Binomial Distribution (Optional), Discrete Random Variables | Statistics

Learning objectives

There are three characteristics of a binomial experiment:

There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials.
There are only two possible outcomes, called success and failure, for each trial. The outcome that we are measuring is defined as a success, while the other outcome is defined as a failure. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1.
The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. Another way of saying this is that for each individual trial, the probability, p, of a success and probability, q, of a failure remain the same. Let us look at several examples of a binomial experiment.
Example 1: Toss a fair coin once and record the result.
This is a binomial experiment since it meets all three characteristics. The number of trials n = 1. There are only two outcomes, a head or a tail, of each trial. We can define a head as a success if we are measuring number of heads. For a fair coin, the probabilities of getting head or tail are both .5. So, p = q − .5. Both p and q remain the same from trial to trial. This experiment is also called a Bernoulli trial, named after Jacob Bernoulli who, in the late 1600s, studied such trials extensively. Any experiment that has characteristics two and three and where n = 1 is called a Bernoulli trial. A binomial experiment takes place when the number of successes is counted in one or more Bernoulli trials.
Example 2: Randomly guess a multiple choice question has A, B, C and D four options.
This is a binomial experiment since it meets all three characteristics. The number of trials n = 1. There are only two outcomes, guess correctly or guess wrong, of each trial. We can define guess correctly as a success. For a random guess (you have no clue at all), the probability of guessing correct should be $\frac{1}{4}$ because there are four options and only one option is correct. So, and $p = \frac{1}{4}$ and $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$ . Both p and q remain the same from trial to trial. This experiment is also a Bernoulli trial. It meets the characteristics two and three and n = 1.
Example 3: Toss a fair coin five times and record the result.
This is a binomial experiment since it meets all three characteristics. The number of trials n = 5. There are only two outcomes, head or tail, of each trial. If we define head as a success, then p = q = 0.5. Both p and q remain the same for each trial. Since n = 5, this experiment is not a Bernoulli trial although it meets the characteristics two and three.
Example 4: Randomly guess 10 multiple choice questions in an exam. Each question has A, B, C and D four options.
This is a binomial experiment since it meets all three characteristics. The number of trials n = 10. There are only two outcomes, guess correctly or guess wrong, of each trial. We can define guess correctly as a success. As we explained in example 2, $p = \frac{1}{4}$ and $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$ . Both p and q remain the same for each guess. Since n = 10, this experiment is not a Bernoulli trial.
The next two experiments are not binomial experiments.
Example 5: Randomly select two balls from a jar with five red balls and five blue balls without replacement. This means we select the first ball, and then without returning the selected ball into the jar, we will select the second ball.
This is not a binomial experiment since the third characteristic is not met. The number of trials n = 2. There are only two outcomes, a red ball or a blue ball, of each trial. If we define selecting a red ball as a success, then selecting a blue ball is a failure. The probability of getting the first ball red is $\frac{5}{10}$ since there are five red balls out of 10 balls. So, $p = \frac{5}{10}$ and $q = 1 - p = 1 - \frac{5}{10} = \frac{5}{10}$ . However, p and q do not remain the same for the second trial. If the first ball selected is red, then the probability of getting the second ball red is $\frac{4}{9}$ since there are only four red balls out of nine balls. But if the first ball selected is blue, then the probability of getting the second ball red is $\frac{5}{9}$ since there are still five red balls out of nine balls.
Example 6: Toss a fair coin until a head appears.
This is not a binomial experiment since the first characteristic is not met. The number of trials n is not fixed. n could be 1 if a head appears from the first toss. n could be 2 if the first toss is a tail and the second toss is a head. So on and so forth.
More examples of binomial and non-binomial experiments will be discussed in this section later.

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials.

There are shortcut formulas for calculating mean μ, variance σ², and standard deviation σ of a binomial probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book.

μ = n p, σ^{2} = n p q, σ = \sqrt{n p q} .

Here n is the number of trials, p is the probability of a success, and q is the probability of a failure.

