Say I have a container with room for B balls. I know that there are black and white balls but I don't know the ratio between them.

Say I pick P balls, and R% are black. How can I use this information to establish an upper bound on the number of white balls, with C% certainty?

To give a specific example:

I have 1000 Balls, I pick 10 and they are all black. If I want to be 98% certain, what is the upper bound on the number of white balls?

I don't know any statistics beyond simple elementary probabilities, so I have no idea how to approach this. Some help with setting up an equation that I can use to solve these kinds of problems would be much appreciated.

Edit:

I've been thinking about it and I believe I can get some of the way towards an answer. For any given number of white balls, I can get the probability for that particular setup. Lets say that there was 100 white balls; the probability of me getting 10 black would then be

[tex]\frac{\binom{900}{10} \binom{100}{0}}{\binom{1000}{10}}[/tex]

I guess I could start at the probability of 990 white balls, add that together with the probability of 989 white balls, and keep going until I get to 98%, but there must be a better solution? This summation solution works for this example, but it gets pretty unfeasible if I have something like 10^31 balls.

k

Say I pick P balls, and R% are black. How can I use this information to establish an upper bound on the number of white balls, with C% certainty?

To give a specific example:

I have 1000 Balls, I pick 10 and they are all black. If I want to be 98% certain, what is the upper bound on the number of white balls?

I don't know any statistics beyond simple elementary probabilities, so I have no idea how to approach this. Some help with setting up an equation that I can use to solve these kinds of problems would be much appreciated.

Edit:

I've been thinking about it and I believe I can get some of the way towards an answer. For any given number of white balls, I can get the probability for that particular setup. Lets say that there was 100 white balls; the probability of me getting 10 black would then be

[tex]\frac{\binom{900}{10} \binom{100}{0}}{\binom{1000}{10}}[/tex]

I guess I could start at the probability of 990 white balls, add that together with the probability of 989 white balls, and keep going until I get to 98%, but there must be a better solution? This summation solution works for this example, but it gets pretty unfeasible if I have something like 10^31 balls.

k

Last edited: