6  Week 3: The Binomial Distribution

6.1 Why This Matters

Last week we defined random variables and their probability distributions in general terms, you had to be handed a table of \(x\) and \(P(X=x)\) values before you could compute anything. Many business situations, though, share a common structure: a fixed number of trials, each ending in one of two outcomes, with the same success probability every time. When that structure holds, we don’t need a custom table at all; a single formula, the binomial distribution, gives us every probability, the mean, and the standard deviation automatically. It’s the first named distribution we’ll study, and one of the most widely used in business analytics: conversion rates, defect rates, and response rates are all, at their core, binomial situations.

6.2 When Does a Binomial Distribution Apply?

A discrete random variable \(X\) counting the number of successes follows a binomial distribution, written \(X \sim \text{Bin}(n, p)\), when four conditions hold:

  1. There is a fixed number \(n\) of trials.
  2. Each trial results in one of exactly two outcomes: success or failure.
  3. The probability of success, \(p\), is the same on every trial.
  4. The trials are independent of one another.
ExampleExample 3.1: Does the binomial model apply?

Applies. A company sends the same promotional email to 500 randomly selected customers on its list, and each has an estimated 8% chance of clicking through, independent of whether any other customer clicks. Let \(X\) = number of clicks. Here \(n=500\) is fixed, each customer either clicks or doesn’t, \(p=0.08\) is the same for each (they were selected the same way), and one customer’s click doesn’t affect another’s. \(X \sim \text{Bin}(500, 0.08)\).

Does not apply. A hiring manager interviews all 12 finalists for a role and counts how many receive an offer. This is not binomial: the number of offers is capped by how many openings exist, and once one finalist is hired, that changes the pool and the effective odds for the rest: the trials are not independent, and the “success probability” doesn’t stay fixed.

6.3 The Binomial Probability Formula

For \(n\) independent trials with success probability \(p\), the probability of exactly \(x\) successes is \[ P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}, \qquad x = 0, 1, 2, \dots, n \] where \(\displaystyle \binom{n}{x} = \frac{n!}{x!(n-x)!}\) counts the number of different ways to arrange \(x\) successes among \(n\) trials.

ExampleExample 3.2: Where the formula comes from

A sales rep closes any given cold call with probability \(p=0.20\), independent of other calls. Over \(n=3\) calls, what’s the probability of exactly 2 closes?

Writing \(S\) for a close and \(F\) for a miss, there are three orderings that give exactly 2 successes: \(SSF\), \(SFS\), \(FSS\). Each has the same probability, since the calls are independent: \[ P(SSF) = P(SFS) = P(FSS) = (0.20)^2(0.80)^1 = 0.032 \] Because these three orderings are mutually exclusive, we add them: \[ P(X=2) = 3 \times 0.032 = 0.096 \] This matches the formula directly: \(\binom{3}{2}=3\) counts the number of orderings, and \((0.20)^2(0.80)^1\) is the probability of any one of them. For larger \(n\), listing every ordering by hand becomes impractical, this is exactly why we use the formula instead.

ExampleExample 3.3: Applying the formula directly

Using the same rep from Example 3.2 (\(p=0.20\), \(n=3\) calls), what is the probability of exactly 1 close? \[ P(X=1) = \binom{3}{1}(0.20)^1(0.80)^2 = 3 \times 0.20 \times 0.64 = 0.384 \]

6.4 Mean and Standard Deviation of the Binomial

For \(X \sim \text{Bin}(n,p)\), the mean and standard deviation have simple closed forms, no need to build a full probability table and sum by hand: \[ \mu = E(X) = np \qquad\qquad \sigma = \sqrt{np(1-p)} \]

ExampleExample 3.4: Defects in a production batch

A production line has a 5% defect rate, and defects occur independently across units. In a batch of 60 units, let \(X\) = number of defective units. Then \(X \sim \text{Bin}(60, 0.05)\), and \[ \mu = np = 60(0.05) = 3 \] \[ \sigma = \sqrt{np(1-p)} = \sqrt{60(0.05)(0.95)} \approx 1.688 \] A quality manager reviewing this batch should expect around 3 defective units, typically varying by about 1.7 units from batch to batch—so seeing 5 or 6 defects in a given batch wouldn’t be especially alarming on its own, but seeing 10 would be well outside the typical range.

NoteA guideline for sampling without replacement

The binomial distribution technically requires independent trials, which breaks down when sampling without replacement from a finite population (much like the dependent-draws example from Chapter 4). In practice, when the sample is small relative to the population (a common rule of thumb is less than 10% of the population) the binomial distribution still approximates the situation well. Most business sampling situations (surveying 500 out of 50,000 customers, inspecting 60 out of thousands of units produced) comfortably satisfy this guideline.