At ABC High School, the withdrawal rate from an elementary physics course is 30 percent for any given term. This implies that, for any given term, 70 percent of the students stay in the class for the entire term. The random variable X = the number of students who withdraw from the randomly selected elementary physics class. Since we are measuring the number of students who withdrew, a success is defined as an individual who withdrew.

Suppose you play a game that you can only either win or lose. The probability that you win any game is 55 percent, and the probability that you lose is 45 percent. Each game you play is independent. If you play the game 20 times, write the function that describes the probability that you win 15 of the 20 times. Here, if you define X as the number of wins, then X takes on the values 0, 1, 2, 3, . . ., 20. The probability of a success is p = 0.55. The probability of a failure is q = .45. The number of trials is n = 20. The probability question can be stated mathematically as P(x = 15). If you define X as the number of losses, then a success is defined as a loss and a failure is defined as a win. A success does not necessarily represent a good outcome. It is simply the outcome that you are measuring. X still takes on the values of 0, 1, 2, 3, . . ., 20. The probability of a success is $p = .45$ . The probability of a failure is $q = .55$ .

A fair coin is flipped 15 times. Each flip is independent. What is the probability of getting more than 10 heads? Let X = the number of heads in 15 flips of the fair coin. X takes on the values 0, 1, 2, 3, . . ., 15. Since the coin is fair, p = .5 and q = .5. The number of trials n = 15. State the probability question mathematically.

P(x > 10)

Approximately 70 percent of statistics students do their homework in time for it to be collected and graded. Each student does homework independently. In a statistics class of 50 students, what is the probability that at least 40 will do their homework on time? Students are selected randomly.

a. This is a binomial problem because there is only a success or a ________, there are a fixed number of trials, and the probability of a success is .70 for each trial.

a. failure

b. If we are interested in the number of students who do their homework on time, then how do we define X?

b. X = the number of statistics students who do their homework on time

c. What values does x take on?

c. 0, 1, 2, . . ., 50

d. What is a failure, in words?

d. Failure is defined as a student who does not complete his or her homework on time.

The probability of a success is p = .70. The number of trials is n = 50.

e. If p + q = 1, then what is q?

e. q = .30

f. The words at least translate as what kind of inequality for the probability question P(x ____ 40)?

f. greater than or equal to (≥)
The probability question is P(x ≥ 40).

Notation for the Binomial: B = Binomial Probability Distribution Function

X ~ B(n, p)

Read this as X is a random variable with a binomial distribution. The parameters are n and p: n = number of trials, p = probability of a success on each trial.

It has been stated that about 41 percent of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?

Let X = the number of workers who have a high school diploma but do not pursue any further education.

X takes on the values 0, 1, 2, . . ., 20 where n = 20, p = .41, and q = 1 – .41 = .59. X ~ B(20, .41)

Find P(x ≤ 12). There is a formula to define the probability of a binomial distribution P(x). We can use the formula to find $P (x \leq 12)$ . But the calculation is tedious and time consuming, and people usually use a graphing calculator, software, or binomial table to get the answer. Use a graphing calculator, you can get $P (x \leq 12) = .9738$ . The instruction of TI-83, 83+, 84, 84+ is given below.

Go into 2^nd DISTR. The syntax for the instructions are as follows:

To calculate the probability of a value $P (x = v a l u e)$ : use binompdf(n, p, number). Here binompdf represents binomial probability density function. It is used to find the probability that a binomial random variable is equal to an exact value. n is the number of trials, p is the probability of a success, and number is the value. If number is left out, which means use binompdf(n, p), then all the probabilities $P (x = 0), P (x = 1), \dots, P (x = n)$ will be calculated.
To calculate the cumulative probability $P (x \leq v a l u e)$ : use binomcdf(n, p, number). Here binomcdf represents binomial cumulative distribution function. It is used to determine the probability of at most type of problem, the probability that a binomial random variable is less than or equal to a value. n is the number of trials, p is the probability of a success, and number is the value. If number is left out, all the cumulative probabilities $P (x \leq 0), P (x \leq 1), \dots, P (x \leq n)$ will be calculated.
To calculate the cumulative probability $P (x \geq v a l u e)$ : use 1 - binomcdf(n, p, number). n is the number of trials, p is the probability of a success, and number is the value. TI calculators do not have a built-in function to find the probability that a binomial random variable is greater than a value. However, we can use the fact that $P (x > v a l u e) = 1 - P (x \leq v a l u e)$ to find the answer.
For this problem: After you are in 2^nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,.41,12). The result is P(x ≤ 12) = .9738.

