The Binomial Distribution

The binomial distribution is one of the most important probability distributions in statistics. It models a simple but powerful scenario: you repeat the same experiment a fixed number of times, each trial has exactly two outcomes (success or failure), and you count the total number of successes. Coin flips, pass/fail inspections, yes/no survey questions, and defective/non-defective product checks all follow this pattern.

The Four Conditions (BINS)

A random variable $X$ follows a binomial distribution only when all four of these conditions are met. The mnemonic BINS makes them easy to remember:

• Binary outcomes — each trial has exactly two outcomes: success or failure
• Independent trials — the result of one trial does not affect any other trial
• Number of trials is fixed — you decide on $n$ before the experiment begins
• Same probability — the probability of success $p$ stays constant from trial to trial

If any condition is violated, the binomial model does not apply. For example, drawing cards without replacement violates independence (the deck composition changes after each draw), so that scenario is not binomial.

The Binomial Formula

If $X$ is the number of successes in $n$ independent trials, each with probability of success $p$, then:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n$ = number of trials
• $k$ = number of successes you are calculating the probability for
• $p$ = probability of success on a single trial
• $(1-p)$ = probability of failure on a single trial
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ = the number of ways to arrange $k$ successes among $n$ trials

The combination $\binom{n}{k}$ is essential because the $k$ successes can occur in any order across the $n$ trials, and each arrangement has the same probability.

Worked Examples

Example 1: Free Throws

A basketball player makes 80% of free throws ($p = 0.80$). In 5 attempts ($n = 5$), what is the probability of making exactly 3?

Identify the parts: $n = 5$, $k = 3$, $p = 0.80$, $(1-p) = 0.20$

Step 1: Calculate the combination.

$\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{5 \times 4}{2 \times 1} = 10$

Step 2: Calculate $p^k$ and $(1-p)^{n-k}$.

$p^k = (0.80)^3 = 0.512$

$(1-p)^{n-k} = (0.20)^2 = 0.04$

Step 3: Multiply everything together.

$P(X = 3) = 10 \times 0.512 \times 0.04 = 10 \times 0.02048 = 0.2048$

Answer: The probability of making exactly 3 out of 5 free throws is 0.2048, or about 20.5%.

Verification: $\binom{5}{3} = 10$ $\checkmark$. $(0.80)^3 = 0.512$ $\checkmark$. $(0.20)^2 = 0.04$ $\checkmark$. $10 \times 0.512 \times 0.04 = 0.2048$ $\checkmark$

Example 2: Quality Control — Zero Defects

A factory has a 5% defect rate ($p = 0.05$). In a batch of 10 items ($n = 10$), what is the probability that none are defective?

$P(X = 0) = \binom{10}{0}(0.05)^0(0.95)^{10}$

$= 1 \times 1 \times (0.95)^{10}$

$= 0.5987$

Answer: There is about a 59.9% chance that all 10 items are defect-free.

Verification: $(0.95)^{10} = 0.95^{10}$. Computing step by step: $0.95^2 = 0.9025$, $0.95^4 = 0.8145$, $0.95^5 = 0.77378$, $0.95^{10} = 0.77378^2 = 0.5987$ $\checkmark$

Example 3: At Least One Defect (Complement Approach)

Using the same scenario, what is the probability of at least one defective item?

The complement approach is much faster than calculating $P(X=1) + P(X=2) + \cdots + P(X=10)$:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5987 = 0.4013$

Answer: There is about a 40.1% chance of finding at least one defective item in a batch of 10.

This is a surprisingly high probability given that the defect rate is only 5%. Even a low per-item defect rate accumulates quickly across multiple items.

Mean and Standard Deviation

For a binomial random variable, the mean and standard deviation have simple formulas:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

These follow from the general expected value and variance formulas for random variables, but the binomial structure makes them especially clean.

Example 4: Free Throws — Mean and Standard Deviation

Using the free throw scenario ($n = 5$, $p = 0.80$):

$\mu = 5 \times 0.80 = 4$

On average, the player makes 4 out of 5 free throws.

