The Binomial Distribution

Think about flipping a coin 10 times. How likely are you to get exactly 6 heads? What about at least 8 heads? Or consider a basketball player who makes 75% of her free throws. If she takes 12 free throws in a game, what is the probability she makes exactly 9? What is her expected number of makes?

These questions all share something in common: they involve repeated trials where each trial has two possible outcomes (heads or tails, make or miss, success or failure), and we want to know how many successes to expect. This pattern shows up constantly in real life - from quality control in manufacturing to medical trials, from survey responses to games of chance.

The binomial distribution gives us a complete framework for answering these questions. Once you learn to recognize when a situation fits the binomial setting and how to apply the formula, you will have a powerful tool for analyzing success and failure in repeated trials.

Core Concepts

The Binomial Setting: When Does Binomial Apply?

Not every situation involving repeated trials follows a binomial distribution. For the binomial distribution to apply, four specific conditions must be met. These conditions form the binomial setting, sometimes remembered by the acronym BINS:

B - Binary outcomes: Each trial has exactly two possible outcomes. We call one outcome “success” and the other “failure.” These labels are arbitrary - “success” just means the outcome we are counting.

I - Independent trials: The outcome of one trial does not affect the outcome of any other trial. Knowing what happened on flip 1 tells you nothing about what will happen on flip 2.

N - Number of trials is fixed: You decide in advance how many trials you will perform. The number of trials, called $n$, is set before you begin.

S - Same probability: The probability of success, called $p$, remains constant from trial to trial. The first flip has the same 50% chance of heads as the tenth flip.

If all four conditions are satisfied, you are in a binomial setting, and the binomial distribution applies.

Example of a binomial setting: Flip a fair coin 20 times and count the number of heads.

• Binary: Each flip results in heads (success) or tails (failure). ✓
• Independent: Each flip is independent of the others. ✓
• Number fixed: We decided in advance to flip 20 times. ✓
• Same probability: $P(\text{heads}) = 0.5$ for every flip. ✓

Example that is NOT binomial: Draw 5 cards from a standard deck without replacement and count the number of hearts.

• Binary: Each card is either a heart or not. ✓
• Independent: No! After drawing a heart, fewer hearts remain, changing the probability for the next draw.
• The trials are not independent, so this is not binomial.

The Binomial Random Variable

When you are in a binomial setting, the count of successes is called a binomial random variable. We use the notation:

$$X \sim B(n, p)$$

This is read as “$X$ follows a binomial distribution with parameters $n$ and $p$,” or simply “$X$ is binomial with $n$ trials and success probability $p$.”

For example, if you flip a fair coin 10 times and let $X$ = number of heads:

$$X \sim B(10, 0.5)$$

The possible values of $X$ are 0, 1, 2, 3, …, up to $n$. You could get anywhere from zero successes (none of the trials succeed) to $n$ successes (every trial succeeds).

The Binomial Coefficient: Counting Arrangements

Before we can write the binomial probability formula, we need to understand how to count the number of ways to arrange successes among trials. This is where the binomial coefficient comes in.

The binomial coefficient $\binom{n}{k}$ (read as “$n$ choose $k$”) counts the number of ways to choose $k$ items from $n$ items, where order does not matter. In our context, it counts the number of ways to arrange $k$ successes among $n$ trials.

The formula is:

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

where $n!$ (read “$n$ factorial”) means $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$.

Example: How many ways can you get exactly 2 heads in 5 coin flips?

$$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(3 \times 2 \times 1)} = \frac{120}{2 \times 6} = \frac{120}{12} = 10$$

There are 10 different sequences of 5 flips that contain exactly 2 heads: HHTTT, HTHTT, HTTHT, HTTTH, THHTT, THTHT, THTTH, TTHHT, TTHTH, TTTHH

The Binomial Probability Formula

Now we can state the formula for calculating binomial probabilities. If $X \sim B(n, p)$, then the probability of getting exactly $k$ successes is:

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

Let’s understand what each part means:

• $\binom{n}{k}$ counts the number of ways to arrange $k$ successes among $n$ trials
• $p^k$ is the probability of $k$ successes (each with probability $p$)
• $(1-p)^{n-k}$ is the probability of the remaining $(n-k)$ failures (each with probability $1-p$)

The formula multiplies the number of arrangements by the probability of each arrangement. Since each arrangement has the same probability, this gives us the total probability.

Example: What is the probability of getting exactly 3 heads in 5 flips of a fair coin?

Here $n = 5$, $k = 3$, and $p = 0.5$.

$$P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{5-3}$$

$$= \binom{5}{3} (0.5)^3 (0.5)^2$$

First, calculate the binomial coefficient: $$\binom{5}{3} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \times 2} = 10$$

Then: $$P(X = 3) = 10 \times (0.5)^3 \times (0.5)^2 = 10 \times 0.125 \times 0.25 = 10 \times 0.03125 = 0.3125$$

There is a 31.25% chance of getting exactly 3 heads in 5 flips.