The probability that at most 12 workers have a high school diploma but do not pursue any further education is .9738.

The graph of X ~ B(20, .41) is as follows.

View original figure (external; image not reproduced in this text edition)

Original figure description: This is a probability distribution histogram, graphing X ~ B(20, .41)

The previous graph is called a probability distribution histogram. It is made of a series of vertical bars. The x-axis of each bar is the value of X = the number of workers who have only a high school diploma, and the height of that bar is the probability of that value occurring.

The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, μ = np = (20)(.41) = 8.2.

The formula for the variance is σ² = npq. The standard deviation is σ = $\sqrt{n p q}$ .
σ = $\sqrt{(20) (.41) (.59)}$ = 2.20.

The following is the interpretation of the mean $μ = 8.2$ and standard deviation $σ = 2.20$ :

If you randomly select 20 adult workers, and do that over and over, you expect around eight adult workers out of 20 to have a high school diploma but do not pursue any further education on average. And you expect that to vary by about two workers on average.

A store releases a 560-page art supply catalog. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists.

What values does x take on?
What is the probability distribution? Find the following probabilities:
1. the probability that two pages feature signature artists
2. the probability that at most six pages feature signature artists
3. the probability that more than three pages feature signature artists
Using the formulas, calculate the (i) mean and (ii) standard deviation.

x = 0, 1, 2, 3, 4, 5, 6, 7, 8
This is a binomial experiment since all three characteristics are met. Each page is a trial. Since we sample 100 pages, the number of trials is n = 100. For each page, there are two possible outcomes, features signature artists or does not feature signature artists. Since we are measuring the number of pages that feature signature artists, a page that features signature artists is defined as a success and a page that does not feature signature artists is defined as a failure. There are 8 out of 560 pages that feature signature artists. Therefore the probability of a success p= 8 560 and the probability of a failure q=1−p=1− 8 560 = 552 560 .
Both p and q remain the same for each page. Therefore, X is a binomial random variable, and it can be written as X~B( 100, 8 560 ) .
We can use a graphing calculator to answer Parts i to iii.
1. P(x = 2) = binompdf $(100, \frac{8}{560}, 2)$ = .2466
2. P(x ≤ 6) = binomcdf $(100, \frac{8}{560}, 6)$ = .9994
3. P(x > 3) = 1 – P(x ≤ 3) = 1 – binomcdf $(100, \frac{8}{560}, 3)$ = 1 – .9443 = .0557
1. mean = np = (100) $(\frac{8}{560})$ = $\frac{800}{560}$ ≈ 1.4286
2. standard deviation = $\sqrt{n p q}$ = $\sqrt{(100) (\frac{8}{560}) (\frac{552}{560})}$ ≈ 1.1867

Try It

According to a poll, 60 percent of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending.

What is the probability distribution for X?
Use your calculator to find the following probabilities:
1. The probability that 25 adults in the sample prefer saving over spending
2. The probability that at most 20 adults prefer saving
3. The probability that more than 30 adults prefer saving
Using the formulas, calculate the (i) mean and (ii) standard deviation of X.

