4.3 Binomial Distribution
SLO 3
Describe and apply probability concepts and distributions.
The binomial is the first named distribution you test an experiment against: check its three conditions, write X ~ B(n, p), and answer exact, 'at most', and 'at least' questions - then read the mean np as how many successes to expect and npq under a root as how far results stray.
Learning Objectives
By the end of this section, you will be able to:
- check whether a real experiment meets the three conditions that make it binomial;
- translate a word problem into the notation \(X \sim B(n, p)\) and state the probability question symbolically;
- compute binomial probabilities with the formula and with technology;
- calculate the mean and standard deviation of a binomial random variable and explain what they tell you.
4.3.1 What Makes an Experiment Binomial
A statistical experiment is a binomial experiment if it meets all three of these conditions:
- 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 letter \(p\) denotes the probability of a success on one trial, and \(q\) denotes the probability of a failure on one trial, so that \(p + q = 1\).
- The \(n\) trials are independent and are repeated under identical conditions. Because the trials are independent, the outcome of one trial does not help you predict the outcome of another. For each individual trial the probability \(p\) of a success and the probability \(q\) of a failure stay the same.
The outcomes of a binomial experiment fit a binomial probability distribution. The random variable \(X\) counts the number of successes obtained in the \(n\) independent trials.
The word success is doing some work here that is worth flagging early. It does not mean "the good outcome." It means "the outcome you decided to count." If you are studying students who withdraw from a course, then withdrawing is the success, because withdrawing is what you are counting. Pick the label first, keep it fixed for the whole problem, and \(p\) is the probability of the thing you labelled.
Consider randomly guessing at a true-false statistics question. It has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Diego always guesses correctly on any statistics true-false question with probability \(p = 0.6\). Then \(q = 0.4\). This means that for every true-false statistics question Diego answers, his probability of success \((p = 0.6)\) and his probability of failure \((q = 0.4)\) remain the same. That constancy is condition 3 doing its job.
Definition 4.3.1 — A binomial experiment is a fixed row of two-outcome trials that all run at the same p.
Any experiment that has characteristics two and three and where \(n = 1\) is called a Bernoulli trial, named after Jacob Bernoulli, who studied them extensively in the late 1600s. A binomial experiment takes place when the number of successes is counted in one or more Bernoulli trials.
A binomial question never asks which trials succeeded. It asks how many did. Whether the three heads came up on flips 1, 4, 7 or on flips 2, 3, 9, they all land on the same value \(x = 3\).
So a Bernoulli trial is a binomial experiment with the counter stuck at one repetition: a single flip, a single question, a single yes-or-no answer. Stack \(n\) of them together under identical conditions and you have rebuilt the binomial.
A binomial setup turns up constantly. You flip a coin ten times and each flip is heads or tails. You ask 50 students whether they finished their homework and each answer is yes or no. You guess at 32 multiple-choice questions and each guess is right or wrong. Every one of those is the same simple action repeated, with only two ways each repetition can come out, which is exactly the setup the binomial family handles.
All three conditions have to hold before you may reach for a binomial formula. If even one of them fails, the experiment is not binomial and none of the results in this section apply to it.
Definition 4.3.2 — A Bernoulli trial is one two-outcome trial; n identical copies rebuild the binomial.
The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a "success" be in this case?
Solution
A success is whichever outcome you have chosen to count, and here the board is tracking schools that offer fruit every day. So a success is a school that offers fruit in its lunches every day, which gives
$$ p = 0.48 \qquad q = 1 - 0.48 = 0.52 $$The random variable \(X\) would then count the number of schools, out of the \(n\) schools sampled, that offer fruit every day.
Answer: A success is a school that offers fruit every day, with \(p = 0.48\).
At ABC College, the withdrawal rate from an elementary physics course is 30% for any given term. This implies that, for any given term, 70% of the students stay in the class for the entire term. A "success" could be defined as an individual who withdrew. Let the random variable \(X\) be the number of students who withdraw from the randomly selected elementary physics class.
Solution
Step 1 — Check the three conditions. Each student either withdraws or does not, which gives two outcomes. Each student's decision is treated as independent of the others, and the withdrawal rate is the same 30% for everyone in the class. If the class has a fixed enrollment, the number of trials \(n\) is fixed. All three conditions hold.
