MATH 1401 · Dr. Barry Monk

The Binomial Distribution

Section 6.2

Section 6.1 was about random variables in general. This one is about a particular random variable: the one that counts how many successes you get in a fixed number of tries.

Binomial Distribution

Your favorite restaurant is giving away a coupon with every meal purchase. Twenty percent of the coupons entitle you to a free milkshake; the rest say “better luck next time.”

Ten of you order lunch. What is the probability that exactly three of you win a free milkshake?

This is not straightforward: the answer depends on the number of ways to get three winners from ten.

If we let $X$ be the number of winners out of ten, then $X$ follows what's called the binomial distribution, one of the most useful probability distributions.

A coupon graphic: FREE MILKSHAKE in bold capitals inside a white oval on a dark red field, a $5 VALUE ribbon below, and a line of fine-print joke terms.

Section 6.2 · Determine whether a random variable is binomial

Conditions for a Binomial Distribution

In this example, each coupon examined is a “trial,” so there are 10 trials. A coupon good for a free milkshake is a “success.” The random variable $X$ represents the number of successes in 10 trials. The word success does not necessarily refer to a desirable outcome.

A random variable that counts successes in a series of trials has a binomial distribution if:

The number of trials is fixed.

Each trial has two possible outcomes: “success” or “failure.”

The probability of success is the same on each trial.

The trials are independent. The outcome of one trial does not affect the others.

Notation

The number of trials is denoted by $n$ . The probability of success is denoted by $p$ , and the probability of failure is $1 - p$ .

A hand fans out a dozen brightly colored scratch-off cards printed WINNER along their tops, held over a cafe table.

Example

Example: Binomial Random Variable

Determine whether $X$ is a binomial random variable:

A quiz has 8 true/false questions. A student guesses randomly on each one. Let $X$ be the number answered correctly.

Solution:

This is a binomial random variable. Each question is a trial with two outcomes, the probability of success is 0.5 for each, and the trials are independent.

A hand holds a pencil over a printed answer sheet with rows of empty check boxes; an eraser and a desk lamp sit beside it.

Example

Example: Binomial Random Variable

Determine whether $X$ is a binomial random variable:

Five basketball players each attempt a free throw. Let $X$ be the number made.

Solution:

This is not a binomial random variable. The probability of success differs from player to player.

Five basketball players in different uniforms stand in a row on a gym court, each holding a ball at the set point of a free throw.

Example

Example: Binomial Random Variable

Determine whether $X$ is a binomial random variable:

A bowl contains 10 notes: 5 for indoor activities, 5 for outdoor. A team draws three notes without replacement. Let $X$ be the number of outdoor activities drawn.

Solution

This is not a binomial random variable. The trials are not independent.

When a simple random sample comprises less than 5% of the population, we will consider the sampled individuals to be independent.

A hand reaches into a glass bowl of folded green and orange paper notes on a wooden table; two unfolded notes lie beside it.

Check your understanding

Quick Round: Binomial or Not?

a) About 52% of never-married U.S. adults have used a dating app or site. A simple random sample of 20 never-married adults is selected. $X$ = number who have used one.

Binomial. The number of trials is fixed at 20, each adult has used one or not, the probability of success is 0.52 for each, and since the sample is less than 5% of the population, the trials are independent.

Two hands scroll a phone showing a grid of short videos, string lights glowing behind.

b) You open TikToks until you find one you like. $X$ = number opened.

Not binomial. The number of trials is not fixed: you keep going until you find one you like.

Five friends sit close together on a couch, each looking down at a phone, mugs on the table in front of them.

c) Five friends each play a different Wordle puzzle. $X$ = number who solve it.

Not binomial. The probability of success differs from puzzle to puzzle.

Section 6.2 · Determine the probability distribution of a binomial random variable

The Binomial Probability Distribution

A biased coin has probability 0.6 of coming up heads. The coin is tossed three times. Let $X$ be the number of heads that come up. Find $P (2)$ , the probability that exactly two of the tosses are heads.

Solution

Since $X$ is the number of heads, coming up heads is a success. Then $X$ is binomial, with $n = 3$ trials and success probability $p = 0.6$ .

