Lets say we have a sample size of n=50 and from this n, 10 times k happens.
k has a probability of 10%.

In our example, sequences doesnt matter and when we draw from n, we put it back, so n and probability always stay the same.

The question is, what is the likelyhood of this even occuring? I´m not exactly sure if i´m doing it right as i get some weird numbers and help would be appreciated. Thanks!

Binomial distribution.

You can do this in Excel. The chance a 10% event occurs at least 10 times in a sample of 50 tries, is 2.5%.

=BINOMDIST(40,50,90%,TRUE) = 2.5%

This is saying, the chance that the 90% event (not K) happens at MOST 40 times. Same thing. You can also use a binomial calculator like this one
http://stattrek.com/online-calculator/binomial.aspx

This isn't a combinatorics question. Combinations refers to ways to choose k things from a set of n. You are choosing just one thing, not a combination of things.

Incidentally, in your scenario the chance k occurs at most 5 times (the mean value) is 57%, and the chance it occurs exactly 5 times is 18.5%.

Thanks guys!

At the risk of being too technical it is, at least partially, a combinatorics question. The chance of a particular combination of 10 successes and 40 failures in 50 trials when success probability = p is

0.p^10 * (1-p)^40.

Then this probabililty has to be multiplied by the number of possible combinations of 10 successes out of 50 trials, which is C(50,10). The product is the binomial distribution formula.

It kind of is. To understand the formula for binomial distribution, you have to know some combinatorics. There are C(50,10) binary arrangements of the 10 successes and 40 failures (you're choosing 10 spots for the successes from a set of 50). Each of those binary permutations has the same probability of (p^10)(1-p)^40. We're adding that probability a bunch of times, and combinatorics tells us how many times so that we can just multiply it by C(50,10) instead of adding it that many times.

Edit -- Lol statman, you might be my clone or something. That's twice in 10 minutes that you posted right before me saying basically the same thing as me.

Well, I can't tell you how many times I thought you were my clone

Well yes, I made a poor choice of words. I should have said that since the binomial distribution is known, we can just use it instead of doing the combinatorics to figure it out, which is quite awkward. The combination formula
C(n,k)*(p^k)*((1-p)^(n-k))

doesn't give the cumulative distribution, merely the exact probability for some K. So we would have to sum up 10 values of k to get the answer.

We could also answer the question using Pascal's triangle.

I'm being nitpicky but I don't see the difference. When you "use the binomial distribution", you're telling Excel to use that formula for you. When you take the integral of 5x, you're using integration regardless of whether you did it by hand or by typing "integrate 5x" into Wolfram.

Yes I know and you're right, but do you think this answer is what the OP was looking for:

For k=10-50 sum C(n,k)*(p^k)*((1-p)^(n-k)).

?

Well first of all I'd do: For k=0 to 9 sum C(n,k)*(p^k)*(1-p)^(n-k)
And then take 1 - that

But I think he's glad to have seen both the math and the binomdist command. In his post he said he tried things himself and got weird numbers. Plus he's not necessarily the only one reading our posts, so if I see an inaccurate comment (esp. bolded :P) I'll try to correct it (and I hope others do the same when I get combinatorially out of line).

Edit: and when p=0.5, it can be viewed completely as a combinatorics problem.

For k=0 to 9 sum C(n,k)
Then 1 - that/(2^n)

For general p, if he doesn't have binomdist handy, instead of the formula he can use inclusion-exclusion to converge on the answer in just a few additions/subtractions if he doesn't need 10 decimal places of accuracy. (Edit #2 -- I take that back, the formula probably converges faster most of the time.)

Maybe I'll catch you one day.