Step 2 — Name the success and read off \(p\). We chose withdrawing as the success, so
$$ p = 0.30 \qquad q = 1 - 0.30 = 0.70 $$Step 3 — Say what \(X\) counts. \(X\) = the number of students out of the \(n\) enrolled who withdraw from the class.
Answer: This is a binomial experiment with \(p = 0.30\) and \(q = 0.70\), and \(X\) counts withdrawals. Notice that if we had instead called staying enrolled the success, we would have \(p = 0.70\) and \(X\) would count the students who stayed. Both are correct; they are just different labellings of the same experiment.
4.3.2 The Binomial Probability Formula
If \(X\) is the number of successes in \(n\) independent trials, each with probability of success \(p\) and probability of failure \(q = 1 - p\), then
$$ P(X = x) = \binom{n}{x} p^{x} q^{\,n-x}, \qquad x = 0, 1, 2, \ldots, n $$where \(\displaystyle\binom{n}{x} = \frac{n!}{x!\,(n-x)!}\) is the number of ways to choose which \(x\) of the \(n\) trials are the successes.
Getting 2 heads in 3 flips can happen as HHT, HTH, or THH. Each pattern has the same probability, and there are \(\binom{3}{2} = 3\) of them. The coefficient just stops you from counting one pattern when three would do.
The formula reads left to right as a sentence. The term \(p^{x}\) is the probability that \(x\) particular trials all succeed. The term \(q^{\,n-x}\) is the probability that the remaining \(n - x\) trials all fail. Multiplying them is legal precisely because condition 3 said the trials are independent. That product covers one specific pattern of successes and failures, so the binomial coefficient \(\binom{n}{x}\) multiplies it by the number of patterns that produce the same count.
One formula produces every probability you could want from a binomial experiment. It has two moving parts: a count of the ways the successes could be arranged, and the probability of any one such arrangement.
Definition 4.3.3 — Every pattern with the same count carries the same probability, and the coefficient counts the patterns.
Aiyana, a marine trainer, is teaching a rescued dolphin to catch live fish before she returns it to the wild. The probability that the dolphin successfully catches a fish is 35%, and the probability that the dolphin does not successfully catch the fish is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.
Solution
Step 1 — Define the random variable. Let \(X\) be the number of successful catches, so \(X\) takes on the values \(0, 1, 2, \ldots, 20\).
Step 2 — Read off the parameters. \(n = 20\), \(p = 0.35\), and \(q = 0.65\).
Step 3 — State the question. "Succeeds 12 times" is an exact count, so we want
$$ P(X = 12) = \binom{20}{12}(0.35)^{12}(0.65)^{8} $$Answer: The probability question is \(P(X = 12)\), and evaluating gives about \(0.0136\).
Suppose you play a game that you can only either win or lose. The probability that you win any game is 55%, and the probability that you lose is 45%. 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.
Solution
Step 1 — Define the random variable. Let \(X\) be the number of wins. Then \(X\) takes on the values \(0, 1, 2, 3, \ldots, 20\).
Step 2 — Read off the parameters. The probability of a success is \(p = 0.55\). The probability of a failure is \(q = 0.45\). The number of trials is \(n = 20\).
Step 3 — State the probability question mathematically. We want \(P(X = 15)\).
Step 4 — Substitute into the formula.
$$ P(X = 15) = \binom{20}{15}(0.55)^{15}(0.45)^{5} $$Answer: \(P(X = 15) = \binom{20}{15}(0.55)^{15}(0.45)^{5}\), which evaluates to about \(0.0365\).
4.3.3 Reading a Binomial Distribution from Its Graph
Writing down every value of \(P(X = x)\) at once gives you the full probability density function, and graphing it turns the distribution into a picture. Once you have the picture, questions like "more than three" become questions about which bars to add up.
A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.
Solution
Step 1 — Define the random variable. Let \(X\) be the number of rolls that come up a one, so \(X\) takes on the values \(0, 1, 2, \ldots, 10\).
Step 2 — Read off the parameters. The die is fair and there are six faces, so \(p = \frac{1}{6}\) and \(q = \frac{5}{6}\). The number of trials is \(n = 10\).