There are three arrangements of two heads in three tosses: HHT, HTH, and THH. Each has probability $(0.6) (0.6) (0.4) = {(0.6)}^{2} {(0.4)}^{1}$ .

The tosses are independent, so we multiply: the Multiplication Rule for Independent Events.

P (2) = P (HHT or HTH or THH) = 3 {(0.6)}^{2} {(0.4)}^{1}

The three arrangements are mutually exclusive, so we add: the Addition Rule for Mutually Exclusive Events.

Formula for Binomial Probabilities

P (2) = 3 {(0.6)}^{2} {(0.4)}^{1}

3 is the number of arrangements of two successes (heads) and one failure (tails). In general, it is the number of ways of arranging $x$ successes among $n$ trials.

0.6 is the success probability, which in general is $p$ . The exponent 2 is the number of successes, which in general is $x$ .

0.4 is the failure probability, which is $1 - p$ , and the exponent 1 is the number of failures, which is $n - x$ .

For a binomial random variable

X

that represents the number of successes in

n

trials with success probability

p

, the probability of obtaining

x

successes is

P (x) = C \cdot p^{x} {(1 - p)}^{n - x}

where $C$ is the number of ways of arranging $x$ successes among $n$ trials. The possible values of $X$ are $0, 1, \dots, n$ .

Interactive

Exploring the Shape of the Binomial Distribution

The shape of the binomial distribution depends on both $n$ and $p$ .

When $p$ is small, the probabilities pile up on the left: the distribution is skewed to the right.

When $p$ is large, they pile up on the right: skewed to the left.

As $n$ grows, for many values of $p$ the shape starts to look bell-shaped.

Section 6.2 · Compute binomial probabilities

Computing Binomial Probabilities

The binomial probability distribution can require tedious calculations. While we can compute simple probabilities by hand, for more involved problems it is better to use technology.

Statistics Calculator

The Binomial tool computes binomial probabilities. Enter the number of trials $n$ , the success probability $p$ , choose the form of the probability to find, and enter $x$ .

https://barrymonk.com/stats-calculator/ The Statistics Calculator's tool buttons, grouped as Describe data, Probability and Inference, with Binomial selected.

The Statistics Calculator's tool buttons, grouped as Describe data, Probability and Inference, with Binomial selected.

TI-84 Plus

Two commands compute binomial probabilities: binompdf and binomcdf. Both are on the DISTR menu (2nd → VARS).

Use binompdf to find the probability that $X$ equals a specific value. Use binomcdf to find the probability that $X$ is less than or equal to a specific value. The c stands for cumulative.

The TI-84 Plus DISTR menu, listing binompdf( as item A and binomcdf( as item B.

Example

Back to the Milkshakes

Ten of you order lunch, and 20% of the coupons win a free milkshake. What is the probability that exactly three of you win a free milkshake?

Solution

Each lunch is a trial and a winning coupon is a success, so $n = 10$ and $p = 0.2$ . We want exactly three, $P (X = 3)$ .

Statistics Calculator

Choose $P (X = x)$ with $x = 3$ .

The Binomial tool with n = 10, p = 0.2, P(X = x) and x = 3; the result is 0.2013.

TI-84 Plus

binompdf(10, 0.2, 3)

The probability that exactly three of the ten win a free milkshake is approximately 0.201.

Example

Example: Binomial Probabilities

Approximately 30% of people use the image sharing website Pinterest. A simple random sample of 15 people is taken. Use the binomial probability distribution to find the following probabilities.

(a)

Find the probability that exactly four of the sampled people use Pinterest.

(b)

Find the probability that fewer than three of the people use Pinterest.

(c)

Find the probability that more than one person uses Pinterest.

(d)

Find the probability that the number of people who use Pinterest is between 1 and 4, inclusive.

Solution

We use the binomial probability distribution with $n = 15$ and $p = 0.3$ .

People at cafe tables working on laptops and phones, seen from above in a bright, high-windowed room.

