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Bernoulli and Binomial Families · Tutorial 113 of 994

Adding Independent Binomial Counts

Combine independent binomial counts with a common success probability, and use the combined trial count to calculate pooled probabilities and summaries.

Beginner 10 min read

What You'll Learn

  • State the independence and common-probability conditions for adding binomial counts.
  • Derive the distribution of the sum using the binomial probability formula.
  • Extend the result from two independent counts to any finite number of counts.
  • Add trial counts correctly when pooling samples or batches.
  • Calculate probabilities, means, and variances for the combined count.
  • Recognize why adding counts is different from averaging success proportions.

Why Independent Binomial Counts Can Be Pooled

A study may record successes separately at several sites, or a quality-control team may inspect products in multiple batches. If every trial has the same success probability and the counts from the groups are independent, the total number of successes has a simple distribution: it is binomial, with the trial counts added together. This lets us calculate a probability for the pooled total without first analyzing each possible allocation of successes among groups.

The conditions matter. Each group must have a fixed number of trials, the success probability must be common across the groups, and the group counts must be independent. The result below concerns this specific setting; it does not say that arbitrary counts can be added and remain binomial.

Theorem: Let \(X_1,\ldots,X_r\) be independent random variables, where \(X_i\sim\operatorname{Bin}(n_i,p)\) for each \(i\), with fixed nonnegative integers \(n_i\) and a common \(p\in[0,1]\). Then \[ \sum_{i=1}^r X_i\sim\operatorname{Bin}\left(\sum_{i=1}^r n_i,p\right). \]

Proof for Two Counts

First consider two independent counts, \(X\sim\operatorname{Bin}(m,p)\) and \(Y\sim\operatorname{Bin}(n,p)\), and let \(S=X+Y\). For now suppose \(0<p<1\). The possible values of \(S\) are the integers from 0 to \(m+n\). For any such \(k\), the event \(S=k\) can occur by having \(j\) successes in the first group and \(k-j\) in the second. Independence allows us to multiply the two group probabilities and sum over the possible values of \(j\):

$$ \begin{aligned} P(S=k) &=\sum_{j=\max(0,k-n)}^{\min(m,k)}P(X=j)P(Y=k-j)\\ &=\sum_{j=\max(0,k-n)}^{\min(m,k)} \binom{m}{j}\binom{n}{k-j} p^k(1-p)^{m+n-k}. \end{aligned} $$

The factors involving \(p\) do not depend on \(j\). The remaining sum is \(\binom{m+n}{k}\): it counts the ways to choose \(k\) objects from two disjoint groups of sizes \(m\) and \(n\), according to how many are chosen from the first group. Equivalently, it follows by comparing the coefficient of \(t^k\) in \((1+t)^m(1+t)^n=(1+t)^{m+n}\). Therefore

$$ P(S=k)=\binom{m+n}{k}p^k(1-p)^{m+n-k},\qquad k=0,\ldots,m+n. $$

This is the probability mass function of \(\operatorname{Bin}(m+n,p)\). If \(p=0\), both counts are zero with probability one; if \(p=1\), they equal \(m\) and \(n\) with probability one. The claimed result also holds in those boundary cases. This proves the two-count result for every \(p\in[0,1]\).

Corollary: The theorem holds for any finite number \(r\) of independent counts with the same success probability. To see this, use induction. It holds for one count. If it holds for \(r-1\), then \(T=\sum_{i=1}^{r-1}X_i\) has distribution \(\operatorname{Bin}(\sum_{i=1}^{r-1}n_i,p)\). Because \(T\) is a function of \(X_1,\ldots,X_{r-1}\), it is independent of \(X_r\). Applying the two-count result to \(T\) and \(X_r\) gives the stated binomial distribution for the sum of all \(r\) counts. \(\square\)

Worked Applications

Worked Example: Pooling Two Independent Study Sites

Two independent study sites each record whether a participant meets a specified outcome. At Site A, 7 participants are observed; at Site B, 5 are observed. Suppose every participant has the same outcome probability, \(p=0.2\), and the site counts are independent. Let \(X\) and \(Y\) be the numbers meeting the outcome at the two sites.

