me several tries to get the given answer to question 1).
That answer is to this question: “What is the probability that exactly one German is in an outside lane”?
BUT the way the question is written it means at least one.

BTW: $\displaystyle \dfrac{4\cdot 5\cdot 5!}{7!}=0.47619$ WHY?

1) $\displaystyle \frac{\binom 2 1 \binom 5 1}{\binom 7 2}$

2) $\displaystyle 1 - \frac{\binom 3 2}{\binom 7 2}$

Ridley that seems to be correct. Could give me a more elaborate explanation? Especially 2)..?