$\sigma = \sqrt{5 \times 0.80 \times 0.20} = \sqrt{5 \times 0.16} = \sqrt{0.80} \approx 0.894$

The standard deviation is about 0.894, meaning most outcomes will be within about 1 make of the average. You would typically see 3, 4, or 5 makes — which lines up with our probability table below.

When to Use the Binomial Distribution

Not every yes/no scenario is binomial. The table below helps you decide:

Rule of thumb: If sampling without replacement but the sample is less than 10% of the population, you can still approximate with the binomial distribution. Drawing 5 items from a warehouse of 10,000 is effectively binomial because removing one item barely changes the composition.

Binomial Probability Table

Here is the complete distribution for the free throw example ($n = 5$, $p = 0.80$):

Verification: $0.00032 + 0.00640 + 0.05120 + 0.20480 + 0.40960 + 0.32768 = 1.00000$ $\checkmark$

Notice that $k = 4$ is the most likely outcome (about 41%), consistent with the mean of 4. Making all 5 has about a 33% probability, while making 0 or 1 is extremely unlikely (less than 1% combined).

Real-World Application: Nursing — Medication Side Effects

A medication has a 15% side effect rate ($p = 0.15$). A nurse administers it to 8 patients ($n = 8$). What is the probability that 2 or fewer patients experience side effects?

We need $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$.

P(X = 0):

$P(X = 0) = \binom{8}{0}(0.15)^0(0.85)^8 = 1 \times 1 \times (0.85)^8$

Computing $(0.85)^8$: $0.85^2 = 0.7225$, $0.85^4 = 0.7225^2 = 0.5220$, $0.85^8 = 0.5220^2 = 0.2725$

$P(X = 0) = 0.2725$

P(X = 1):

$P(X = 1) = \binom{8}{1}(0.15)^1(0.85)^7 = 8 \times 0.15 \times (0.85)^7$

$(0.85)^7 = (0.85)^8 / 0.85 = 0.2725 / 0.85 = 0.3206$

$P(X = 1) = 8 \times 0.15 \times 0.3206 = 8 \times 0.04809 = 0.3847$

P(X = 2):

$P(X = 2) = \binom{8}{2}(0.15)^2(0.85)^6 = 28 \times 0.0225 \times (0.85)^6$

$(0.85)^6 = (0.85)^4 \times (0.85)^2 = 0.5220 \times 0.7225 = 0.3771$

$P(X = 2) = 28 \times 0.0225 \times 0.3771 = 28 \times 0.008485 = 0.2376$

Total:

$P(X \leq 2) = 0.2725 + 0.3847 + 0.2376 = 0.8948$

Answer: There is about an 89.5% chance that 2 or fewer of the 8 patients will experience side effects.

Verification: $0.2725 + 0.3847 + 0.2376 = 0.8948$ $\checkmark$. This makes sense because the expected number of patients with side effects is $\mu = 8 \times 0.15 = 1.2$, and 2 or fewer is above the mean, so the cumulative probability should be well above 50%.

Why this matters in nursing: If a nurse is monitoring 8 patients and 4 or 5 experience side effects from a drug with a 15% side effect rate, the binomial model tells us that outcome is very unlikely (the probability of 4 or more is only about $1 - P(X \leq 3) \approx 0.02$). This would signal a potential issue — perhaps a bad batch, a drug interaction, or a patient population with elevated risk factors.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The binomial distribution models the number of successes in $n$ fixed, independent trials with constant success probability $p$.
• Remember the BINS checklist: Binary outcomes, Independent trials, fixed Number of trials, Same probability.
• Binomial formula: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
• Mean: $\mu = np$ gives the expected number of successes.
• Standard deviation: $\sigma = \sqrt{np(1-p)}$ measures the typical variation around the mean.
• Use the complement $P(X \geq 1) = 1 - P(X = 0)$ to find “at least one” probabilities efficiently.
• The binomial distribution appears in quality control, medical studies, survey analysis, retail conversion rates, and any context where you count successes in repeated trials.

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