Example

Example: Part a)

Recall

$n = 15$

$p = 0.3$

(a) Find the probability that exactly four of the sampled people use Pinterest.

Solution

We want exactly four: $P (X = 4)$ .

Statistics Calculator

Choose $P (X = x)$ with $x = 4$ .

The Binomial tool with n = 15, p = 0.3, P(X = x), x = 4; the result is 0.2186.

TI-84 Plus

binompdf(15, 0.3, 4)

The probability that exactly four people use Pinterest is approximately 0.219.

Example

Example: Part b)

Recall

$n = 15$

$p = 0.3$

(b) Find the probability that fewer than three of the people use Pinterest.

Solution

The possible numbers of people that are fewer than three are 0, 1, and 2. So “fewer than three” is $P (X \leq 2)$ .

Statistics Calculator

Choose $P (X \leq x)$ with $x = 2$ .

The Binomial tool with n = 15, p = 0.3, P(X ≤ x), x = 2; the result is 0.1268.

TI-84 Plus

binomcdf(15, 0.3, 2)

Entering 3 would give $P (X \leq 3)$ , which is wrong here.

The probability that fewer than three people use Pinterest is approximately 0.127.

Example

Example: Part c)

Recall

$n = 15$

$p = 0.3$

Solution

The complement of “more than 1” is “1 or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ .

Statistics Calculator

The calculator finds it directly: choose $P (X > x)$ with $x = 1$ .

The Binomial tool with n = 15, p = 0.3, P(X > x), x = 1; the result is 0.9647.

TI-84 Plus

binomcdf(15, 0.3, 1)1 − Ans

The probability that more than one person uses Pinterest is approximately 0.965.

Example

Example: Part d)

Recall

$n = 15$

$p = 0.3$

(d) Find the probability that the number of people who use Pinterest is between 1 and 4, inclusive.

Solution

Between 1 and 4 inclusive means 1, 2, 3, or 4. So $P (1 \leq X \leq 4) = P (X \leq 4) - P (0)$ .

Statistics Calculator

Choose Between $a$ and $b$ with $a = 1$ and $b = 4$ .

The Binomial tool with n = 15, p = 0.3, Between a and b, 1 ≤ X ≤ 4; the result is 0.5107.

TI-84 Plus

binomcdf(15, 0.30, 4) −binompdf(15, 0.30, 0)

The probability that between 1 and 4 people, inclusive, use Pinterest is approximately 0.511.

Check your understanding

Check Your Understanding

An environmentalist plants three types of trees, each with probability 0.4 of adapting to the local climate.

(a)

Find the probability that none of these successfully adapt to the environment.

$n = 3, p = 0.4, x = 0$ : $P (X = 0) = 0.216$

(b)

Find the probability that at least two of the trees adapt to the local climate.

$P (X \geq 2) = P (2) + P (3) = 0.352$

A seedling with two leaves, its roots exposed below the soil line, against a white background.

Check your understanding

Check Your Understanding

A 2017 government investigation found that airport screeners failed to detect prohibited items approximately 80% of the time. If someone attempts to sneak 3 prohibited items through security, find the probability that at least one is caught.

Let a success be an item that is caught: $n = 3$ , $p = 0.2$ .

P (at least one caught) = 1 - P (none caught)

= 1 - 0.512 = 0.488

An airport security checkpoint: travelers wait with bags at an X-ray scanner while an officer watches a monitor.

Section 6.2 · Compute the mean and standard deviation of a binomial random variable

Mean and Standard Deviation

Let $X$ be a binomial random variable with $n$ trials and success probability $p$ .

If we toss a fair coin 10 times, we expect to get 5 heads, on the average: $5 = 10 \cdot 0.5$ .

In Section 6.1, the mean was $μ_{X} = Σ [x \cdot P (x)]$ .

The mean of

X

μ_{X} = n p

The standard deviation of

X

σ_{X} = \sqrt{n p (1 - p)}

A carnival ring-toss booth: colored rings scattered over a pegboard and the ground, stuffed-animal prizes hanging above.