Step 3 — Translate the words. "More than three times" is strictly greater than 3.
Answer: The probability question is \(P(X > 3)\), where \(X \sim B\left(10, \frac{1}{6}\right)\). Evaluating gives about \(0.0697\).
A coin has been altered to weight the outcome from 0.5 to 0.25 and flipped 5 times. Each flip is independent. What is the probability of getting more than 3 heads? Let \(X\) be the number of heads in 5 flips of the coin, so \(X\) takes on the values 0, 1, 2, 3, 4, 5. Since the coin is altered to result in \(p = 0.25\), \(q\) is 0.75. The number of trials is \(n = 5\).
Solution
Step 1 — State the question mathematically. "More than 3 heads" means strictly more than three, so we want \(P(X > 3)\).
Step 2 — Develop the probability density function. Every value comes from the same formula:
$$ P(X = x) = \binom{5}{x}(0.25)^{x}(0.75)^{5-x} $$Figure 4.3.1 — Probability density function for the altered coin, X ~ B(5, 0.25), skewed right with a mean of 1.25.
Step 3 — Identify which bars we need. Above 3 there are exactly two values, 4 and 5, so
$$ P(X > 3) = P(X = 4) + P(X = 5) $$We add the two individual probabilities because of the addition rule from Probability Topics.
Step 4 — Evaluate each one.
$$ P(X = 4) = \binom{5}{4}(0.25)^{4}(0.75)^{1} = 5(0.00390625)(0.75) = 0.0146 $$ $$ P(X = 5) = \binom{5}{5}(0.25)^{5}(0.75)^{0} = 0.0010 $$Step 5 — Add them.
$$ P(X > 3) = 0.0146 + 0.0010 = 0.0156 $$Answer: \(P(X > 3) = 0.0156\), or about a 1.6% chance. Figure 4.3.1 shows why: the two bars to the right of 3 are the shortest on the graph, so the area they contribute is tiny.
Figure 4.3.1 also shows the link between the probability density function and probability as area: each bar's height is the probability of that value, so adding bars is the same as accumulating area. The graph also makes the skew visible. When \(p\) is not equal to 0.5 the distribution is lopsided, and here it leans right because the mean \(\mu = np = 1.25\) sits well to the left of the centre of the range 0 through 5.
4.3.4 Translating a Word Problem into a Binomial Question
Most of the work in a binomial problem happens before any arithmetic. You have to decide what one trial is, what counts as a success, how many trials there are, and which inequality the English is asking for. Get those four things right and the calculation is mechanical.
"At least 40" means \(X \ge 40\). "At most 12" means \(X \le 12\). "More than 3" means \(X > 3\). Mistranslating one of these is the single most common way a correct formula produces a wrong answer.
Sixty-five percent of people pass the state driver's exam on the first try. A group of 50 individuals who have taken the driver's exam is randomly selected. Give two reasons why this is a binomial problem.
Solution
Any two of the three defining conditions count as a correct answer. Here they all are:
Reason 1 — Two outcomes per trial. Each person either passes the exam on the first try or does not. Calling passing the success gives \(p = 0.65\) and \(q = 0.35\).
Reason 2 — A fixed number of trials. The group has exactly 50 people, so \(n = 50\) and it does not change partway through.
Reason 3 — Independent, identical trials. The individuals were selected randomly, so one person's result does not affect another's, and the 65% pass rate applies to each of them.
Answer: It is binomial because each of the 50 randomly selected people gives exactly one of two outcomes (pass or not) and the probability of passing stays at 0.65 for every one of them.
Approximately 70% 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 0.70 for each trial.
b. If we are interested in the number of students who do their homework on time, then how do we define \(X\)?
c. What values does \(X\) take on?
d. What is a "failure," in words?
e. If \(p + q = 1\), then what is \(q\)?
f. The words "at least" translate as what kind of inequality for the probability question \(P(X \ \_\_\_\_\ 40)\)?
Solution
a. failure
b. \(X\) = the number of statistics students who do their homework on time
c. \(0, 1, 2, \ldots, 50\)
d. Failure is defined as a student who does not complete their homework on time.
The probability of a success is \(p = 0.70\). The number of trials is \(n = 50\).
e. \(q = 1 - 0.70 = 0.30\)
f. greater than or equal to \((\ge)\)
Answer: The probability question is \(P(X \ge 40)\), with \(X \sim B(50, 0.70)\).
