Additivity of Independent Binomial Variables

When Can Binomial Counts Be Combined?

A binomial random variable counts successes in a fixed number of independent trials, each having the same probability of success. If two independent groups use the same success probability, their combined count can be treated as one binomial count. This can make a probability calculation much shorter: instead of listing all ways to split a total between the groups, combine their trial counts and use one binomial probability.

The condition on the success probability is essential. Independence alone does not make the sum binomial. If the two groups have different success probabilities, the combined count generally has a different distribution, even though each group separately has a binomial distribution.

Definition: $X\sim\operatorname{Bin}(m,p)$ means that $X$ counts successes in $m$ independent Bernoulli trials, each with success probability $p$. Its probability mass function is $$ P(X=j)=\binom{m}{j}p^j(1-p)^{m-j},\qquad j=0,1,\ldots,m. $$ Here $m$ is a nonnegative integer and $0\le p\le 1$.

For two counts to be combined using the additivity result, the counts must be independent, and they must have the same success probability. Their numbers of trials may differ. For example, an independent $\operatorname{Bin}(4,0.3)$ count and $\operatorname{Bin}(6,0.3)$ count can be combined, but $\operatorname{Bin}(4,0.3)$ and $\operatorname{Bin}(6,0.4)$ do not meet the common-probability condition.

Why the Common-Probability Result Holds

Let $X\sim\operatorname{Bin}(m,p)$ and $Y\sim\operatorname{Bin}(n,p)$, with $X$ and $Y$ independent. Set $S=X+Y$. For $S$ to equal $k$, the first group must have $j$ successes and the second group $k-j$ successes, for some $j$ that is possible in both groups. Independence lets us multiply the two probabilities and add over all such splits.

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

The remaining sum is $\binom{m+n}{k}$. To see why, imagine choosing $k$ objects from two disjoint groups, one containing $m$ objects and the other $n$. If exactly $j$ objects are chosen from the first group, there are $\binom{m}{j}\binom{n}{k-j}$ such selections. Summing over every feasible value of $j$ counts every selection of $k$ objects exactly once. There are $\binom{m+n}{k}$ selections in total. Substitution gives the binomial probability formula for $m+n$ trials.

Theorem: If $X\sim\operatorname{Bin}(m,p)$ and $Y\sim\operatorname{Bin}(n,p)$ are independent, then $$ X+Y\sim\operatorname{Bin}(m+n,p). $$ The result extends to any finite collection of mutually independent binomial variables with a common success probability: add their numbers of trials and keep the shared probability $p$. The cases $p=0$ and $p=1$ also satisfy the result, with the sum constant at zero or at the total number of trials, respectively.

The proof is a convolution argument: each possible split of the total count contributes one product of probabilities. The common value of $p$ allows the same factor $p^k(1-p)^{m+n-k}$ to be taken out of every term. The combinatorial identity then combines the remaining coefficients. If the probabilities differ, that factorization does not work in this form.

Applying the Result to Probability Calculations

Worked Example: Combining Two Trial Groups

In a hypothetical quality-control process, two independent batches are checked for a specified defect. In the first batch, the number of defective items is $X\sim\operatorname{Bin}(4,0.3)$; in the second it is $Y\sim\operatorname{Bin}(6,0.3)$. Find the probability that there are exactly three defectives in total, and the probability that there is at most one.

Method. The batch counts are independent and share the same defect probability, $0.3$. Therefore, for $S=X+Y$, $$ S\sim\operatorname{Bin}(4+6,0.3)=\operatorname{Bin}(10,0.3). $$

For exactly three defectives, $$ \begin{aligned} P(S=3) &=\binom{10}{3}(0.3)^3(0.7)^7\\ &=120(0.027)(0.0823543)\\ &=0.266827932. \end{aligned} $$ To four decimal places, $P(S=3)=0.2668$.

“At most one” includes zero defectives and one defective: $$ \begin{aligned} P(S\le 1) &=P(S=0)+P(S=1)\\ &=(0.7)^{10}+\binom{10}{1}(0.3)(0.7)^9\\ &=0.0282475249+10(0.3)(0.040353607)\\ &=0.0282475249+0.121060821\\ &=0.1493083459. \end{aligned} $$ To four decimal places, $P(S\le 1)=0.1493$. Both probabilities are between zero and one, as required.

Common mistake avoided: using only the first batch’s four trials, or treating the two batch counts as a single observation. The combined model has ten trials because it counts every item in both independent batches.

Worked Example: A Tail Probability for a Combined Count

Suppose two independent groups of hypothetical policy applications are reviewed. The number flagged for additional review is $X\sim\operatorname{Bin}(3,0.4)$ in the first group and $Y\sim\operatorname{Bin}(5,0.4)$ in the second. Find the probability that six or more applications are flagged in total.

Method. The groups are independent and have the same flag probability, $0.4$, so $S=X+Y\sim\operatorname{Bin}(8,0.4)$. Since $S$ cannot exceed eight, “six or more” consists of $S=6$, $S=7$, and $S=8$.

