Suppose we have 77 red and black balls in a urn. And we draw without replacement 24. We get 11 red and the rest are black, giving us the proportion of the red balls to all balls - 11/24. What can we infer for the proportion of red balls in the urn? Can we give our answer as an interval which covers 95% of the possible scenarios?

There is similar but inappropriate procedure (I think) in Wikipedia, that concerns estimation of proportions in big samples with replacement:

Something similar in our case. I saw the page in Wikipedia of the hypergeometric distribution. The example there assumes that we know in advance the count of the red balls in the urn. It seems that if we take this as unknown, and using the fact that all the probabilities sum to one we can get some quiteEstimating proportions and means

A relatively simple situation is estimation of a proportion. For example, we may wish to estimate the proportion of residents in a community who are at least 65 years old.

The estimator of a proportion is , whereXis the number of 'positive' observations (e.g. the number of people out of thensampled people who are at least 65 years old). When the observations are independent, this estimator has a (scaled) binomial distribution (and is also the sample mean of data from a Bernoulli distribution). The maximum variance of this distribution is 0.25/n, which occurs when the true parameter isp= 0.5. In practice, sincepis unknown, the maximum variance is often used for sample size assessments.

For sufficiently largen, the distribution of will be closely approximated by a normal distribution with the same mean and variance.^{[1]}Using this approximation, it can be shown that around 95% of this distribution's probability lies within 2 standard deviations of the mean. Because of this, an interval of the form

will form a 95% confidence interval for the true proportion. If this interval needs to be no more thanWunits wide, the equation

can be solved forn, yielding^{[2]}^{[3]}n= 4/W^{2}= 1/B^{2}whereBis the error bound on the estimate, i.e., the estimate is usually given aswithin ± B. So, forB= 10% one requiresn= 100, forB= 5% one needsn= 400, forB= 3% the requirement approximates ton= 1000, while forB= 1% a sample size ofn= 10000 is required. These numbers are quoted often in news reports of opinion polls and other sample surveys.

weird polynomial, describing m. But I can't understand what the value of m will mean, even if we solve it.

EDIT: No, we don't know how many terms we must sum to get to 1. So no polynomial, no nothing...