4.3.5 Notation for the Binomial
The notation
$$ X \sim B(n, p) $$is read "\(X\) is a random variable with a binomial distribution." The parameters are \(n\) and \(p\), where \(n\) is the number of trials and \(p\) is the probability of a success on each trial. The random variable \(X\) counts the number of successes in \(n\) independent trials and takes on the values \(x = 0, 1, 2, 3, \ldots, n\). The letter \(q\) denotes the probability of a failure for any trial, so \(p + q = 1\) and \(q = 1 - p\).
The tilde is not an equals sign. It means "is distributed as." Writing \(X \sim B(20, 0.41)\) tells a reader every single thing they need to compute any probability about \(X\), in eleven characters.
Statisticians pack an entire binomial setup into a few symbols rather than restate the three conditions every time.
About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school.
Solution
Step 1 — State the distribution. Let \(X\) be the number of the 30 students who participate. Then \(n = 30\), \(p = 0.32\), \(q = 0.68\), and
$$ X \sim B(30, 0.32) $$Step 2 — Translate "at most 14." At most 14 means \(P(X \le 14)\), which is a cumulative probability.
Step 3 — Evaluate. A cumulative probability like this one is what the At Most tab of rāSHio's Calculators → Binomial panel is for: set \(n = 30\) and \(p = 0.32\), then enter 14.
$$ P(X \le 14) = 0.9695 $$Answer: \(P(X \le 14) = 0.9695\). This is unsurprising once you notice the mean is \(\mu = np = 9.6\), so 14 is already well above the centre of the distribution.
It has been stated that about 41% 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?
Solution
Step 1 — Define the random variable and state its distribution. Let \(X\) be the number of workers who have a high school diploma but do not pursue any further education. Then \(X\) takes on the values \(0, 1, 2, \ldots, 20\), where \(n = 20\), \(p = 0.41\), and \(q = 1 - 0.41 = 0.59\). So
$$ X \sim B(20, 0.41) $$Step 2 — Translate "at most 12." At most means less than or equal to, so we want \(P(X \le 12)\).
Step 3 — Evaluate the cumulative probability.
$$ P(X \le 12) = 0.9738 $$Figure 4.3.2 — The distribution of X ~ B(20, 0.41), centred near 8.2 with most of its area at or below 12.
Step 4 — Find the expected number. The number of adult workers you expect is the mean:
$$ \mu = np = (20)(0.41) = 8.2 $$Answer: The probability that at most 12 workers have a high school diploma but do not pursue further education is \(0.9738\), and you expect about 8.2 such workers out of 20.
One panel answers every binomial question
You do not need a graphing calculator for any of this. In rāSHio, choose Calculators → Binomial and you get one panel with four tabs, one per question type: Exact for \(P(X = k)\), At Most for \(P(X \le k)\), At Least for \(P(X \ge k)\), and Range for \(P(a \le X \le b)\). Fill in Trials (n) and Success probability (p) once and they stay put as you switch tabs. For Example 4.3.5, set \(n = 20\) and \(p = 0.41\), pick At Most, enter 12, and the panel shades every bar at or below 12 and reports 0.9738.
Figure 4.3.3 — The rāSHio Binomial Distribution Calculator: Calculators → Binomial. One panel, four tabs — Exact, At Most, At Least, Range — and it prints the complement beside the region it shades. The walkthrough is shown with \(n = 10\) and \(p = 0.5\) as an illustration, not with this section’s numbers.
Four tabs cover every binomial question you will meet, because every question is either an exact value, a running total, a tail, or a stretch in the middle. The one worth pointing out is the pair the panel prints above the chart: alongside the shaded region it also reports the complement. So the At Most tab answers \(P(X > 12)\) at the same time it answers \(P(X \le 12)\) — 0.0262 sits right beside 0.9738, because "more than 12" is exactly what is left over after "at most 12" is removed. That saves you doing the subtraction by hand, and it is why you will rarely need to write \(1 - P(X \le k)\) yourself. The chart the panel draws is the same probability density function graphed in Figure 4.3.1, built for whatever \(n\) and \(p\) you typed.