X ∼ B(50, 0.6)
Using the TI-83, 83+, 84 calculator with instructions as provided in example5:
1. P(x = 25) = binompdf(50, 0.6, 25) = 0.0405
2. P(x ≤ 20) = binomcdf(50, 0.6, 20) = 0.0034
3. P(x > 30) = 1 - binomcdf(50, 0.6, 30) = 1 – 0.5535 = 0.4465
1. mean = np = 50(0.6) = 30
2. standard deviation = $\sqrt{n p q}$ = $\sqrt{50 (0.6) (0.4)}$ ≈ 3.4641

The lifetime risk of developing a specific disease is about 1 in 78 (1.28 percent). Suppose we randomly sample 200 people. Let X = the number of people who will develop the disease.

What is the probability distribution for X?
Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
Use your calculator to find the probability that at most eight people develop the disease.
Is it more likely that five or six people will develop the disease? Justify your answer numerically.

This is a binomial experiment since all three characteristics are met. Each person is a trial. Since we sample 200 people, the number of trials is n = 200. For each person, there are two possible outcomes: will develop the disease or not. Since we are measuring the number of people who will develop the disease, a person who will develop the disease is defined as a success and a person who will not develop the disease is defined as a failure. The risk of developing the disease is 1.28 percent. Therefore the probability of a success, $p = 1.28$ percent, $.0128$ , and the probability of a failure, $q = 1 - p = 1 - .0128 = .9872$ . Both p and q remain the same for each person. Therefore, X is a binomial random variable and it can be written as $X ~ B (200, .0128)$ .
We can use a graphing calculator to answer Questions c and d.
1. Mean = np = 200(.0128) = 2.56
2. Standard Deviation = $\sqrt{n p q} = \sqrt{(200)(0 .128)(.9872)} \approx 1. 5897$
Using the TI-83, 83+, 84 calculator with instructions as provided in example5:
P(x ≤ 8) = binomcdf(200, .0128, 8) = .9988
P(x = 5) = binompdf(200, .0128, 5) = .0707
P(x = 6) = binompdf(200, .0128, 6) = .0298
So P(x = 5) > P(x = 6); it is more likely that five people will develop the disease than six.

Try It

During the 2013 regular basketball season, a player had the highest field goal completion rate in the league. This player scored with 61.3 percent of his shots. Suppose you choose a random sample of 80 shots made by this player during the 2013 season. Let X = the number of shots that scored points.

What is the probability distribution for X?
Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
Use your calculator to find the probability that this player scored with 60 of these shots.
Find the probability that this player scored with more than 50 of these shots.

X ~ B(80, 0.613)
1. Mean = np = 80(0.613) = 49.04
2. Standard Deviation = $\sqrt{n p q} = \sqrt{80 (0.613) (0.387)} \approx 4.3564$
Using the TI-83, 83+, 84 calculator with instructions as provided in example5:
P(x = 60) = binompdf(80, 0.613, 60) = 0.0036
P(x > 50) = 1 – P(x ≤ 50) = 1 – binomcdf(80, 0.613, 50) = 1 – 0.6282 = 0.3718

The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC High School has a student advisory committee made up of 10 staff members and six students. The committee wishes to choose a chairperson and a recorder. What is the probability that the chairperson and recorder are both students? The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is $\frac{6}{16}$ because there are six students out of 16 members (10 staff members + six students). If the first draw selects a student, then the probability of a student on the second draw is $\frac{5}{16}$ because there are only five students out of 15 members. If the first draw selects a staff member, then the probability of a student on the second draw is $\frac{6}{15}$ because there are still six students out of 15 members. The probability of drawing a student's name changes for each of the trials and, therefore, violates the condition of independence.

