How would you work out the answer to this:

There is a ball pit. 20% of the balls are blue. If you are blindfolded and select 5 balls at random, what are the chances that at least one is blue?


Sorry, it seems like a bit of a pointless thread but it's really bothering me. If someone can tell me the answer and how you would find it, then I'd be eternally grateful.

Damn, Im to lazy to think now, but there is actually a mathematical furmula for that.... If I fond it ill post it...

To me it still seems like 20%.

It'd be 20% chance if you only tookone ball, but what if you took 5? It can't simply be 100% as most people would initially think.

Basically, you have to detract from 100% the probability that none of the balls you took is blue. That is (4/5)^5 = approx. 0.33. So 1-0.33=0.67= 67%

Written by -tom- on 01.06.2008 at 17:12

How would you work out the answer to this:

There is a ball pit. 20% of the balls are blue. If you are blindfolded and select 5 balls at random, what are the chances that at least one is blue?

Sorry, it seems like a bit of a pointless thread but it's really bothering me. If someone can tell me the answer and how you would find it, then I'd be eternally grateful.

I just had an entire exam about shit like this. That was last week, and seriously, I am completely baffled. (Ok, I failed the test anyway, but still, I should remember this).
But let me think for a second...

For the record: P = chance
P(at least one is blue) = P(a maximum of 4 balls is non-blue)

Which can be calculated with a Cumulative Binomial Distribution. So let me get my calculator.
Now we should fill in:

N = 5 (number of tries)
P = 0.8 (chance of succes, i.e. a non-blue ball)
X = 4 (the number of succeses, i.e. a non-blue ball)

binomcdf(5, 0.8, 4) = 0.67232

So I think the answer you are looking for is 0.67232. But as I said, I am not sure. So don't pin me down on it.

EDIT: It comes quite close to Buggy's reasoning, so I guess what I did was the (one of) the right way(s).

It was just the long way around The tiny difference in results is merely due to the fact that I used approximation.

woooow im so glad i didnt take statistics haha! i dont really get any of it....