AP Statistics Topic 2.10: The Binomial Distribution

01 · Start here

The concept in 60 seconds

A binomial random variable counts how many successes occur in a fixed number of independent trials, with the same success probability on every trial.

Imagine $6$ practice attempts. Each attempt is either successful or unsuccessful. If the attempts are independent and each has success probability $0.30$ , the number of successful attempts has a binomial distribution.

Separate $1$ trial from the whole experiment:
$1$ trial: $1$ attempt, giving success or failure.
$1$ repetition of the experiment: all $6$ attempts, giving a count from $0$ to $6$ .

“Success” is a label for the outcome you are counting. It can mean an error, a faulty item or a missed shot; it does not have to mean something desirable.

Quick check: does the count itself have only

2

possible values?

No. Each trial has $2$ outcome categories. The count across $6$ trials can be $0, 1, 2, 3, 4, 5,\text{ or }6$ .

02 · Check the setting before calculating

Check the binomial conditions

Use BINS to remember the conditions: Binary outcomes, Independent trials, a fixed Number of trials, and the Same success probability.

Visual guide 1: BINS in the

6

-attempt setting

B · Binary $2$ categories per trial

Each attempt is either successful or unsuccessful. $X$ counts the successful category.

I · IndependentNo outcome changes another

For this fictional model, outcomes are assumed independent. $1\text{ result}$ does not change another attempt’s success chance.

N · Number fixed $6$ attempts in advance

Complete $n = 6$ attempts. Do not stop early when a particular outcome occurs.

S · Same probability $p = 0.30$ every time

Every attempt has $30\%$ success probability and $70\%$ failure probability.

All four conditions concern the individual trials and how the experiment is conducted. The resulting count $X$ can take $7$ values, from $0\text{ through }6$ .

For our original fictional $6$ -attempt model, let $X$ be the number of successful practice attempts. The setting explicitly assumes $6$ attempts, independent outcomes, and success probability $0.30$ on every attempt. Those assumptions justify a binomial model.

A realistic story does not automatically establish independence or a constant probability. Learning, fatigue or a change in task difficulty may make that model unsuitable unless the question supplies appropriate assumptions.

Two common settings that need a different model

Stop at the first success: the number of trials is not fixed in advance. A waiting-time variable is not a binomial count.
Draw without replacement from a small collection: the remaining mix changes. The trials are dependent and the success probability changes after a draw.

For example, draw $3$ tokens without replacement from $10$ tokens, $4\text{ of which}$ are Green. Initially $P(\text{Green}) = \frac{4}{10}$ . After a Green draw it is $\frac{3}{9}$ , and after a non-Green draw it is $\frac{4}{9}$ . This is not exactly binomial. Replacing the token and mixing before each independent uniform draw would keep $p = \frac{4}{10}$ .

Sampling a small fraction of a large population can sometimes support an approximation to independence. An approximation must be justified from the sampling setting; without-replacement draws are not exactly independent.

Improve this justification: “It is binomial because there are

2

outcomes.”

$2$ categories alone are not enough. State all four conditions in context: “ $X$ counts successes in $6$ attempts. Each attempt is success/failure, the attempts are independent, $6\text{ is fixed}$ in advance, and each attempt has success probability $0.30$ .”

03 · Define the count and its model

Key ideas and notation

$X$ : the success count

$X$ records the number of successes across all $n$ trials. Its possible values are the integers $0\text{ through }n$ .

Here $X$ counts successful attempts out $\text{of }6$ .

$n$ : fixed trial count

$n$ is the number of trials in $1$ complete experiment, fixed before observing outcomes.

Here $n = 6$ attempts, even if the first attempt succeeds.

$p$ : $1\text{-trial}$ probability

$p$ is the chance of success on each trial. It stays the same across trials.

Here $p = 0.30$ ; failure probability $1 – p = 0.70$ .

$x$ : the requested count

A lowercase $x$ names a particular possible count in a calculation.

For “exactly $2$ successes,” use $x = 2$ .

We can write $X \sim \operatorname{Binomial}(n = 6, p = 0.30)$ . The distribution below gives a probability to each possible success count.

Visual guide 2: the full

6

-attempt distribution

Exact model probabilities for $X \sim \operatorname{Binomial}(6, 0.30)$ .
Successful attempts $x$	$P(X = x)$
$0$	$0.117649$
$1$	$0.302526$
$2$	$0.324135$
$3$	$0.185220$
$4$	$0.059535$
$5$	$0.010206$
$6$	$0.000729$

Original fictional independent-attempt model. Separate bars show probabilities at integer counts. All probabilities use the same $0\text{-}1$ scale; direct chart labels are rounded to $3$ decimals, while the table keeps $6$ decimal places.

