What is the probability x type of item will be drawn in the first 50 attempts?

[Skelbo]

There are a total of 3326 items, you can call them balls, of various colours. 171 of them are blue. You can either consider this as drawing one ball at a time or a handfull of 50 balls (just imagine you have a giant's hand). What is the probability that at least one of the first 50 balls will be blue?

This isn't for homework or anything, I was just trying to figure out the probability of this x event happening and I thought it would be some quick math but I've been at it for a few days now so I'm quite obsessed at this point. Been asking around and no one can solve this properly. I even tried to make a program in python to speed up calculations but I'm failing at the logic of approaching this problem I think, so the results are all wrong. I got 2.58% using one method, 5.18 using another, 93% using python...
I looked at is as permutation, so used formula [MATH]\frac{n!}{((n-k!)*k!)}[/MATH]where
n = total number of outcomes (3326)
k = favourable outcomes (171)

But no calculator can do 3326! so... got stuck

I tried looking at it as each draw is an event, and the formula for that being P(a or b) = P(a) + P(b) - P(a and b)
and P(a and b) = P(a) x P(b)

So I applied it to my problem, replacing a and b with each draw, so there will be 50 events:
P(_e1 or _e2 or _e3 ... or _e50) = P(_e1) + P(_e2) ... + P(_e50) - (P_e1 and _e2 ... and _e50)

Just to clarify, I took each event to mean the probability of drawing a blue ball on first draw, second draw, third draw etc. So first draw would be [MATH]\frac{171}{3326}[/MATH], second draw would be [MATH]\frac{171}{3325}[/MATH], third draw [MATH]\frac{171}{3324}[/MATH] etc.
This is a situation of choosing without replacement.

After a loooong time calculating each event's probability and adding/multiplying them, I got the 2.58 which I assumed it represented percentage. I don't know if I got it right. Feels like it's too high maybe.
Can someone just tell me the answer but also what was the right way to calculate it?

 

[Romsek]

[MATH]P[\text{at least 1 blue ball out of 50 drawn}] = 1 - P[\text{0 blue balls out of 50 drawn}] = \\~\\ 1-\dfrac{\dbinom{171}{0}\dbinom{3326-171}{50}}{\dbinom{3326}{50}} =1- \dfrac{\dbinom{3155}{50}}{\dbinom{3326}{50}} \approx 0.93 = 93\% [/MATH]
[MATH]\text{In case you're not familiar with it $\dbinom{n}{k}$ is known as n choose k and is the number of ways you can pick k distinct items out of n}\\ \text{$\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$}[/MATH]

 

[Skelbo]

Thank you very much! So it was 93% after all. Why does that seem so strangely high to me? Feels like it can't be almost guaranteed to have a blue ball in first 50. I expected much lower. At least I have an answer now, I am at peace.