MATH 1401 · Dr. Barry Monk

The Binomial Distribution

Section 6.2

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.

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.

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.

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

Example

Example: Binomial Experiment

Decide if the following scenario describes a binomial experiment:

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 experiment. 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 Experiment

Decide if the following scenario describes a binomial experiment:

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

Solution:

This is not a binomial experiment. 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 Experiment

Decide if the following scenario describes a binomial experiment:

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 experiment. The trials are not 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) A dating app shows your profile to 20 randomly selected users. $X$ = number who swipe right.

Binomial. The number of trials is fixed at 20, each user swipes right or not, the probability of success is the same for randomly selected users, and the decisions 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.

The Binomial Probability Distribution

Consider tossing a biased coin 3 times, where the probability of heads is 0.6. To find $P (2)$ , note that there are 3 arrangements of two heads: HHT, HTH, THH. Each has probability $(0.6) (0.6) (0.4) = {(0.6)}^{2} (0.4)$ .

So, $P (2) = P (HHT or HTH or THH) = 3 {(0.6)}^{2} (0.4)$

3

the number of ways of arranging exactly two heads among the three tosses

{(0.6)}^{2} = p^{x}

the probability of a success, raised to the second power — two is the number of successes wanted

{(0.4)}^{1} = {(1 - p)}^{n - x}

the probability of a failure — the complement of the probability of a success — raised to the first power even though the exponent is not written, because $n - x = 3 - 2 = 1$

For a binomial random variable $X$ with $n$ trials and success probability $p$ :

P (x) = C \cdot p^{x} {(1 - p)}^{n - x}, x = 0, 1, 2, 3, \dots, n

where $C$ is the number of ways of arranging $x$ successes among $n$ trials.

Interactive

Exploring the Shape of the Binomial Distribution

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

Binomial Probabilities on the Statistics Calculator

On the 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 Binomial tool

The Statistics Calculator's tool strip: Summarize, Scatterplot, Regression, Confidence interval, Hypothesis test, Frequency, Binomial, Normal, Sampling, Discrete, Probability, with Binomial selected.

Binomial Probabilities on the TI-84 Plus

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

Example

Example: Part a) (Statistics Calculator)

We have $n = 15$ and $p = 0.3$ .

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

Since we are finding the probability that $x$ equals 4, we choose $P (X = x)$ with $n = 15$ , $p = 0.3$ , and $x = 4$ , .

We find the probability that exactly four people use Pinterest is approximately 0.219.

Result

The form, filled in

The Binomial form filled in: Trials n 15, Success probability p 0.3, Find P(X = x), x 4.

Example

Example: Part b) (Statistics Calculator)

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

The event “fewer than three” is equivalent to “less than or equal to two”.

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

The probability that fewer than three people who use Pinterest is 0.1268.

Result

The form, filled in

The Binomial form filled in: n 15, p 0.3, Find P(X <= x), x 2.

Example

Example: Part b) (TI-84 Plus)

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

The TI-84 Plus display showing binomcdf(15,0.3,2) and its result, .1268277146.

On the TI-84 Plusbinomcdf(15, 0.3, 2)

The binomcdf command computes the probability that there are less than or equal to $x$ successes — two or fewer. Entering 3 would give $P (X \leq 3)$ , which is wrong here.

Example

Example: Part c) (Statistics Calculator)

Find the probability that more than one person uses Pinterest.

We choose $P (X > x)$ with $x = 1$ .

The result is approximately 0.965.

We could also use the Rule of Complements: the complement of “more than 1” is “1 or fewer,” so $P (X > 1) = 1 - P (X \leq 1)$ . Choosing $P (X \leq x)$ with $x = 1$ gives 0.0353, and $1 - 0.0353 = 0.9647$ .

P (X > 1) = 1 - P (X \leq 1) = 1 - 0.0353 = 0.9647

Result

The form, filled in

The Binomial form filled in: n 15, p 0.3, Find P(X > x), x 1.

Example

Example: Part d) (Statistics Calculator)

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

Because $P (Between 1 and 4) = P (4 or less) - P (0)$ , we choose Between a and b with $a = 1$ and $b = 4$ .

The result is approximately 0.511.

Result

The form, filled in

The Binomial form filled in: Trials n 15, Success probability p 0.3, Find Between a and b, and a Region row reading P(1 <= X <= 4).

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.

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.

Mean and Standard Deviation

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

Then the mean of $X$ is

μ_{x} = n p

The standard deviation of $X$ is

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 $X$ .

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

The mean tells us that 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

(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

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The Binomial Distribution

Binomial Distribution

Determine whether a random variable is binomial

Conditions for a Binomial Distribution

Example: Binomial Experiment

Example: Binomial Experiment

Example: Binomial Experiment

Quick Round — Binomial or Not?

Determine the probability distribution of a binomial random variable

The Binomial Probability Distribution

Exploring the Shape of the Binomial Distribution

Compute binomial probabilities

Binomial Probabilities on the Statistics Calculator

Binomial Probabilities on the TI-84 Plus

Example: Binomial Distribution (Statistics Calculator)

Example: Part a) (Statistics Calculator)

Example: Part a) (TI-84 Plus)

Example: Part b) (Statistics Calculator)

Example: Part b) (TI-84 Plus)

Example: Part c) (Statistics Calculator)

Example: Part c) (TI-84 Plus)

Example: Part d) (Statistics Calculator)

Example: Part d) (TI-84 Plus)

Check Your Understanding

Check Your Understanding

Compute the mean and variance of a binomial random variable

Mean and Standard Deviation

Example: Mean and Standard Deviation

Check Your Understanding

Marketing Madness

6.2 HW: The Binomial Distribution