Calculate each term: $$ \begin{aligned} P(S=6)&=\binom{8}{6}(0.4)^6(0.6)^2 =28(0.004096)(0.36)=0.04128768,\\ P(S=7)&=\binom{8}{7}(0.4)^7(0.6) =8(0.0016384)(0.6)=0.00786432,\\ P(S=8)&=\binom{8}{8}(0.4)^8 =0.00065536. \end{aligned} $$ Therefore, $$ P(S\ge 6)=0.04128768+0.00786432+0.00065536=0.04980736. $$ To four decimal places, the probability is $0.0498$.

Check. The three terms are nonnegative, and their sum is less than $P(S\ge 1)$, as expected. The calculation includes the endpoint $S=8$, which corresponds to all eight applications being flagged.

Common mistake avoided: interpreting “six or more” as “more than six.” The event includes six, seven, and eight, not just seven and eight.

What Changes When the Probabilities Differ?

When independent groups have different success probabilities, each group still has a binomial distribution, but the common-probability theorem does not apply. The sum is often called a Poisson-binomial count when it is viewed as the total of independent Bernoulli trials with potentially different success probabilities. In particular, one should not generally replace the different probabilities by their average and declare the total binomial.

Worked Example: Showing Why Averaging Probabilities Fails

Two independent one-trial checks have different success probabilities. Let $X\sim\operatorname{Bin}(1,0.2)$ and $Y\sim\operatorname{Bin}(1,0.8)$. Find the distribution of $S=X+Y$, and determine whether it can be treated as $\operatorname{Bin}(2,q)$ for some $q$.

Method. Enumerate the ways to obtain zero, one, or two successes. For zero successes, both checks must fail: $$ P(S=0)=(0.8)(0.2)=0.16. $$ For two successes, both must succeed: $$ P(S=2)=(0.2)(0.8)=0.16. $$ Exactly one success can occur in either of two ways, so $$ \begin{aligned} P(S=1) &=P(X=1,Y=0)+P(X=0,Y=1)\\ &=(0.2)(0.2)+(0.8)(0.8)\\ &=0.04+0.64=0.68. \end{aligned} $$ The probabilities sum to $0.16+0.68+0.16=1$, confirming the full distribution.

If $S$ were $\operatorname{Bin}(2,q)$, then $P(S=2)=q^2=0.16$, which forces $q=0.4$. But this would give $P(S=0)=(1-q)^2=(0.6)^2=0.36$, not the actual value $0.16$. Thus no binomial model with two trials fits this distribution.

Common mistake avoided: averaging the two probabilities to get $q=(0.2+0.8)/2=0.5$, then using $\operatorname{Bin}(2,0.5)$. That incorrect model gives $P(S=0)=0.25$, whereas the actual probability is $0.16$. Independence permits multiplication of probabilities for particular outcomes; it does not make unequal trial probabilities equal.

Limits of the Result and an Exam Trap

There are two distinct requirements: independence and a common success probability. If counts are dependent, the convolution multiplication used in the proof is not justified. If the counts are independent but their probabilities differ, the common factor in the proof cannot be taken out of the convolution sum. In either case, do not apply the additivity theorem without checking its assumptions.

A frequent trap is to notice that two counts are both described as binomial and conclude that their sum must be binomial. Being binomial separately is not enough. Another trap is to use a weighted or unweighted average success probability as if it made the individual trials identically distributed. An average can summarize the probabilities, but it does not change the probability of each pattern of successes and failures.

Key takeaway: For independent binomial counts with a common success probability $p$, add the trial counts and retain $p$. With unequal probabilities, calculate the distribution using the actual group probabilities or another justified method; do not assume the sum is binomial.

Check Your Understanding

For each problem, first check whether the independence and common-probability conditions permit direct use of the additivity result.

Independent counts have distributions $\operatorname{Bin}(2,0.25)$ and $\operatorname{Bin}(4,0.25)$. Find the probability of exactly two successes in total. (Answer: $0.2966$, to four decimal places.)
Independent counts have distributions $\operatorname{Bin}(3,0.6)$ and $\operatorname{Bin}(2,0.6)$. Find the probability of at least four successes in total. (Answer: $0.3370$, to four decimal places.)
Independent counts have distributions $\operatorname{Bin}(2,0.2)$ and $\operatorname{Bin}(3,0.5)$. Calculate the probability of exactly one success in total. (Answer: $0.2800$.)
Independent groups produce counts $\operatorname{Bin}(4,0.35)$ and $\operatorname{Bin}(6,0.35)$. Find the probability that the combined count is either zero or ten. (Answer: $0.0135$, to four decimal places.)
Independent counts have distributions $\operatorname{Bin}(5,0.4)$ and $\operatorname{Bin}(7,0.4)$. Find the probability of at most two successes in total. (Answer: $0.0834$, to four decimal places.)

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