Sums of Independent Bernoulli and Binomial Variables

Why These Sums Are Useful

Many probability models count how many times an event occurs across a group of trials: how many independent devices pass a test, how many applications meet a criterion, or how many hypothetical claims fall into a specified category. A binomial model is appropriate when each trial has two outcomes, the trials are independent, and every trial has the same probability of success. These conditions also explain when separately modelled counts can be combined.

The key result is simple but conditional: independent Bernoulli variables with a common success probability sum to a binomial variable, and independent binomial variables with a common success probability add to another binomial variable. Independence alone is not enough; the common probability matters too. These results are aligned with the SOA Exam P (Probability) syllabus.

Bernoulli Trials and the Binomial Model

Definition: A Bernoulli random variable \(I\) takes the value 1 for success and 0 for failure. If \(P(I=1)=p\), then \(P(I=0)=1-p\), where \(0\le p\le1\). A binomial random variable \(X\sim\operatorname{Bin}(n,p)\) counts successes in \(n\) independent Bernoulli trials, each with success probability \(p\). Its probability mass function is \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\), for integers \(k=0,1,\ldots,n\).

The binomial assumptions concern the trials, not merely the range of possible values of a count. The trials must be independent, and the success probability must be the same for each. A Bernoulli variable is a binomial variable with \(n=1\). When independent Bernoulli variables all have the same success probability, their sum counts successes across all of those trials.

Worked Example: Counting Successful Device Tests

A hypothetical quality check tests eight devices. For each device, the test result is success or failure, with success probability \(0.25\). Assume the results for the eight devices are independent and have the same probability of success. Find the probability that exactly three devices pass.

Method and model. Let \(I_j=1\) if device \(j\) passes and \(I_j=0\) otherwise, for \(j=1,\ldots,8\). Each \(I_j\) is Bernoulli with \(p=0.25\), and the variables are independent. Therefore \(T=I_1+\cdots+I_8\sim\operatorname{Bin}(8,0.25)\).

Use the binomial probability formula with \(n=8\), \(k=3\), and \(1-p=0.75\):

P(T=3)=\binom{8}{3}(0.25)^3(0.75)^5 =56\cdot\frac{1}{64}\cdot\frac{243}{1024} =\frac{13608}{65536} =0.2076416015625.

Thus the probability is 0.2076 to four decimal places. The result is between 0 and 1, and it is plausible for exactly three successes when the expected count is \(8(0.25)=2\). The common mistake to avoid is multiplying the probability of one particular arrangement of three successes by nothing: there are \(\binom{8}{3}=56\) arrangements, and all must be counted.

Why a Common Success Probability Produces a Binomial Sum

Let \(I_1,\ldots,I_n\) be independent Bernoulli variables, each with success probability \(p\), and define \(T=\sum_{j=1}^{n} I_j\). For exactly \(k\) successes, a particular arrangement of successes and failures has probability \(p^k(1-p)^{n-k}\), by independence. There are \(\binom{n}{k}\) possible arrangements of the \(k\) successful trials. These arrangements are disjoint and together describe the event \(T=k\), so their probabilities add.

P(T=k)=\binom{n}{k}p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n.

This is the probability mass function of \(\operatorname{Bin}(n,p)\), proving that the sum has that distribution. The argument explains both the coefficient and the assumptions: independence gives the probability of each arrangement, while the shared \(p\) makes that probability the same for every arrangement with \(k\) successes.

Theorem: If \(I_1,\ldots,I_n\) are independent Bernoulli random variables with the same success probability \(p\), then \(\sum_{j=1}^{n} I_j\sim\operatorname{Bin}(n,p)\). If the variables are not independent or do not share a success probability, this theorem does not establish a binomial distribution for their sum.

Adding Independent Binomial Variables

Suppose \(X_1,\ldots,X_r\) are independent and \(X_j\sim\operatorname{Bin}(n_j,p)\) for every \(j\), with the same \(p\). Each count represents \(n_j\) Bernoulli trials. The total count \(T=X_1+\cdots+X_r\) combines all the trials, so the total number of trials is \(N=n_1+\cdots+n_r\). Independence of the counts and their common binomial success probability give

X_1+\cdots+X_r\sim\operatorname{Bin}\left(\sum_{j=1}^{r} n_j,p\right).

The same result can be checked using moment-generating functions. For \(X\sim\operatorname{Bin}(n,p)\), the MGF is \(M_X(t)=(1-p+pe^t)^n\). Independence makes the MGF of a sum the product of the individual MGFs. Consequently,

M_T(t)=\prod_{j=1}^{r}(1-p+pe^t)^{n_j} =(1-p+pe^t)^{\sum_{j=1}^{r}n_j}.

This is the MGF of the stated binomial distribution. Since these variables have finite support, their MGFs exist and uniquely identify their distributions. The common \(p\) is what lets the factors combine into a single power of the same expression. If the success probabilities differ, this calculation does not produce that binomial MGF.

