Binomial Model Pitfalls: Dependence and Unequal Probabilities

Why a Fixed Number of Trials Is Not Enough

A count of successes from \(n\) trials is binomial only when the trials have the right probability structure. Having a fixed number of trials and recording each as a success or failure is not sufficient: dependence between outcomes can change the distribution of the count, and unequal success probabilities can change it even when trials are independent.

Let \(X_i\) indicate whether trial \(i\) is a success: \(X_i=1\) for success and \(X_i=0\) for failure. The total number of successes is \(S=\sum_{i=1}^n X_i\). For \(S\) to be binomial under the standard trial scheme, the indicators \(X_1,\ldots,X_n\) must be independent Bernoulli random variables with the same success probability \(p\). Then \(S\sim\operatorname{Bin}(n,p)\).

Definition: A binomial count with parameters \(n\) and \(p\) is the sum of \(n\) independent Bernoulli indicators, each with success probability \(p\). Its probability mass function is \(P(S=k)=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k=0,\ldots,n\).

The shared probability and independence are separate assumptions. Independence concerns how outcomes occur together; equal probabilities concern how likely success is on each trial. A model can satisfy one assumption and violate the other. Either violation can make a binomial calculation wrong.

When Dependence Changes the Count

Dependence matters because the chance of several successes together affects the distribution of their sum. The covariance identity below makes one consequence precise. It applies to any indicators with finite second moments, so no independence assumption is needed for the identity itself.

Result: If \(X_1,\ldots,X_n\) are indicators and \(S=\sum_{i=1}^n X_i\), then \(E[S]=\sum_{i=1}^n E[X_i]\) and \(\operatorname{Var}(S)=\sum_{i=1}^n\operatorname{Var}(X_i)+2\sum_{i<j}\operatorname{Cov}(X_i,X_j)\).

Proof. Linearity of expectation gives \(E[S]=\sum_i E[X_i]\). To obtain the variance, write \(S-E[S]=\sum_i(X_i-E[X_i])\), square this expression, and take expectations. Each squared term has expectation \(\operatorname{Var}(X_i)\); each cross-term has expectation \(\operatorname{Cov}(X_i,X_j)\), and each pair appears twice. Therefore the stated identity follows. \(\square\)

If the indicators are independent, every covariance in the formula is zero. Positive covariance adds variability to the count; negative covariance reduces it. Either way, the dependence can change the count’s distribution, even if each trial has the same marginal success probability.

Worked Example: Two Trials That Always Agree

Let \(B\) be Bernoulli with success probability \(1/2\), and set \(X_1=B\) and \(X_2=B\). Each trial has success probability \(1/2\), but the outcomes are perfectly dependent: they always agree. Find the distribution of \(S=X_1+X_2\) and compare it with a binomial count with two trials and success probability \(1/2\).

If \(B=0\), then \(S=0\); if \(B=1\), then \(S=2\). Thus \(P(S=0)=1/2\), \(P(S=1)=0\), and \(P(S=2)=1/2\). In contrast, a \(\operatorname{Bin}(2,1/2)\) count has probabilities \(1/4,1/2,1/4\) at \(0,1,2\), respectively. These distributions are different.

Both counts have mean \(1\), but their variances differ. Here \(\operatorname{Var}(S)=\operatorname{Var}(2B)=4(1/4)=1\), whereas a \(\operatorname{Bin}(2,1/2)\) count has variance \(2(1/2)(1/2)=1/2\). The dependence increases the variance because \(\operatorname{Cov}(X_1,X_2)=1/4\).

Independent Trials With Unequal Success Probabilities

Independence alone does not justify the binomial formula. Suppose the indicators are independent but \(P(X_i=1)=p_i\), where the probabilities need not agree. For a particular set \(A\) of successful trial indices, independence gives probability \(\prod_{i\in A}p_i\prod_{i\notin A}(1-p_i)\). To have exactly \(k\) successes, sum these probabilities over all \(k\)-element sets.

Result: For independent Bernoulli indicators with probabilities \(p_1,\ldots,p_n\), the count \(S=\sum_i X_i\) has \(P(S=k)=\sum_{A\subseteq\{1,\ldots,n\},\,|A|=k}\prod_{i\in A}p_i\prod_{i\notin A}(1-p_i)\). This is the Poisson-binomial distribution. If all \(p_i=p\), the terms for each \(k\)-element set are identical, and there are \(\binom{n}{k}\) such sets, giving the binomial formula.

The formula follows by partitioning the event \(\{S=k\}\) into disjoint events specifying exactly which \(k\) trials succeeded. Independence gives the probability of each event as the displayed product, and adding over the disjoint possibilities gives the result. In general, those products differ from one set of successful indices to another, so they cannot be replaced by \(\binom{n}{k}p^k(1-p)^{n-k}\) for a common \(p\).

Worked Example: Independent Trials With Probabilities 0.2 and 0.8

Two trials are independent. The first succeeds with probability \(0.2\), and the second with probability \(0.8\). Calculate the count distribution and check whether it is binomial.

There are two ways to get zero successes, namely both fail, so \(P(S=0)=(0.8)(0.2)=0.16\). Exactly one success can happen as success then failure or failure then success: \(P(S=1)=(0.2)(0.2)+(0.8)(0.8)=0.04+0.64=0.68\). Finally, \(P(S=2)=(0.2)(0.8)=0.16\). The probabilities sum to \(0.16+0.68+0.16=1\).