Example

Example: Mean and Standard Deviation

The probability that a new car of a certain model will require repairs during the warranty period is 0.15. A dealership sells 80 such cars. Let $X$ be the number that will require repairs during the warranty period.

Find the mean and standard deviation of $X$ .

Solution

There are $n = 80$ trials, with success probability $p = 0.15$ . The mean is $μ_{X} = n p = 80 (0.15) = 12$ .

The standard deviation is

σ_{X} = \sqrt{n p (1 - p)} = \sqrt{(80) (0.15) (0.85)} = 3.194

In the long run, 12 out of every 80 cars will require repairs.

The standard deviation measures the spread in the number of cars requiring repairs.

Rows of new cars parked outside a dealership service center at dusk; a man with a clipboard walks between them.

Check your understanding

Check Your Understanding

Gregor Mendel studied heredity in pea plants. When parent plants carry genes for both green and yellow pods, offspring have a 0.75 probability of green pods. Assume 80 such plants are produced.

(a)

Find the mean number of plants with green pods.

$μ_{X} = n p = 80 (0.75) = 60$ . In the long run, 60 out of every 80 plants have green pods.

(b)

Find the standard deviation.

$σ_{X} = \sqrt{n p (1 - p)} = \sqrt{(80) (0.75) (0.25)} = \sqrt{15} \approx 3.873$

Close-up of bright green peas inside an open pod.

Case study

Marketing Madness

Imagine that you work in the advertising department of a fast-growing social media tech firm. Your team has been tasked with an ad campaign with a budget of $10,000.

Each ad costs $100. A better ad costs more to make: the creative team on it, research into what this audience responds to, a stronger image. That work raises the chance somebody buys. Increasing the probability of a sale by 1% (or 0.01) costs $500.

A cartoon of a young man at a laptop marked AD, beside a poster reading AD CAMPAIGN with a rising bar chart, and social-media icons around him.

More ads, or better ads? Which way of spending the $10,000 brings the most sales?

Here are three ways to spend all of it.

Plan	Ads, $n$	Sale probability, $p$	Spent on ads	Spent on sale probability	Total
Reach	90	0.02	$9,000	$1,000	$10,000
Balance	50	0.10	$5,000	$5,000	$10,000
Premium	25	0.15	$2,500	$7,500	$10,000

Case study

Expected Sales for Each Plan

Each ad is one trial and a sale is a success. The number of trials is fixed at $n$ , each ad has the same probability of a sale, $p$ , and the ads are independent. So for each plan, the number of sales $X$ is binomial.

Reach: $μ_{X} = n p = 90 (0.02) = 1.80$

Balance: $μ_{X} = n p = 50 (0.10) = 5.00$

Premium: $μ_{X} = n p = 25 (0.15) = 3.75$

Balance gives the most expected sales.

Premium buys the best ads and still comes out behind, because every dollar spent on sale probability is a dollar not spent on ads.

largest: 5.00 at

p = 0.10

Sale probability,

p

Expected sales for every split of the budget

Case study

What Does 5.00 Expected Sales Mean?

The Balance plan expects 5.00 sales. No single campaign makes exactly 5.00 sales. What does the number mean?

In the long run, the average number of sales per campaign is 5.00.

μ_{X} = 5.00

No campaigns run yet.

Early on, with only a handful of campaigns, the average swings around a good deal. As the number of campaigns grows, it settles down onto the mean.

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The Binomial Distribution

Binomial Distribution

Conditions for a Binomial Distribution

Example: Binomial Random Variable

Example: Binomial Random Variable

Example: Binomial Random Variable

Quick Round: Binomial or Not?

The Binomial Probability Distribution

Formula for Binomial Probabilities

Exploring the Shape of the Binomial Distribution

Computing Binomial Probabilities

Back to the Milkshakes

Example: Binomial Probabilities

Example: Part a)

Example: Part b)

Example: Part c)

Example: Part d)

Check Your Understanding

Check Your Understanding

Mean and Standard Deviation

Example: Mean and Standard Deviation

Check Your Understanding

Marketing Madness

Expected Sales for Each Plan

What Does 5.00 Expected Sales Mean?

6.2 HW: The Binomial Distribution