The Binomial Distribution
Verify BINS conditions (Binary, Independent, Number fixed, Same probability). Compute P(X = k) = C(n,k)pᵏ(1−p)ⁿ⁻ᵏ, mean μ = np, and standard deviation σ = √(np(1−p)). Use the normal approximation to the binomial.
A lesson of AP Statistics, in Probability, about 10 minutes.
By the end of this lesson, you will be able to identify binomial settings, calculate binomial probabilities, and find the mean and standard deviation of a binomial random variable.
The Binomial Setting (BINS)
A random variable X follows a binomial distribution if all four conditions hold:
B — Binary: each trial has exactly two outcomes (success or failure)
I — Independent: trials are independent of each other
N — Number of trials: a fixed number $n$ of trials
S — Same probability: the probability of success $p$ is the same on every trial
If all four hold, we write $X \sim B(n, p)$.
Check the conditions explicitly on the AP exam. Stating "B, I, N, S" and verifying each earns points.
The 10% Condition
Trials are technically only independent when sampling with replacement. However, if the sample is less than 10% of the population, we treat the trials as approximately independent. This is the 10% condition.
Binomial Probability Formula
$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the number of ways to choose k successes from n trials.
Example: A free throw shooter makes 70% of shots. Find the probability of making exactly 4 out of 5 attempts.
$P(X = 4) = \binom{5}{4}(0.7)^4(0.3)^1 = 5 \cdot 0.2401 \cdot 0.3 = 5 \cdot 0.07203 = 0.3602$
Computing Cumulative Probabilities
P(X ≤ k): sum the binomial formula for x = 0, 1, 2, ..., k. Use calculator functions: binomcdf(n, p, k).
P(X ≥ k) = 1 − P(X ≤ k−1): use the complement rule.
P(X = k): use binompdf(n, p, k) or the formula.
Mean and Standard Deviation of a Binomial
$$\mu_X = np \qquad \sigma_X = \sqrt{np(1-p)}$$
Example: Out of 20 free throw attempts at 70%:
$\mu_X = 20(0.7) = 14$ makes expected on average
$\sigma_X = \sqrt{20(0.7)(0.3)} = \sqrt{4.2} \approx 2.05$
Normal Approximation to the Binomial
When the sample size is large enough, the binomial distribution is approximately normal:
$X \sim B(n,p)$ is approximately $N(\mu, \sigma) = N(np, \sqrt{np(1-p)})$
Condition for approximation: $np \geq 10$ AND $n(1-p) \geq 10$.