If this were \(\operatorname{Bin}(2,p)\), its probability of zero successes would be \((1-p)^2=0.16\), implying \(p=0.6\) (since \(p\) is a probability). Its probability of two successes would then be \(p^2=0.36\), not \(0.16\). So the count is not binomial.

Its mean is \(0.2+0.8=1\), the same as the mean of \(\operatorname{Bin}(2,1/2)\), but the distributions differ. Its variance is \(0.2(0.8)+0.8(0.2)=0.32\), while \(\operatorname{Bin}(2,1/2)\) has variance \(0.5\). A matching mean alone is not enough to identify a binomial model.

Dependence in Sampling Without Replacement

A common source of dependence is sampling without replacement. After one item is selected, the composition of the remaining pool changes, which can alter the next success probability. The following example also shows why knowing each trial’s marginal success probability does not settle the question.

Worked Example: Drawing Two Items Without Replacement

A box contains two items labeled success and two labeled failure. Draw two items uniformly at random without replacement, and let \(S\) be the number of successes. Find the distribution and compare it with the binomial model that would apply to two independent trials each having success probability \(1/2\).

There are \(\binom{4}{2}=6\) equally likely pairs of items. One pair contains both successes, one contains both failures, and four contain one success and one failure. Therefore \(P(S=0)=1/6\), \(P(S=1)=4/6\), and \(P(S=2)=1/6\). The independent \(\operatorname{Bin}(2,1/2)\) model instead assigns probabilities \(1/4,1/2,1/4\), so the count is not binomial.

Each draw position has marginal success probability \(1/2\). However, the indicators are dependent: after a success on the first draw, the probability of success on the second is \(1/3\), not \(1/2\). The mean count is \(1\), but its variance is \(1/3\): the count differs from its mean by one only when it is zero or two, events whose total probability is \(1/3\). The binomial variance would be \(1/2\).

What the Assumptions Do—and Do Not—Tell You

The standard binomial trial scheme is a reliable sufficient condition: independent Bernoulli trials with a common success probability produce a binomial count. The counterexamples show that violating either condition can change the count distribution. But it is important not to turn this into an incorrect converse claim. Dependence among trial indicators does not always prevent their sum from having a binomial distribution; the count alone does not reveal every feature of the joint trial model.

Worked Example: A Binomial Count From Dependent Indicators

Suppose two indicators have joint probabilities \(P(X_1=0,X_2=0)=0.25\), \(P(X_1=1,X_2=0)=0.40\), \(P(X_1=0,X_2=1)=0.10\), and \(P(X_1=1,X_2=1)=0.25\). Their probabilities sum to \(1\). Determine the distribution of their sum and whether the indicators are independent.

The sum \(S=X_1+X_2\) equals zero with probability \(0.25\), one with probability \(0.40+0.10=0.50\), and two with probability \(0.25\). These are precisely the probabilities of a \(\operatorname{Bin}(2,1/2)\) count.

But \(P(X_1=1)=0.40+0.25=0.65\) and \(P(X_2=1)=0.10+0.25=0.35\). If the indicators were independent, \(P(X_1=1,X_2=1)\) would be \((0.65)(0.35)=0.2275\), not \(0.25\). Thus they are dependent even though their sum has a binomial distribution. This example is why the safe claim is that the standard assumptions guarantee a binomial count, not that they are logically necessary whenever the count happens to be binomial.

Common Pitfalls

When deciding whether a binomial model is appropriate, check the mechanism producing the data, not just the way the outcomes are recorded. In particular:

Wrong reasoning: “There are \(n\) trials, so the success count is binomial.” Correction: A fixed trial count is only one part of the model. Check independence and a common success probability.
Wrong reasoning: “Each trial has probability \(p\) on average, so the count is binomial.” Correction: Equal average or marginal probabilities do not remove dependence, as sampling without replacement illustrates.
Wrong reasoning: “The trials are independent, so the count is binomial.” Correction: Independent trials with unequal success probabilities generally produce a Poisson-binomial count instead.
Wrong reasoning: “The mean matches \(np\), so the model is binomial.” Correction: Different distributions can share a mean. Check the trial assumptions or calculate probabilities that distinguish the models.
Wrong reasoning: “If the indicators are dependent, their sum cannot be binomial.” Correction: Dependence generally changes the count distribution, but the sum alone can still have a binomial distribution in special cases.

The practical diagnostic is to write down what each indicator represents, then ask whether the trials are independent and whether each has the same success probability. If an assumption fails, do not use the binomial probability formula without separately establishing that it applies to the count at hand.

Check Your Understanding

For each scenario, identify the assumption at issue and explain whether a binomial model is justified by the standard trial scheme.

Three independent tests have success probabilities \(0.2\), \(0.5\), and \(0.8\). Which binomial assumption fails?
Two indicators always have the same value, and each has success probability \(0.4\). Explain why the count is not generally binomial under the standard two-trial scheme.
Why is sampling two items without replacement from a small population a reason to check for dependence?
Can dependent indicators have a binomially distributed sum? Explain what the two-indicator example demonstrates.
Why does knowing that a count has mean \(np\) not by itself establish a \(\operatorname{Bin}(n,p)\) distribution?