References

American Cancer Society. (2013). What are the key statistics about pancreatic cancer? Retrieved from http://www.cancer.org/cancer/pancreaticcancer/detailedguide/pancreatic-cancer-key-statistics

Central Intelligence Agency. (n.d.). The world factbook. Retrieved from https://www.cia.gov/library/publications/theworld-factbook/geos/af.html

ESPN NBA. (2013). NBA statistics 2013. Retrieved from http://espn.go.com/nba/statistics/_/seasontype/2

Newport, F. (2013). Americans still enjoy saving rather than spending: Few demographic differences seen in these views other than by income. GALLUP Economy. Retrieved from http://www.gallup.com/poll/162368/americansenjoy-saving-rather-spending.aspx

Pryor, J. H., et al. (2011). The American freshman: National norms fall 2011. Los Angeles, CA: Cooperative Institutional Research Program, Higher Education Research Institute. Retrieved from http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf

Wikipedia. (n.d.). Distance education. Retrieved from http://en.wikipedia.org/wiki/Distance_education

World Bank Group. (2013). Access to electricity (% of population). Retrieved from http://data.worldbank.org/indicator/EG.ELC.ACCS.ZS?order=wbapi_data_value_2009%20wbapi_data_value%20wbapi_data_value-first&sort=asc

Chapter Review

A statistical experiment can be classified as a binomial experiment if the following conditions are met:

There are a fixed number of trials, n
There are only two possible outcomes, called success and failure, for each trial; the letter p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial
The n trials are independent and are repeated using identical conditions

The outcomes of a binomial experiment fit a binomial probability distribution. The random variable X = the number of successes obtained in the n independent trials. The mean of X can be calculated using the formula μ = np, and the standard deviation is given by the formula σ = $\sqrt{n p q}$ .

Formula Review

X ~ B(n, p) means that the discrete random variable X has a binomial probability distribution with n trials and probability of success p.

X = the number of successes in n independent trials

n = the number of independent trials

X takes on the values x = 0, 1, 2, 3, . . . , n

p = the probability of a success for any trial

q = the probability of a failure for any trial

p + q = 1

q = 1 – p

The mean of X is μ = np. The standard deviation of X is σ = $\sqrt{n p q}$ .

Use the following information to answer the next eight exercises: Researchers collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the United States. Of those students, 71.3 percent replied that, yes, they agreed with a recent federal law that was passed.

Suppose that you randomly pick eight first-time, full-time freshmen from the survey. You are interested in the number who agreed with that law.

In words, define the random variable X.

X = the number that reply yes

X ~ _____(_____,_____)

What values does the random variable X take on?

0, 1, 2, 3, 4, 5, 6, 7, 8

Construct the probability distribution function (PDF).

x	P(x)

On average (μ), how many would you expect to answer yes?

5.7

What is the standard deviation (σ)?

What is the probability that at most five of the freshmen reply yes?

.4151

What is the probability that at least two of the freshmen reply yes?

HOMEWORK

According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery.

Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf.

Use the following information to answer the next four exercises: Recently, a nurse commented that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 4 percent. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu.

Define the random variable and list its possible values.

X = the number of patients calling in claiming to have the flu, who actually have the flu.

X = 0, 1, 2, ...25

State the distribution of X.

Find the probability that at least four of the 25 patients actually have the flu.

.0165

On average, for every 25 patients calling in, how many do you expect to have the flu?

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video to Go is given M04_Ch04_tbl005. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.

x	P(x)
0	.03
1	.50
2	.24
3
4	.07
5	.04

Describe the random variable X in words.
Find the probability that a customer rents three DVDs.
Find the probability that a customer rents at least four DVDs.
Find the probability that a customer rents at most two DVDs.

X = the number of DVDs a Video to Go customer rents
.12
.11
.77

A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18 percent of students attend Tet festivities. We are interested in the number of students who will attend the festivities.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
How many of the 12 students do we expect to attend the festivities?
Find the probability that at most four students will attend.
Find the probability that more than two students will attend.

Use the following information to answer the next three exercises: The probability that a local hockey team will win any given game is 0.3694 based on a 13-year win history of 382 wins out of 1,034 games played (as of a certain date). An upcoming monthly schedule contains 12 games.

What is the expected number of wins for that upcoming month?

1.67
12
$\frac{382}{1043}$
4.43

d. 4.43

Let X = the number of games won in that upcoming month.

What is the probability that the team wins six games in that upcoming month?