The model assumptions give \(X\sim\operatorname{Bin}(7,0.2)\) and \(Y\sim\operatorname{Bin}(5,0.2)\). Their total is \(S=X+Y\), with \(7+5=12\) trials and the same success probability:

$$ S\sim\operatorname{Bin}(12,0.2). $$

For example, the probability of exactly 3 successes in the pooled sample is

$$ \begin{aligned} P(S=3) &=\binom{12}{3}(0.2)^3(0.8)^9\\ &=220(0.008)(0.134217728)\\ &=0.23622320128. \end{aligned} $$

The single pooled calculation accounts for every possible split of those 3 successes between the sites. It would be incorrect to use 7 or 5 as the pooled trial count; all 12 participants contribute one trial.

Worked Example: Combining Independent Inspection Batches

A manufacturer inspects two independent batches of products. The first batch contains 18 products and the second contains 32. Suppose each product independently has probability \(0.04\) of being defective, and let \(D_1,D_2\) be the defect counts.

The counts have distributions \(\operatorname{Bin}(18,0.04)\) and \(\operatorname{Bin}(32,0.04)\). Since the success event here is “is defective,” the common success probability is \(0.04\). Thus the total \(D=D_1+D_2\) has 50 trials:

$$ D\sim\operatorname{Bin}(50,0.04). $$

The probability that the inspection finds no defective products is

$$ P(D=0)=\binom{50}{0}(0.04)^0(0.96)^{50}=(0.96)^{50}\approx0.1299. $$

As a parameter check, the combined count has mean \(50(0.04)=2\) and variance \(50(0.04)(0.96)=1.92\). These are the mean and variance for a binomial count with the combined trial total, not quantities obtained by adding the two sample sizes to the probability.

Worked Example: Three Cohorts With a Common Success Probability

Three independent cohorts have sizes 4, 6, and 10. In each cohort, every individual has the same probability \(p=0.5\) of meeting a criterion. Let \(C_i\) denote the number meeting it in cohort \(i\).

The cohort counts are independent binomial variables with trial counts 4, 6, and 10. The pooled total \(C=C_1+C_2+C_3\) therefore has \(4+6+10=20\) trials and success probability \(0.5\):

$$ C\sim\operatorname{Bin}(20,0.5). $$

The probability of exactly 10 successes in the combined cohorts is

$$ \begin{aligned} P(C=10) &=\binom{20}{10}(0.5)^{10}(0.5)^{10}\\ &=184756(0.5)^{20}\\ &=\frac{184756}{1048576}\\ &\approx0.1762. \end{aligned} $$

The finite-sum result justifies pooling all three cohorts at once. It also shows why success allocations such as 2, 3, and 5 do not need separate probability calculations when the question concerns only the total.

Parameter Checks and a Common Mistake

When counts are pooled under the theorem's assumptions, add the numbers of trials and retain the common success probability. In particular, if \(S=\sum_{i=1}^r X_i\), then its mean and variance follow from its binomial distribution:

$$ E[S]=p\sum_{i=1}^r n_i,\qquad \operatorname{Var}(S)=p(1-p)\sum_{i=1}^r n_i. $$

These expressions provide a quick check on the pooled trial count: both are based on the total number of trials, \(\sum_i n_i\). A related way to verify them is to view each binomial count as a sum of Bernoulli indicators. The mean of the pooled count is the sum of the indicator means, and independence makes the variance the sum of their variances.

A common mistake is to add the success probabilities along with the trial counts, or to average the probabilities without considering the sample sizes. If two independent groups each have probability \(p=0.2\), the pooled count does not have success probability \(0.4\); the probability for each trial remains \(0.2\), while the number of trials increases. If the group probabilities are not common, or the group counts are not independent, the theorem's assumptions are not met. Those conditions should be checked before applying the pooling rule.

Key takeaway: For independent binomial counts with a common success probability, add the trial counts and keep that probability unchanged. The resulting total is binomial, so its probabilities can be calculated directly from the binomial formula.

Check Your Understanding

Use the assumptions stated in each question; distinguish the total number of trials from the success probability.

  1. Independent counts have distributions \(\operatorname{Bin}(9,0.3)\) and \(\operatorname{Bin}(11,0.3)\). What is the distribution of their sum?
  2. For the sum in Question 1, write the binomial formula for the probability of exactly 4 successes.
  3. Three independent batches have sizes 5, 8, and 12, with a common success probability of \(0.1\). Find the mean and variance of their combined count.
  4. Two counts have the same success probability but are not independent. Which condition of the theorem is missing?
  5. A pooled count combines 20 trials from one group and 30 from another. If the common success probability is \(p\), what are the parameters of the pooled binomial distribution?

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