Statistics and probability

[Smiskfisk]

Every single pick has is a 5% chance, or even better probability, to be a brown ball. Just the same with the probability it will be a white ball on every individual pick, 95%.
Another way to attack this problem is to turn it around. Ask the question this way:
How large is the probability that you pick only white balls for 10 straight picks. Same rules. You pick up a ball, look at it and then put it pack. So every single pick will be the same.
Maybe that's a better way to look at it?

The probability to ALWAYS get a white ball is
0.95^10 = 0.5987
That equal 59.87%

Therefore, the probability to get a brown at least once, would be 40.13%

[SirTeoHero]

Pleh, maybe like this: 0,05^(0.95^10)=16%

[Cos]

there is so many changing X

like:
if brown is bottom of bucket and whites are at top so u got 100% white? etc

[Nightmare]

Yeah Fish got it correct. But how about the follow-up question? Getting atleast 2 browns?

[Smiskfisk]

I'd say 22.37% to get at least 2 brown in 10 picks
But I'm really not sure about that.

[Nightmare]

But how do you get that result? Like said I can't remember how to calculate these myself, but when I see someone else do it I'm pretty sure it all comes back to me (like it did with the first question).

[Brummy]

hehe sorry that i deleted my post but i didnt see the second page. Now i know that someone explained him. And yes i need my english for that I dont know the right answer but i know that Teo's answer is wrong. and in german i would make an example to promise him that he is wrong but i cant cause he didnt understand my sad english

[mayhem]

Chance to get at least 2 browns is 36,975059%

EDIT: I am 1100100% sure about my answers

[Oynamak]

Chance to get at least 2 browns is 36,975059%

EDIT: I am 1100100% sure about my answers

Could u tell us how you calculated your answers? (:

[mayhem]

Ok so here it is

1 - (19/20)^10 - (19:20)^9 * (1:20)

And now you think WTF?!?

So its 1 - chance to get only whites - chance to get 1 brown

[Waihirere]

Actually, that is not correct Mayhem.

This is because there are 10!/(1!(10 − 1)!) = 10 ways to choose 1 brown ball, whereas there is only one way to choose 0 brown balls.

Or C(10,1) "10 choose 1" ways, unfortunately difficult to use clear notation for this on the forums.

See, you may first take 9 white balls and then the one brown, or you may take 8 white balls, one brown and then one white again, etc.

So the actual probability is

1.0 - C(10,0) * (19/20)^10 - C(10,1) * (19/20)^9 * (1/20) =
1.0 - (19/20)^10 - 10 * (19/20)^9 * (1/20) =
0.0861

that is, 8,61%

...

There is easy way to understand why this must be so.

Think of the case where we are picking up only two balls. Now there are three cases, one where all balls are white, one where one is white and one is brown and one where all balls are brown.

The probability of first case happening is (19/20)^2
The probability of one of the second cases is (19/20)*(1/20)
The probability of third case is (1/20)^2

These are all the possible cases, right? So when we add up the probability masses, the total should be 1.0, right?

Now try summing them directly: (19/20)^2 + (19/20)*(1/20) + (1/20)^2 = 0.9525

Uh oh, we're missing probability mass. However, since there are two ways to pick one brown ball (first take white, then take brown; or first take brown, then take white), we should add one more of the second case, the probability mass of which is 0.0475. Add this and, ta dah, the sum becomes 1.0.

(19/20)^2 + 2*(19/20)*(1/20) + (1/20)^2 = 1.0

...

As you can see by combining Mayhem's result to mine, the difference is huge. From 36.97% to 8.61%. You need to get lucky to get at least two brown balls.

If you want to know more about fun (and frustrating) field of combinatorics, see the Wikipedia article for a starting point: http://en.wikipedia.org/wiki/Combinatorics.

[mayhem]

Damit

I was thinking that does it matter in which order you take those balls out of the box, but then I just thought that it would work just fine if it goes like that (19:20)^9*(1:20)