When Dependence Breaks a Binomial Model

When Is a Binomial Model Appropriate?

A count of successes may look binomial because it comes from a fixed number of yes-or-no observations. But a fixed number of trials and two possible outcomes are not enough. A binomial model also requires the trials to be independent and the probability of success to be the same on every trial. When either condition fails, the binomial probability formula may give the wrong answer, even if the difference is small.

A useful diagnostic is to imagine observing the trials in order. Does knowing one outcome change the probability of success on a later trial? And, even if outcomes do not affect one another, is the success probability identical on every trial? These questions separate two distinct ways a binomial model can fail.

Definition: A count $X$ has a binomial distribution with parameters $n$ and $p$ if it counts successes in a fixed number $n$ of independent trials, each of which has two outcomes and the same probability $p$ of success. For $x=0,1,\ldots,n$, $$ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}. $$ The independence and constant-probability conditions are part of the model, not optional conveniences.

In this tutorial, a success indicator records one trial: it is 1 for success and 0 for failure. The count of successes is the sum of the indicators. This viewpoint makes it possible to examine directly whether trial outcomes influence one another.

Sampling Without Replacement From a Finite Population

Suppose a population contains $N$ objects, of which $K$ are successes and $N-K$ are failures. A sample of $n$ objects is selected uniformly at random, without replacement. The probability that the first draw is a success is $K/N$. But after that draw, the composition of the remaining population depends on its outcome. Thus, for samples with at least two draws, the chance of success on a later draw generally changes according to what was drawn before.

Definition: If $n$ objects are sampled without replacement from a population of $N$ objects containing $K$ successes, the success count $X$ has a hypergeometric distribution. Its probability of exactly $x$ successes is $$ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}, $$ for feasible values $\max(0,n-(N-K))\le x\le \min(n,K)$. The numerator counts samples with $x$ successes and $n-x$ failures; the denominator counts all samples of size $n$.

The hypergeometric formula applies under the stated uniform-sampling conditions. It does not require approximating a changing sequence of probabilities by one fixed value. The formula counts the possible samples directly.

Worked Example: Drawing Without Replacement

A container has 10 cards: 4 marked “success” and 6 marked “failure.” Three cards are drawn at random without replacement. Find the probability of exactly 2 successes, then compare it with the binomial calculation that would use the initial success probability.

There are $\binom{10}{3}=120$ equally likely samples of three cards. To get exactly 2 successes, choose 2 of the 4 success cards and 1 of the 6 failure cards. Therefore,

P(X=2)=\frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} =\frac{6\cdot 6}{120} =0.3000.

The initial success probability is $p=4/10=0.4$. If the three draws were independent with that same probability, a binomial model would give

P(X=2)=\binom{3}{2}(0.4)^2(0.6) =3(0.16)(0.6) =0.2880.

The results differ because sampling without replacement does not produce independent draws with a constant success probability. In this small example the difference is 0.0120, or 1.20 percentage points. The binomial answer is not the exact probability for the sampling design.

Why Distinct Draws Are Negatively Dependent

The dependence can be quantified. Let $I_i$ be the success indicator for draw $i$, and write $p=K/N$. For two distinct draws, the probability that both are successes is the probability the first is a success, multiplied by the chance that the second is a success after one success has been removed.

P(I_i=1,I_j=1)=\frac{K}{N}\frac{K-1}{N-1}, \qquad i\ne j.

For $N>1$, the covariance of two indicators is their joint success probability minus the product of their individual success probabilities. Substituting the probabilities gives

\begin{aligned} \operatorname{Cov}(I_i,I_j) &=\frac{K}{N}\frac{K-1}{N-1}-\left(\frac{K}{N}\right)^2\\ &=-\frac{p(1-p)}{N-1}. \end{aligned}

If $0<K<N$, then $0<p<1$, so this covariance is negative. A success on one draw makes a success on another draw slightly less likely: one success has been removed from the remaining population. This explains why the indicators are negatively dependent.

The qualification matters. For $n\ge 2$, any pair of distinct draw indicators is negatively dependent when $N>1$ and $0<K<N$. If $n=1$, there is no pair of distinct draws to be dependent. In that case the count is both hypergeometric and $\operatorname{Binomial}(1,K/N)$. If $K=0$ or $K=N$, all outcomes are fixed, so the count is degenerate rather than a nontrivial binomial count.

How Dependence Changes the Count’s Variance

The same covariance calculation shows how dependence affects the spread of the total count $X=I_1+\cdots+I_n$. Each indicator has variance $p(1-p)$. The variance of a sum is the sum of the individual variances plus twice the covariance for each distinct pair. There are $n(n-1)/2$ such pairs, so

\begin{aligned} \operatorname{Var}(X) &=np(1-p)+n(n-1)\left(-\frac{p(1-p)}{N-1}\right)\\ &=np(1-p)\frac{N-n}{N-1}. \end{aligned}

For a binomial count with the same $n$ and $p$, the variance is $np(1-p)$. The finite-population variance is smaller when $1<n<N$, because negative dependence reduces the count’s variability. When $n=N$, the sample includes the whole population, so the count is fixed at $K$ and the variance is zero, as the formula also shows.

Worked Example: A Short Sequential Sampling Calculation

A box contains 8 components, 3 of which are defective. Three components are selected at random without replacement. Find the probability that exactly 2 selected components are defective, and show how the success probability changes as the draws proceed.

For the first draw, the probability of a defective component is $3/8$. If the first component is defective, the second-draw probability is $2/7$; if the first is not defective, it is $3/7$. The conditional probability therefore depends on the earlier outcome, so the draws are not independent with one fixed success probability.

