When can you NOT use the binomial distribution?

You cannot use it when trials are not independent (e.g., drawing cards without replacement from a small deck), when the probability of success changes from trial to trial, or when there are more than two possible outcomes per trial. For sampling without replacement, the hypergeometric distribution is more appropriate, though the binomial can approximate it when the population is much larger than the sample.

Binomial Distribution — Definition, Formula & Examples

A binomial distribution models the number of successes in a fixed number of independent trials, where each trial has exactly two outcomes (success or failure) and the same probability of success.

A discrete random variable $X$ follows a binomial distribution with parameters $n$ and $p$ , written $X \sim B(n, p)$ , if $X$ counts the number of 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 $k = 0, 1, 2, \ldots, n$ , with mean $\mu = np$ and variance $\sigma^2 = np(1-p)$ .

Key Formula

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

Where:

$n$ = Total number of independent trials
$k$ = Number of successes you want the probability for
$p$ = Probability of success on a single trial
$\binom{n}{k}$ = Binomial coefficient, equal to n! / (k!(n−k)!)

How It Works

To use the binomial distribution, first verify four conditions: (1) there is a fixed number of trials

n

, (2) each trial is independent, (3) each trial has only two outcomes, and (4) the probability of success

p

stays constant across trials. Once confirmed, plug

n

p

, and the desired number of successes

k

into the binomial probability formula. The binomial coefficient

\binom{n}{k}

counts how many different orderings of

k

successes can occur among

n

trials, while

p^k(1-p)^{n-k}

gives the probability of any single such ordering. For cumulative questions like "at most 3 successes," sum the individual probabilities from

k = 0

k = 3

Worked Example

Problem: You flip a fair coin 10 times. What is the probability of getting exactly 6 heads?

Identify parameters: Each flip is an independent trial with two outcomes. Here

n = 10

p = 0.5

, and

k = 6

Compute the binomial coefficient: Calculate the number of ways to choose 6 heads out of 10 flips.

\binom{10}{6} = \frac{10!}{6!\cdot 4!} = 210

Apply the formula: Substitute into the binomial probability formula.

P(X = 6) = 210 \cdot (0.5)^{6} \cdot (0.5)^{4} = 210 \cdot (0.5)^{10}

Calculate: Since

(0.5)^{10} = \frac{1}{1024}

, multiply to get the final probability.

P(X = 6) = \frac{210}{1024} \approx 0.2051

Answer: The probability of getting exactly 6 heads in 10 flips is approximately 0.205, or about 20.5%.

Another Example

Problem: A multiple-choice quiz has 8 questions, each with 4 choices. If you guess randomly on every question, what is the probability of getting exactly 3 correct?

Identify parameters: Each question is an independent trial with

p = 0.25

(one correct choice out of four). We have

n = 8

and

k = 3

Binomial coefficient: Count the number of ways to choose which 3 questions are answered correctly.

\binom{8}{3} = \frac{8!}{3!\cdot 5!} = 56

Apply the formula: Plug everything into the binomial formula.

P(X = 3) = 56 \cdot (0.25)^{3} \cdot (0.75)^{5}

Calculate: Evaluate each piece:

(0.25)^3 = 0.015625

and

(0.75)^5 \approx 0.23730

P(X = 3) = 56 \cdot 0.015625 \cdot 0.23730 \approx 0.2076

Answer: The probability of guessing exactly 3 out of 8 correct is approximately 0.208, or about 20.8%.

Visualization

Why It Matters

The binomial distribution is one of the most heavily tested topics in AP Statistics, appearing in both multiple-choice and free-response sections. Quality-control engineers use it to decide how many defective items to expect in a production batch. Medical researchers rely on it when analyzing the number of patients who respond to a treatment in a clinical trial of fixed size.

Common Mistakes

Mistake: Forgetting the binomial coefficient and only computing

p^k(1-p)^{n-k}

Correction: The term

p^k(1-p)^{n-k}

gives the probability of one specific ordering. You must multiply by

\binom{n}{k}

to account for all possible arrangements of successes among the

n

trials.

Mistake: Using the binomial distribution when trials are not independent.

Correction: Check independence before applying the formula. If outcomes affect each other (e.g., sampling without replacement from a small population), the binomial model does not apply.