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The probability that the first ball is white is $x$. Partitioning that event according to the second colour shows that $P(WB)=x-p_w$. Uniform sampling without replacement makes the two draw positions exchangeable, so $P(BW)=P(WB)$. ThereforeFor a direct count, the same value is $2(xN)((1-x)N)/[N(N-1)]$.

There are ten balls of each colour. Under uniform sampling without replacement, all $\left(\genfrac{}{}{0ex}{}{20}{12}\right)$ subsets of twelve labelled balls are equally likely. A favourable subset chooses four of the ten white balls and eight of the ten black balls, giving $\left(\genfrac{}{}{0ex}{}{10}{4}\right)\left(\genfrac{}{}{0ex}{}{10}{8}\right)$ choices. The hypergeometric distribution therefore givesThus the requested integers are $\boxed{N_1=N_2=10,m_1=4,m_2=8}$, with $N=20$ and $k=12$.

With equally many balls of each colour, uniform sampling without replacement givesHere an admissible population producing positive probability for three white balls has even $N\ge 6$, so the cancellation is legitimate. Equating this to $2/19$ gives $19(N-4)=16(N-1)$, hence

Write $N=g+r$. Successive conditional probabilities account for the changing bag under sampling without replacement:On this event no equality of the two colour counts has occurred, so the game's stopping rule does not interrupt these three draws.

The equality must be a first return time: an order starting $GR$ or $RG$ would already have stopped at draw two. The only successful four-draw orders are therefore $GGRR$ and $RRGG$. Their probabilities under sampling without replacement add toHere $N=g+r$. Counting all balanced four-draw orders would include games that have already stopped.

Success after two draws occurs on the two mutually exclusive events $GR$ and $RG$. With $N=g+r$, sampling without replacement gives

The specified colour tells us that one green ball has been removed. Thus sampling without replacement again leaves nine green balls among nineteen:The numerical label supplies no further change once this first colour is known.

Let $B$ count the red balls among the $n$ added balls. By independence of random variables, $B$ has a binomial distribution with parameters $n,1/2$. The total red count is $R=1+B$, and the box has $n+1$ balls. Conditional on $R=r$, sampling without replacement giveswhere $D_1,D_2$ denote red first and second draws. Averaging these conditional probabilities and taking their ratio givesThe binomial distribution moments give $\mathbb{E}[B]=n/2$ and $\mathbb{E}[B^2]=n/4+n^2/4$. ConsequentlyTherefore the red-ball prediction with one certain red ball isIt reduces to $2/3$ at $n=1$ and approaches $1/2$ as the number of randomly coloured extra balls grows. Taking an unweighted average of the remaining red fractions would miss the change in weights caused by taking a conditional probability given the first red draw.

Every particular card is second with probabilityHence $X$ and $Y$ have the same distribution, with means zero and variances one. They are not independent, since a card cannot be drawn twice.

Every ordered pair of distinct cards has probability $1/(1000\cdot 999)$. ThereforeThis negative covariance is the effect of sampling without replacement, not a contradiction of their identical marginals.

Let $R_1,R_2$ be the events that the first and second birds are ringed. Under sampling without replacement,Thus the conditional probability is $\mathbb{P}(R_1\mid R_2)=\frac{2/15}{2/5}=\frac{1}{3}$. Alternatively, once a particular ringed second bird is specified, three of the nine possible first birds are ringed.

The two captures form a uniform ordered sample of distinct birds, that is, sampling without replacement. By exchangeability, each of the ten birds has the same chance of occupying the second position, so the probability is $4/10=2/5$. Equivalently, conditioning on the first capture gives $(4/10)(3/9)+(6/10)(4/9)=2/5$.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

Sampling without replacement selects distinct labelled units: after a unit is selected it is unavailable for the remaining draws. Distinct labels can carry equal numerical values, so repeated values do not necessarily imply a repeated unit. The rule alone does not specify selection probabilities; a procedure favoring particular units is still sampling without replacement but need not be simple random sampling.

In uniform sequential sampling without replacement, the next label is chosen uniformly among those remaining. For a population of $N$ labels, any ordered sequence of $n$ distinct labels has probability $1/[N(N-1)\cdots (N-n+1)]$. Each size-$n$ subset has $n!$ possible orders, so its probability is ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. This gives the fixed-size simple random sampling design. Equivalently, take the first $n$ labels of a uniformly random permutation. Restricting a uniformly random permutation to a specified label set preserves uniformity; biased modulo mappings should not replace uniform selection.

The draw positions have identical marginal probability distributions, but exhibit statistical dependence. If the first label is $i$, the conditional probability that the second label is $i$ is zero, whereas its marginal probability is $1/N$. Thus these draws are not independent and identically distributed when $N>1$ and $n>1$. For a marked subset of labels, counting size-$n$ subsets with binomial coefficients gives the hypergeometric distribution. The resulting negative covariance explains the finite population correction for both counts and sample means.

Draw two different labels uniformly from $\{1,2,3,4\}$ using sampling without replacement. Each position has a uniform distribution on a finite set, but $\mathbb{P}(X_1=X_2)=0$. Two independent random variables with those same marginal distributions would match with probability $1/4$. Equal marginal laws therefore do not make this sample an IID sample. Mark some of those four labels as successes. Their count in the two draws has a hypergeometric distribution; independent Bernoulli trials with a common success probability give a binomial distribution.

Here simple random sampling means a random unordered subset $A$ of exactly $n$ distinct labels from $\{1,\dots ,N\}$, with every size-$n$ subset equally likely. The sample size is fixed in advance, $1\le n\le N$, and sampling without replacement prevents duplicate units. This is SRSWOR; some texts also use "simple random sampling" for independent uniform draws with replacement, which is a different design.^[ref]

Every label has inclusion probability $n/N$, but equal inclusion probabilities alone are insufficient. For example, from six labels choose either $\{1,2,3\}$ or $\{4,5,6\}$ with probability $1/2$. Each label is included with probability $1/2$, yet eighteen of the twenty size-three subsets are impossible: this is a two-group cluster sampling design. Unequal probability sampling deliberately changes inclusion probabilities, often to improve precision for a particular estimand, and generally requires an appropriately weighted estimator rather than the plain sample mean.

Consider the original mathematical population with labelled values $(2,4,6,8,10,12)$, and let $n=2$. The fifteen equiprobable subsets, followed by their sample means, areTheir sample means sum to $105$, giving expected value $105/15=7$. Their squared deviations from $7$ sum to $70$, giving variance $70/15=14/3$. This is an enumeration over hypothetical selections from six stipulated numbers, not measured survey data.

With independent uniform sampling with replacement, the $6^2=36$ ordered outcomes each have probability $1/36$. The two orders $(2,4)$ and $(4,2)$ are distinct outcomes, and $(2,2)$ is now possible. Forgetting order does not make the resulting unordered multisets equiprobable: $\{2,4\}$ has probability $2/36$ but $\{2,2\}$ has probability $1/36$. The thirty-six sample means still have expected value $7$, but their squared deviations sum to $210$, giving variance $210/36=35/6$. Avoiding duplicate labels improves precision for this fixed population.