The probabilities in the table total $1$ exactly. The chart labels are rounded to $3$ decimal places for readability; use unrounded values during calculations.

The value $2$ is the most likely single count in this model. That does not mean every experiment has $2$ successes, and it is not the same as the mean $1.8$ .

04 · Count the arrangements as well as their probability

Calculate an exact probability

P(X=x)=\binom{n}{x}p^x(1-p)^{n-x},\qquad x=0,1,\ldots,n.

The combination factor

\binom{n}{x}=\frac{n!}{x!(n-x)!}

counts the arrangements of $x$ successes among $n$ trial positions.

The symbol $\binom{n}{x}$ , also written “ $n$ choose $x$ ,” counts combinations. A factorial such as $4!$ means $4 \times 3 \times 2 \times 1$ , and $0! = 1$ .

What each factor does

$p^x$ : accounts for $x$ successful trials.
$(1 – p)^{n – x}$ : accounts for the remaining failures.
$\binom{n}{x}$ : accounts for all ways those successes can occupy the trial positions.

Independence lets us multiply the probabilities along $1$ ordered sequence. Different sequences are mutually exclusive, so we add their probabilities.

$\text{Exactly }2\text{ successes in }6\text{ attempts}$

$1\text{ particular sequence}$ , $\texttt{SSFFFF}$ , has probability $0.30^2 \times 0.70^4 = 0.021609$ . The event “exactly $2$ successes” includes that sequence and $14\text{ others}$ .

Visual guide 3:

15

arrangements,

1

success count

Every sequence below contains exactly $2$ successes ( $\texttt{S}$ ) and $4$ failures ( $\texttt{F}$ ).

\texttt{SSFFFF}

\texttt{SFSFFF}

\texttt{SFFSFF}

\texttt{SFFFSF}

\texttt{SFFFFS}

\texttt{FSSFFF}

\texttt{FSFSFF}

\texttt{FSFFSF}

\texttt{FSFFFS}

\texttt{FFSSFF}

\texttt{FFSFSF}

\texttt{FFSFFS}

\texttt{FFFSSF}

\texttt{FFFSFS}

\texttt{FFFFSS}

Each sequence:

0.30^2\times0.70^4=0.021609

All $15$ sequences:

15\times0.021609=0.324135

Each letter is $1$ attempt position. Independence and the same $p$ make all these $2\text{-success}$ sequences have the same probability. They are mutually exclusive, so add their probabilities to get $P(X = 2)$ .

\begin{aligned}P(X=2)&=\binom{6}{2}(0.30)^2(0.70)^4\\&=15\times0.021609\\&=0.324135.\end{aligned}

Interpretation: under the model, about $32.4\%$ of complete $6$ -attempt experiments produce exactly $2$ successes in the long run.

This probability concerns the full count $X$ . It is not the chance that $\text{attempt }2$ succeeds, and it is not the $1\text{-trial}$ success probability $p = 0.30$ .

Check the support: a count outside $0\text{ through }6$ has probability $0$ . A single $6$ -attempt experiment cannot have $7$ successes or $2.5$ successes.

05 · Translate the words before using a formula

Tails and intervals

For an integer count, a boundary decides which bars belong to the event. “ $\text{At least }2$ ” $\text{includes }2$ ; “ $\text{more than }2$ ” does not.

Translate the wording for $X \sim \operatorname{Binomial}(6, 0.30)$ before calculating.
Wording	Event	Counts included	Calculation
Exactly $2$	$X = 2$	$2$	Point probability at $2$
At most $2$	$X \le 2$	$0, 1, 2$	Cumulative probability through $2$
Fewer than $2$	$X < 2$	$0, 1$	Cumulative probability through $1$
At least $2$	$X \ge 2$	$2, 3, 4, 5, 6$	$1 – P(X \le 1)$
More than $2$	$X > 2$	$3, 4, 5, 6$	$1 – P(X \le 2)$
Between $1$ and $3$ , inclusive	$1 \le X \le 3$	$1, 2, 3$	$P(X \le 3) – P(X \le 0)$

$\text{At least }2$ : use the complement

$P(X \ge 2)$ includes counts $2, 3, 4, 5,\text{ and }6$ . Its complement contains counts $0$ and $1$ , so:

\begin{aligned}P(X\ge2)&=1-P(X\le1)\\&=1-(0.117649+0.302526)\\&=0.579825.\end{aligned}

Visual guide 4: include the boundary in “

\text{at least }2

”

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The Binomial Distribution

By the end of this lesson, you should be able to:

The concept in 60 seconds

Check the binomial conditions

Two common settings that need a different model