Mean and Standard Deviation of a Binomial Distribution

One of the beautiful things about the binomial distribution is that the mean (expected value) and standard deviation have simple formulas.

Mean (Expected Value): $$\mu = E(X) = np$$

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

These formulas make intuitive sense. If you flip a fair coin ($p = 0.5$) 100 times, you would expect about $np = 100 \times 0.5 = 50$ heads on average. The standard deviation tells you how much the actual count typically varies from this expectation.

Example: A basketball player makes 80% of her free throws. In a game where she attempts 25 free throws:

$$\mu = np = 25 \times 0.8 = 20$$

$$\sigma = \sqrt{np(1-p)} = \sqrt{25 \times 0.8 \times 0.2} = \sqrt{4} = 2$$

On average, she makes 20 free throws, with a standard deviation of 2. Most games, she will make somewhere between 18 and 22 free throws (within one standard deviation of the mean).

“At Least” and “At Most” Problems

Often we want to find the probability of getting at least a certain number of successes, or at most a certain number. These problems require adding up multiple binomial probabilities.

$P(X \geq k)$ means “at least $k$ successes”: Add up the probabilities for $k$, $k+1$, $k+2$, …, up to $n$.

$P(X \leq k)$ means “at most $k$ successes”: Add up the probabilities for 0, 1, 2, …, up to $k$.

The complement rule is often easier. Instead of adding many probabilities, use:

$$P(X \geq k) = 1 - P(X \leq k-1) = 1 - P(X < k)$$

$$P(X \leq k) = 1 - P(X \geq k+1) = 1 - P(X > k)$$

Example: In 10 coin flips, what is the probability of getting at least 8 heads?

We need $P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$.

$$P(X = 8) = \binom{10}{8}(0.5)^8(0.5)^2 = 45 \times 0.5^{10} = 45 \times \frac{1}{1024} \approx 0.0439$$

$$P(X = 9) = \binom{10}{9}(0.5)^9(0.5)^1 = 10 \times 0.5^{10} = 10 \times \frac{1}{1024} \approx 0.0098$$

$$P(X = 10) = \binom{10}{10}(0.5)^{10}(0.5)^0 = 1 \times 0.5^{10} = \frac{1}{1024} \approx 0.00098$$

$$P(X \geq 8) \approx 0.0439 + 0.0098 + 0.00098 = 0.0547$$

There is about a 5.5% chance of getting 8 or more heads in 10 flips.

Using Technology for Binomial Calculations

While the binomial formula is not difficult, the calculations can become tedious for large values of $n$. Most calculators and software have built-in binomial functions.

Calculator notation (varies by model):

• binompdf(n, p, k) gives $P(X = k)$ (the probability of exactly $k$ successes)
• binomcdf(n, p, k) gives $P(X \leq k)$ (the cumulative probability up to and including $k$)

Using the cumulative function for different probability types:

• $P(X = k)$: Use binompdf(n, p, k)
• $P(X \leq k)$: Use binomcdf(n, p, k)
• $P(X < k)$: Use binomcdf(n, p, k-1)
• $P(X \geq k)$: Use 1 - binomcdf(n, p, k-1)
• $P(X > k)$: Use 1 - binomcdf(n, p, k)

Example with technology: Find $P(X \geq 8)$ when $X \sim B(10, 0.5)$.

Using the complement: $P(X \geq 8) = 1 - P(X \leq 7) = 1 - \text{binomcdf}(10, 0.5, 7)$

On a TI calculator: 1 - binomcdf(10, 0.5, 7) $\approx 0.0547$

This matches our hand calculation and takes only seconds.

When Binomial Does NOT Apply

It is just as important to recognize when the binomial distribution does NOT apply. Here are common situations that look like they might be binomial but are not:

Sampling without replacement from a small population: If you draw cards from a deck without replacing them, the probabilities change with each draw. However, if the population is very large compared to the sample (at least 20 times larger), the changing probabilities are negligible and binomial is a good approximation.

Number of trials not fixed: If you keep flipping until you get 3 heads, the number of trials is random, not fixed. This requires a different distribution (the negative binomial).

More than two outcomes: If you roll a die and care about getting a 1, 2, or 3+ (three categories), that is not binomial. Binomial requires exactly two outcomes per trial.

Changing probabilities: If the probability of success changes from trial to trial (like a shooter getting tired or a machine wearing out), the binomial distribution does not apply.