.1476
.2336
.7664
.8903

What is the probability that the team wins at least five games in that upcoming month

.3694
.5266
.4734
.2305

A student takes a 10-question true-false quiz, but did not study and randomly guesses each answer. Find the probability that the student passes the quiz with a grade of at least 70 percent of the questions correct.

A student takes a 32-question multiple choice exam, but did not study and randomly guesses each answer. Each question has three possible choices for the answer. Find the probability that the student guesses more than 75 percent of the questions correctly.

X = number of questions answered correctly
X ~ B $(32, \frac{1}{3})$
We are interested in MORE THAN 75 percent of 32 questions correct. 75 percent of 32 is 24. We want to find P(x > 24). The event more than 24 is the complement of less than or equal to 24.
Using your calculator's distribution menu: 1 – binomcdf $(32, \frac{1}{3}, 24)$
P(x > 24) = 0
The probability of getting more than 75 percent of the 32 questions correct when randomly guessing is very small and practically zero.

Six different colored dice are rolled. Of interest is the number of dice that show a one.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
On average, how many dice would you expect to show a one?
Find the probability that all six dice show a one.
Is it more likely that three or that four dice will show a one? Use numbers to justify your answer numerically.

More than 96 percent of the very largest colleges and universities (more than 15,000 total enrollments) have some online offerings. Suppose you randomly pick 13 such institutions. We are interested in the number that offer distance learning courses.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
On average, how many schools would you expect to offer such courses?
Find the probability that at most 10 offer such courses.
Is it more likely that 12 or that 13 will offer such courses? Use numbers to justify your answer numerically and answer in a complete sentence.

X = the number of college and universities that offer online offerings.
0, 1, 2, …, 13
X ~ B(13, 0.96)
12.48
.0135
P(x = 12) = .3186 P(x = 13) = 0.5882 More likely to get 13.

Suppose that about 85 percent of graduating students attend their graduation. A group of 22 graduating students is randomly chosen.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
How many are expected to attend their graduation?
Find the probability that 17 or 18 attend.
Based on numerical values, would you be surprised if all 22 attended graduation? Justify your answer numerically.

At the Fencing Center, 60 percent of the fencers use the foil as their main weapon. We randomly survey 25 fencers at the Fencing Center. We are interested in the number of fencers who do not use the foil as their main weapon.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
How many are expected to not to use the foil as their main weapon?
Find the probability that six do not use the foil as their main weapon.
Based on numerical values, would you be surprised if all 25 did not use foil as their main weapon? Justify your answer numerically.

X = the number of fencers who do not use the foil as their main weapon
0, 1, 2, 3,... 25
X ~ B(25,.40)
10
.0442
The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

Approximately 8 percent of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number who participated in after-school sports all four years of high school.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
How many seniors are expected to have participated in after-school sports all four years of high school?
Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
Based upon numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.

The chance of an IRS audit for a tax return reporting more than $25,000 in income is about 2 percent per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
How many audits are expected in a 20-year period?
Find the probability that a person is not audited at all.
Find the probability that a person is audited more than twice.

X = the number of audits in a 20-year period
0, 1, 2, …, 20
X ~ B(20, .02)
.4
.6676
.0071

It has been estimated that only about 30 percent of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
What is the probability that at least eight have adequate earthquake supplies?
Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why?
How many residents do you expect will have adequate earthquake supplies?

There are two similar games played for Chinese New Year and Vietnamese New Year. In the Chinese version, fair dice with numbers 1, 2, 3, 4, 5, and 6 are used, along with a board with those numbers. In the Vietnamese version, fair dice with pictures of a gourd, fish, rooster, crab, crayfish, and deer are used. The board has those six objects on it, also. We will play with bets being $1. The player places a bet on a number or object. The house rolls three dice. If none of the dice show the number or object that was bet, the house keeps the $1 bet. If one of the dice shows the number or object bet (and the other two do not show it), the player gets back his or her $1 bet, plus $1 profit. If two of the dice show the number or object bet (and the third die does not show it), the player gets back his or her $1 bet, plus $2 profit. If all three dice show the number or object bet, the player gets back his or her $1 bet, plus $3 profit. Let X = number of matches and Y = profit per game.