Important condition: The sum rule requires independence between the binomial variables and a shared success probability \(p\). The trial counts \(n_j\) may differ. If either condition is absent, do not apply this binomial closure result without further analysis.

Worked Example: Combining Two Independent Groups

In a hypothetical screening program, group A has a count \(X\sim\operatorname{Bin}(5,0.20)\) of positive results, and group B has a count \(Y\sim\operatorname{Bin}(7,0.20)\). Assume the group counts are independent. Find the probability that there are at most one positive result across both groups.

Method and model. The counts are independent and share \(p=0.20\), so the sum \(T=X+Y\) has a binomial distribution with \(5+7=12\) trials: \(T\sim\operatorname{Bin}(12,0.20)\). At most one result means \(T=0\) or \(T=1\).

For zero positive results:

P(T=0)=\binom{12}{0}(0.20)^0(0.80)^{12} =(0.80)^{12}=0.068719476736.

For exactly one positive result:

P(T=1)=\binom{12}{1}(0.20)(0.80)^{11} =12(0.20)(0.08589934592) =0.206158430208.

Add the disjoint probabilities:

P(T\le1)=0.068719476736+0.206158430208 =0.274877906944.

The answer is 0.2749 to four decimal places. As a check, the probability is below 1 and includes the zero-result case. A common mistake is to use \(5\) or \(7\) as the total number of trials; the two independent groups together contribute \(12\) trials.

Combining Individual Trials and Grouped Counts

A total may include both individual Bernoulli variables and already-aggregated binomial counts. If every component is independent and has the same success probability \(p\), each Bernoulli variable contributes one trial, while a \(\operatorname{Bin}(n,p)\) variable contributes \(n\) trials. The total is binomial with the sum of those trial counts.

Worked Example: Combining Individual and Aggregated Counts

A hypothetical monitoring system records a count \(X\sim\operatorname{Bin}(5,0.30)\) from one set of trials. It also records three additional Bernoulli results \(I_1,I_2,I_3\), each with success probability \(0.30\). Assume all five underlying trials for \(X\) and all three individual results are mutually independent. Find the probability that the combined total is exactly four.

Method and model. Define \(T=X+I_1+I_2+I_3\). There are \(5+3=8\) independent trials, all with \(p=0.30\), so \(T\sim\operatorname{Bin}(8,0.30)\). Therefore,

\begin{aligned} P(T=4) &=\binom{8}{4}(0.30)^4(0.70)^4\\ &=70(0.0081)(0.2401)\\ &=70(0.00194481)\\ &=0.1361367. \end{aligned}

The probability is 0.1361 to four decimal places. The combined trial count is eight, not four: four is the required number of successes. This avoids the common mistake of confusing the target count \(k\) with the total number of trials \(n\).

Assumptions and Common Exam Traps

Before combining counts, state what is independent and what success probability applies. In particular, two counts can each be binomial while their sum is not justified as binomial by the common-\(p\) rule if their success probabilities differ. Likewise, matching probabilities do not fix dependence: if one count affects another, the independent-sum argument cannot be used as stated.

Ignoring independence. The formula for the probability of a particular pattern multiplies trial probabilities only under independence.
Overlooking unequal probabilities. A binomial sum with the stated common parameter requires the same \(p\) for every constituent trial.
Adding probabilities instead of trial counts. For independent \(\operatorname{Bin}(n_j,p)\) counts, add the \(n_j\) values; retain the shared \(p\).
Using the wrong event. “At most \(k\)” includes every count from 0 through \(k\); “at least \(k\)” includes \(k\) and all larger possible counts.

These checks are useful in multi-step problems because the numerical calculation is usually straightforward once the distribution has been identified correctly. If the independence or common-probability condition fails, pause rather than forcing the sum into a binomial model.

Check Your Understanding

For each problem, identify whether the binomial sum rule applies before calculating.

Six independent Bernoulli trials each have success probability \(0.20\). Find the probability of exactly two successes. (Answer: 0.2458)
Independent counts have distributions \(X\sim\operatorname{Bin}(3,0.50)\) and \(Y\sim\operatorname{Bin}(5,0.50)\). Find \(P(X+Y\ge7)\). (Answer: 0.0352)
Two independent Bernoulli variables and an independent \(\operatorname{Bin}(4,0.30)\) count all have success probability \(0.30\). Find the probability that the combined total is zero. (Answer: 0.1176)
Independent counts have distributions \(X\sim\operatorname{Bin}(2,0.20)\) and \(Y\sim\operatorname{Bin}(3,0.50)\). Find \(P(X+Y=0)\), and state whether the common-probability binomial sum rule applies. (Answer: \(0.0800\); the rule does not apply because the success probabilities differ.)
Independent counts have distributions \(X\sim\operatorname{Bin}(4,0.25)\) and \(Y\sim\operatorname{Bin}(6,0.25)\). Find the probability of at least one success in the combined count. (Answer: 0.9437)

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