For the count, use the finite-population formula with $N=8$, $K=3$, and $n=3$:

P(X=2)=\frac{\binom{3}{2}\binom{5}{1}}{\binom{8}{3}} =\frac{3\cdot 5}{56} =\frac{15}{56} \approx 0.2679.

A binomial calculation using the initial probability $p=3/8$ would instead give

\binom{3}{2}\left(\frac{3}{8}\right)^2\left(\frac{5}{8}\right) =\frac{135}{512} \approx 0.2637.

The exact sampling probability is about 0.0042 greater. The small numerical difference does not remove the conceptual issue: the conditional success probability changes after each draw.

Changing Probabilities Without Dependence

A changing success probability and dependence are related possibilities, but they are not the same condition. Trials can be independent while having different success probabilities. The count from such trials is not binomial unless those probabilities are equal (apart from special degenerate cases). Its probability can be found by adding the probabilities of the separate outcome sequences that produce the desired count.

Worked Example: Independent Trials With Unequal Probabilities

Suppose three independent trials have success probabilities $0.2$, $0.5$, and $0.8$, respectively. Find the probability of exactly one success.

There are three disjoint ways to have exactly one success: success only on trial 1, only on trial 2, or only on trial 3. Independence allows the probabilities within each sequence to be multiplied:

\begin{aligned} P(X=1) &=(0.2)(0.5)(0.2)+(0.8)(0.5)(0.2)+(0.8)(0.5)(0.8)\\ &=0.020+0.080+0.320\\ &=0.420. \end{aligned}

The trials are independent, but their success probabilities are not identical, so the binomial conditions are not met. Using their average probability, $p=(0.2+0.5+0.8)/3=0.5$, would give the binomial value $\binom{3}{1}(0.5)(0.5)^2=0.375$, not 0.420.

Sequential Probabilities That Depend on Earlier Outcomes

In a sequential process, a later success probability may explicitly depend on what happened earlier. In that case the trials are dependent, even if the average success probability on each trial happens to be the same. The following example distinguishes that situation from independent trials with unequal fixed probabilities.

Worked Example: A Success Probability That Responds to the First Outcome

Consider two trials. The first succeeds with probability $0.5$. On the second trial, the success probability is $0.9$ if the first trial succeeded and $0.1$ if it failed. Find the distribution of the total number of successes.

The four possible outcome sequences have probabilities obtained by multiplying the first-outcome probability by the relevant conditional probability:

\begin{aligned} P(SS)&=(0.5)(0.9)=0.45,\\ P(SF)&=(0.5)(0.1)=0.05,\\ P(FS)&=(0.5)(0.1)=0.05,\\ P(FF)&=(0.5)(0.9)=0.45. \end{aligned}

Thus $P(X=2)=0.45$, $P(X=1)=0.05+0.05=0.10$, and $P(X=0)=0.45$. The second trial’s marginal success probability is also $0.5$, since $(0.5)(0.9)+(0.5)(0.1)=0.5$. However, the outcomes are dependent: $P(SS)=0.45$, whereas independence with two marginal probabilities of $0.5$ would require $P(SS)=(0.5)(0.5)=0.25$. A $\operatorname{Binomial}(2,0.5)$ model would also give $P(X=2)=0.25$, not 0.45.

A Practical Diagnostic

Before using a binomial model for a count, check the process that generated the observations. These questions focus on the assumptions rather than on whether the data happen to look like a familiar example.

Is the number of trials fixed?
A binomial count refers to a specified number $n$ of trials, not a process that stops according to the outcomes.

Does each trial have two relevant outcomes?
Identify which outcome is being counted as success and which as failure.

Are the trials independent?
Ask whether learning an earlier outcome changes the probability distribution of a later one. Without-replacement sampling from a mixed finite population generally fails this condition when there are at least two draws.

Is the success probability the same on every trial?
Independent trials with unequal success probabilities still do not meet the binomial definition.

If sampling is without replacement from a finite population, the hypergeometric model gives the exact count probabilities under uniform sampling. If trials are independent but their success probabilities differ, calculate count probabilities by considering the relevant outcome sequences. If a later probability depends on earlier outcomes, retain the conditional structure when calculating sequence probabilities; do not treat the trials as independent merely because their marginal probabilities match.

Common Mistake

Wrong reasoning: “There are three draws and each item is either a success or failure, so the count is binomial.” This checks the fixed number of trials and two outcomes but ignores how the selection process changes the probabilities.

Correction: A binomial model also requires independence and one common success probability. For a sample without replacement from a population containing both successes and failures, those conditions fail when there are at least two draws. Use the hypergeometric model for the exact count. Do not overstate the exception: with just one draw, the count is both hypergeometric and $\operatorname{Binomial}(1,K/N)$.

Key takeaway: A fixed count of binary outcomes is binomial only when the trials are independent and share one success probability. Sampling without replacement from a mixed finite population creates negative dependence across distinct draws; changing probabilities can also break the common-probability condition, even when trials remain independent.

Check Your Understanding

For each situation, identify whether the binomial conditions hold and briefly explain why.

A fair coin is tossed 12 times. What features of the process support a binomial model?
Two cards are drawn from a deck without replacement, and success means drawing a heart. Why does a binomial model with $p=1/4$ not give the exact count distribution?
One item is drawn uniformly from a population of $N$ items, $K$ of which are successes. What binomial distribution does this one-draw count also have?
Four independent tests have success probabilities $0.1$, $0.3$, $0.6$, and $0.8$. Which binomial condition fails?
In a two-stage process, the second-stage success probability changes according to the first-stage outcome. Which binomial condition is threatened, and why?