Probability - Athletics Scenario

[Joey Barber]

Hi, this is not a question for school, but just one that occurred to me, and I'm curious to know the answer. I do not have a maths background...

Imagine an athletics race which involves 13 competitors, and imagine 4 of them are British (nationality is not relevant, but I'm choosing British cos I'm British ).
Assuming each competitor has an equal chance of finishing in any position (disregarding favourites and underdogs etc.), what are the chances of at least one Brit finishing in a medal-winning position (one of the top three positions)?

I've tried to work it out but am probably on completely the wrong track. Here are my notes:

Brit A can finish in 13 positions
Brit B can finish in 12
Brit C can finish in 11
Brit D can finish in 10

13x12x11x10 = 17,160. So, there are 17,160 combinations which contain at least one Brit in the top four positions. But what about the top three? Here's where I'm struggling...

Do I calculate 13x12x11x2 (because there are 13 possibilities for one of the top three, 12 for another, and only 11 possibilities for the final place of the top three, but it could be either C or D who gets it, therefore I multiply it by 2)?

If that's the right thing to do, then 13x12x11x2 = 3432. So, there are 3432 combinations which contain at least one Brit in the top three positions (?)

Now, non-medal-winning positions:
A has 10 non-medal-winning positions
B has 9
C has 8
D has 7

10x9x8x7 = 5040.

So the chances of no Brit winning a medal are 5040/3432 ??

Even if my calculations so far are correct (which I doubt), I still don't know how to get from there to what the chances are that one of them actually wins a medal.

If someone could show me how to get to the right answer, I would really appreciate it. As I said, this isn't for homework or anything, so there's no need to respond right away. Just when you have time

Thank you for your time,
Joey

 

[Dr.Peterson]

I'd use permutations and combinations. (You say you don't know a lot of math(s), but it sounds like you know a little more than you admit!)

I'd also use complements: that is, first find the probability that none of your 4 gets a medal. And you can ignore all the non-Brits.

In fact, what you've done involves a lot of this, but you've mixed in a few wrong ideas (as you should know when you get a probability greater than 1).

How many ways are there to place 4 people in a row of 13 chairs? That's your denominator.

How may ways are there to put A, B, C, D in only the last 10 places? That's your numerator.

If you want, you can consider all 13 people, and you'll get the same result.

 

[Steven G]

I would use the method called Hypergeometric. Look it up as it fits your problem nicely.