In words, define the random variable X.
List the values that X may take on.
Give the distribution of X. X ~ _____(_____,_____)
List the values that Y may take on. Then, construct one PDF table that includes both X and Y and their probabilities.
Calculate the average expected matches over the long run of playing this game for the player.
Calculate the average expected earnings over the long run of playing this game for the player.
Determine who has the advantage, the player or the house.

X = the number of matches
0, 1, 2, 3
X ~ B $(3, \frac{1}{6})$
In dollars: −1, 1, 2, 3
$\frac{1}{2}$
Multiply each Y value by the corresponding X probability from the PDF table. The answer is −.0787. You lose about eight cents, on average, per game.
The house has the advantage.

According to the World Bank, only 9 percent of the population of Uganda had access to electricity as of 2009. Suppose we randomly sample 150 people in Uganda. Let X = the number of people who have access to electricity.

What is the probability distribution for X?
Using the formulas, calculate the mean and standard deviation of X.
Use your calculator to find the probability that 15 people in the sample have access to electricity.
Find the probability that at most 10 people in the sample have access to electricity.
Find the probability that more than 25 people in the sample have access to electricity.

The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate in Afghanistan is 28.1 percent. Suppose you choose 15 people in Afghanistan at random. Let X = the number of people who are literate.

Sketch a graph of the probability distribution of X.
Using the formulas, calculate the (i) mean and (ii) standard deviation of X.
Find the probability that more than five people in the sample are literate. Is it more likely that three people or four people are literate?

X ~ B(15, .281)

View original figure (external; image not reproduced in this text edition)
Original figure description: This histogram shows a binomial probability distribution. It is made up of bars that are fairly normally distributed. The x-axis shows values from 0 to 15, with bars from 0 to 9. The y-axis shows values from 0 to 0.25 in increments of 0.05.
1. Mean = μ = np = 15(.281) = 4.215
2. Standard Deviation = σ = $\sqrt{n p q}$ = $\sqrt{15 (.281) (.719)}$ = 1.7409
P(x > 5) = 1 – P(x ≤ 5) = 1 – binomcdf(15, .281, 5) = 1 – 0.7754 = .2246
P(x = 3) = binompdf(15, .281, 3) = .1927
P(x = 4) = binompdf(15, .281, 4) = .2259
It is more likely that four people are literate than three people are.

Glossary

binomial experiment

a statistical experiment that satisfies the following three conditions:

There are a fixed number of trials, n
There are only two possible outcomes, called success and, failure, for each trial; the letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial
The n trials are independent and are repeated using identical conditions

Bernoulli trials

an experiment with the following characteristics:

There are only two possible outcomes called success and failure for each trial
The probability p of a success is the same for any trial (so the probability q = 1 − p of a failure is the same for any trial)

binomial probability distribution

a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of independent trials
Independent means that the result of any trial (for example, trial one) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV X is defined as the number of successes in n trials. The notation is: X ~ B(n, p). The mean is μ = np and the standard deviation is σ =

\sqrt{n p q}

. The probability of the following exactly x successes in n trials is
P(X = x) =

(\begin{matrix} n \\ x \end{matrix})

p^xq^{n − x}

Attribution. Statistics, OpenStax, Rice University, and the contributing authors credited in the book. Licensed under Creative Commons Attribution 4.0 International. Source revision 7dea80ae28 of openstax/osbooks-statistics (2026-04-23). CC BY 4.0. Original source ↗. Text edition of a pinned CC BY 4.0 source revision of the OpenStax book, converted from CNXML by the EvidenceHelix importer. Original prose, learning objectives, tables, exercises, glossary and captions are preserved; images are not copied (figures link to the original source and keep their credits). No new content was added. Later revisions of this book can carry different licence terms and are not used.