Notation and Terminology

Term	Meaning	Example
Binomial setting	$n$ independent trials, probability $p$ of success	Flipping a coin 10 times
$n$	Number of trials	10 coin flips
$p$	Probability of success on each trial	0.5 for a fair coin
$k$	Number of successes	Getting 6 heads
$X \sim B(n,p)$	$X$ follows binomial distribution	$X \sim B(10, 0.5)$
$\binom{n}{k}$	Binomial coefficient, “$n$ choose $k$”	$\binom{10}{6} = 210$
binompdf	Probability of exactly $k$ successes	$P(X = k)$
binomcdf	Cumulative probability up to $k$	$P(X \leq k)$

Examples

Key Properties and Rules

The Binomial Probability Formula

For $X \sim B(n, p)$, the probability of exactly $k$ successes:

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

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Mean and Standard Deviation

Measure	Formula	Interpretation
Mean	$\mu = np$	Expected number of successes
Variance	$\sigma^2 = np(1-p)$	Measure of spread (squared units)
Standard Deviation	$\sigma = \sqrt{np(1-p)}$	Typical deviation from the mean

Cumulative Probability Rules

Probability Type	How to Calculate	Calculator
$P(X = k)$	Use binomial formula	binompdf(n, p, k)
$P(X \leq k)$	Sum $P(X = 0)$ through $P(X = k)$	binomcdf(n, p, k)
$P(X < k)$	Same as $P(X \leq k-1)$	binomcdf(n, p, k-1)
$P(X \geq k)$	$1 - P(X \leq k-1)$	1 - binomcdf(n, p, k-1)
$P(X > k)$	$1 - P(X \leq k)$	1 - binomcdf(n, p, k)

Conditions for a Binomial Setting (BINS)

1. Binary: Two outcomes per trial (success/failure)
2. Independent: Trials do not affect each other
3. Number: Fixed number of trials $n$
4. Same probability: Constant probability $p$ for each trial

When Binomial Approximation Works

Even if independence is slightly violated (like sampling without replacement), the binomial is a good approximation when: $$\text{Sample size} \leq \frac{\text{Population size}}{20}$$

In other words, if your sample is less than 5% of the population, the binomial approximation works well.

Real-World Applications

Quality Control in Manufacturing

Manufacturers use the binomial distribution to monitor product quality. If a machine typically produces 2% defective items, quality control can use binomial probability to determine:

• What is the expected number of defects in a batch of 500?
• What is the probability of finding more than 15 defects in a batch?
• How large should a sample be to have a 95% chance of detecting at least one defect if the defect rate has increased to 5%?

This helps set inspection procedures and identify when a manufacturing process needs adjustment.

Medical Trials and Drug Effectiveness

Clinical trials often measure success rates - did the treatment work or not? The binomial distribution helps researchers:

• Determine how many patients to include in a study
• Assess whether an observed success rate is significantly different from the expected rate
• Calculate confidence intervals for drug effectiveness

For example, if a new drug claims 80% effectiveness, and 45 out of 50 patients improve, researchers can use binomial probability to evaluate whether this result supports or contradicts the claim.

Survey Sampling and Polling

Political polls and market research surveys often involve yes/no questions. If 40% of a population supports a policy, and you survey 500 people:

• What is the expected number who will say “yes”?
• What range of responses should we expect with 95% confidence?
• Is our sample result significantly different from what we expected?

The binomial distribution (or its normal approximation for large samples) underlies the margin of error reported in polls.

Sports Analytics

In sports, many events can be modeled as success/failure trials:

• Free throws in basketball (make or miss)
• At-bats in baseball (hit or out)
• Penalty kicks in soccer (goal or save)

If a player has a known success rate, the binomial distribution predicts the probability of various outcomes. How likely is a 75% free throw shooter to go 6-for-10 in a game? What about 10-for-10? These probabilities help analysts evaluate player performance and make in-game decisions.

Genetics and Biology

Many genetic traits follow binomial patterns. If two carriers of a recessive gene have children, each child has a 25% chance of expressing the trait. For a family with 4 children:

• What is the probability none have the trait?
• What is the probability at least one has the trait?
• What is the expected number with the trait?

Binomial probability is fundamental to understanding how traits are distributed across generations.

Self-Test Problems

Summary

• The binomial setting requires four conditions (BINS): Binary outcomes, Independent trials, fixed Number of trials, and Same probability for each trial.

• A binomial random variable $X \sim B(n, p)$ counts the number of successes in $n$ independent trials, each with success probability $p$.

• The binomial probability formula gives the probability of exactly $k$ successes: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$

• The binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ counts the number of ways to arrange $k$ successes among $n$ trials.

• The mean of a binomial distribution is $\mu = np$. The standard deviation is $\sigma = \sqrt{np(1-p)}$.

• For “at least” or “at most” problems, use the complement rule: $P(X \geq k) = 1 - P(X \leq k-1)$.

• Technology functions binompdf(n, p, k) and binomcdf(n, p, k) calculate exact probabilities and cumulative probabilities efficiently.

• The binomial distribution does NOT apply when: trials are not independent, the number of trials is not fixed, there are more than two outcomes, or the probability changes between trials.

• Applications include quality control, medical trials, survey sampling, sports analytics, and genetics - anywhere you count successes in repeated independent trials with constant probability.