Math Core

Many random processes boil down to counting successes in a fixed number of tries: how many of 10 free throws go in, how many of 50 surveyed voters support a measure, how many of 20 parts are defective. When the tries are independent and the success probability stays the same, the count has a binomial distribution, one of the most important models in statistics.

The binomial setting

Definition

Binomial setting

A binomial setting arises when you perform $n$ trials and all four of these conditions hold (remember them as BINS):

Binary: each trial has two outcomes, "success" or "failure".
Independent: the result of one trial doesn't affect any other.
Number: the number of trials $n$ is fixed in advance.
Same probability: each trial has the same probability of success $p$ .

The count $X$ of successes is a binomial random variable with parameters $n$ and $p$ .

"Success" is just the outcome you are counting. It doesn't have to be good; a defective part can be a success.

Sampling without replacement and the 10% condition

When you sample people without replacement, the trials aren't exactly independent: removing one person changes the proportions slightly for the next pick. But if the sample is small relative to the population, the change is negligible. The 10% condition says you may treat the trials as independent if $n \le 0.10N$ , where $N$ is the population size.

The binomial probability formula

How likely is exactly $k$ successes? Any one particular arrangement of $k$ successes and $n - k$ failures has probability $p^k (1-p)^{n-k}$ , by the multiplication rule for independent trials. The number of different arrangements is $\binom{n}{k}$ , read " $n$ choose $k$ ".

Binomial probability

If $X$ is binomial with $n$ trials and success probability $p$ , then for $k = 0, 1, \dots, n$ ,

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}, \qquad \binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}.$

On a TI-84, $P(X = k)$ is binompdf(n, p, k) and $P(X \le k)$ is binomcdf(n, p, k). On the AP exam, name the distribution and its parameters, for example " $X$ is binomial with $n = 10$ , $p = 0.2$ ", and show the formula or the calculator command with labeled inputs.

Worked example: Exactly three

A quiz has 10 multiple-choice questions, each with 5 choices. A student guesses randomly on every question. Find the probability of exactly 3 correct answers.

Solution. Check BINS: each question is right or wrong, guesses are independent, $n = 10$ is fixed and $p = \dfrac{1}{5} = 0.2$ every time. So $X$ is binomial with $n = 10$ , $p = 0.2$ .

$P(X = 3) = \binom{10}{3}(0.2)^3(0.8)^7 = 120(0.008)(0.2097152) \approx 0.2013.$

Cumulative probabilities

For "at most", "at least" and "between", add individual probabilities or use the complement.

Worked example: At least two

For the guessing student, find $P(X \ge 2)$ .

Solution. It's quicker to subtract the two excluded values from 1:

\begin{aligned} P(X \ge 2) &= 1 - P(X = 0) - P(X = 1) \\ &= 1 - (0.8)^{10} - \binom{10}{1}(0.2)(0.8)^9 \\ &= 1 - 0.1074 - 0.2684 \\ &\approx 0.6242. \end{aligned}

With a calculator: $1 - \texttt{binomcdf}(10, 0.2, 1) \approx 0.6242$ .

Common mistake

Watch the endpoint when using the complement. $P(X \ge 2) = 1 - P(X \le 1)$ , not $1 - P(X \le 2)$ . Write the values that are excluded before you subtract.

Mean, standard deviation and shape

Mean and standard deviation of a binomial random variable

$\mu_X = np, \qquad \sigma_X = \sqrt{np(1 - p)}.$

For the guessing student, $\mu_X = 10(0.2) = 2$ correct answers and $\sigma_X = \sqrt{10(0.2)(0.8)} = \sqrt{1.6} \approx 1.26$ . Over many such quizzes, a guessing student would average 2 correct, typically varying from 2 by about 1.26 questions.

The shape depends on $p$ and $n$ . With $p$ below $0.5$ the distribution is skewed right, with $p$ above $0.5$ it's skewed left, and with $p = 0.5$ it's symmetric. As $n$ grows, the distribution becomes more symmetric and bell-shaped. A common guideline says it is approximately normal when $np \ge 10$ and $n(1-p) \ge 10$ , which you will use heavily in Unit 5.

Binomial distribution with n = 5 and p = 0.2. It is skewed right. P(X = 5) = 0.0003 is too small to see.

Tip

Check that your answer is reasonable using the mean. If $\mu_X = 2$ , then $P(X = 3)$ being about $0.2$ is sensible; a result like $0.9$ would signal an error.

Practice

Practice 1

Which random variable has a binomial distribution?

Practice 2

$X$ is binomial with $n = 8$ and $p = 0.3$ . Find $P(X = 2)$ . Round to 4 decimal places.