6.5 Computing Binomial Probabilities in R and Excel

Once \(n\) and \(p\) are known, R and Excel both provide the binomial formula as a built-in function—there’s no need to compute \(\binom{n}{x}\) by hand.

ExampleExample 3.5: A/B test conversions in R and Excel

A retailer runs a checkout redesign as an A/B test on 40 randomly selected sessions, and historical data suggests a 15% baseline conversion probability per session (\(p=0.15\), assumed unchanged by the redesign for now). What’s the probability that exactly 8 of the 40 sessions convert?

In R:

dbinom(x = 8, size = 40, prob = 0.15)
[1] 0.1086647

The mean and standard deviation can be computed directly from the formulas, or by summarizing the full distribution:

n <- 40
p <- 0.15
n * p                    # mean
[1] 6
sqrt(n * p * (1 - p))    # standard deviation
[1] 2.258318

Now suppose the team also wants to know the probability that 8 or fewer sessions convert, a cumulative, “at most” question. R’s dbinom() only gives the probability of one exact value, so we use pbinom() instead, which adds up \(P(X=0) + P(X=1) + \dots + P(X=8)\) for us:

pbinom(q = 8, size = 40, prob = 0.15)
[1] 0.864598

The q argument is the cutoff value, and pbinom() always returns \(P(X \le q)\) by default. If instead we wanted the probability of more than 8 conversions, we can use the complement rule from Chapter 4:

1 - pbinom(q = 8, size = 40, prob = 0.15)
[1] 0.135402

In Excel, the equivalent single-probability calculation is:

=BINOM.DIST(8, 40, 0.15, FALSE)

The final argument, FALSE, requests the probability of exactly 8 successes. Setting it to TRUE instead returns the same cumulative probability that pbinom() gives us, \(P(X \le 8)\):

=BINOM.DIST(8, 40, 0.15, TRUE)

As in R, “more than 8” in Excel is found with the complement: =1 - BINOM.DIST(8, 40, 0.15, TRUE).

6.6 Recap

Keyword Definition
Binomial distribution The distribution of the number of successes in \(n\) independent trials, each with the same success probability \(p\); written \(X \sim \text{Bin}(n,p)\).
Binomial conditions Fixed \(n\), two outcomes per trial, constant \(p\), independent trials.
Binomial probability formula \(P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}\).
Binomial mean \(\mu = np\).
Binomial standard deviation \(\sigma = \sqrt{np(1-p)}\).

6.7 Check Your Understanding

NoteProblems

  1. A call center’s historical data shows 30% of inbound calls result in a completed sale, independent from call to call. Of the next 10 calls, let \(X\) be the number resulting in a sale.

    1. Confirm the binomial conditions are reasonably satisfied.
    2. Compute \(P(X=4)\).
    3. Compute \(E(X)\) and \(\sigma\).

  2. A hiring platform reports that 12% of applicants who reach the final interview stage receive an offer, independent across applicants. Out of 25 finalists this quarter, what is the probability that exactly 3 receive an offer?

  3. A batch of 200 components is drawn from a very large supplier population with a known 4% defect rate. Would the binomial distribution be a reasonable model for the number of defective components in the batch? Explain using the 10% guideline.

  4. Explain, in your own words, why a binomial model would not be appropriate for counting the number of a company’s top 5 salespeople who hit their target this quarter, if only one bonus pool exists and the top 5 are competing for a fixed number of bonus slots that depend on each other’s performance.

    1. There’s a fixed \(n=10\) calls, each call either results in a sale or not, the sale probability is stated as constant at 30%, and calls are reasonably assumed independent—the conditions are satisfied, so \(X \sim \text{Bin}(10, 0.30)\).
    2. \(P(X=4) = \binom{10}{4}(0.30)^4(0.70)^6 = 210 \times 0.0081 \times 0.117649 \approx 0.200\).
    3. \(E(X) = 10(0.30) = 3\). \(\sigma = \sqrt{10(0.30)(0.70)} = \sqrt{2.1} \approx 1.449\).

  2. Let \(X \sim \text{Bin}(25, 0.12)\). \(P(X=3) = \binom{25}{3}(0.12)^3(0.88)^{22} = 2300 \times 0.001728 \times 0.0688 \approx 0.2735\).

  3. Yes, this is reasonable. The batch of 200 is drawn from a “very large” supplier population, and 200 is almost certainly less than 10% of that population, so treating the draws as approximately independent (and thus the count as approximately binomial) is a reasonable approximation.

  4. Because only a fixed number of bonus slots exist and the salespeople are compared against each other rather than against an independent fixed threshold, one salesperson hitting the target changes the effective chances for the others (e.g., if bonuses are capped or relative rank matters). This violates the independence condition required for a binomial model—the trials (each salesperson’s outcome) are not independent of one another.