The formula for the variance is \(\sigma^{2} = npq\), and the standard deviation is \(\sigma = \sqrt{npq}\). For the workers in Example 4.3.5,
$$ \sigma = \sqrt{(20)(0.41)(0.59)} = 2.20 $$4.3.6 The Mean and Standard Deviation of a Binomial
For a binomial random variable \(X \sim B(n, p)\) with \(q = 1 - p\), the mean, variance, and standard deviation are
$$ \mu = np \qquad \sigma^{2} = npq \qquad \sigma = \sqrt{npq} $$Expecting 8.2 workers does not mean anyone saw two-tenths of a worker. The mean is the long-run average over many samples of 20, and averages land between whole numbers all the time.
The mean formula is the one that should feel obvious once you say it out loud: if you run 20 trials and each one succeeds 41% of the time, you expect \(20 \times 0.41 = 8.2\) successes. The standard deviation tells you how far a typical result strays from that centre.
The binomial mean and standard deviation are shortcuts worth appreciating. For most distributions you would have to work out the mean by summing \(x \cdot P(x)\) across every value; the binomial gets it in a single multiplication.
Definition 4.3.5 — The whole x times P(x) column collapses into a single product, n times p.
According to a Gallup poll, 60% of American adults prefer saving over spending. Let \(X\) be the number of American adults out of a random sample of 50 who prefer saving to spending.
a. What is the probability distribution for \(X\)?
b. Find the following probabilities:
i. the probability that 25 adults in the sample prefer saving over spending
ii. the probability that at most 20 adults prefer saving
iii. the probability that more than 30 adults prefer saving
c. Using the formulas, calculate the (i) mean and (ii) standard deviation of \(X\).
Solution
a. With \(n = 50\), \(p = 0.60\), and \(q = 0.40\),
$$ X \sim B(50, 0.60) $$b. i. An exact count, so use the pdf:
$$ P(X = 25) = \text{binompdf}(50, 0.60, 25) = 0.0405 $$ii. "At most 20" is cumulative:
$$ P(X \le 20) = \text{binomcdf}(50, 0.60, 20) = 0.0034 $$iii. "More than 30" is the complement of "at most 30":
$$ P(X > 30) = 1 - P(X \le 30) = 1 - 0.5535 = 0.4465 $$c. i. Mean:
$$ \mu = np = (50)(0.60) = 30 $$ii. Standard deviation:
$$ \sigma = \sqrt{npq} = \sqrt{(50)(0.60)(0.40)} = \sqrt{12} \approx 3.4641 $$Answer: \(X \sim B(50, 0.60)\), \(P(X = 25) = 0.0405\), \(P(X \le 20) = 0.0034\), \(P(X > 30) = 0.4465\), \(\mu = 30\), \(\sigma \approx 3.46\). Notice how part (ii) confirms the mean: 20 is nearly three standard deviations below 30, so almost none of the distribution's area sits there.
In the 2013 art supplies catalog there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let \(X\) be the number of pages that feature signature artists.
a. What values does \(X\) take on?
b. What is the probability distribution? Find the following probabilities:
i. the probability that two pages feature signature artists
ii. the probability that at most six pages feature signature artists
iii. the probability that more than three pages feature signature artists
c. Using the formulas, calculate the (i) mean and (ii) standard deviation.
Solution
a. \(x = 0, 1, 2, 3, 4, 5, 6, 7, 8\)
b. The probability that any one sampled page features a signature artist is \(\frac{8}{560}\), and we sample \(n = 100\) pages, so
$$ X \sim B\left(100, \frac{8}{560}\right) $$i. An exact count uses the pdf:
$$ P(X = 2) = \text{binompdf}\left(100, \frac{8}{560}, 2\right) = 0.2466 $$ii. "At most six" is cumulative:
$$ P(X \le 6) = \text{binomcdf}\left(100, \frac{8}{560}, 6\right) = 0.9994 $$iii. "More than three" is the complement of "at most three":
$$ P(X > 3) = 1 - P(X \le 3) = 1 - \text{binomcdf}\left(100, \frac{8}{560}, 3\right) = 1 - 0.9443 = 0.0557 $$c. i. Mean:
$$ \mu = np = (100)\left(\frac{8}{560}\right) = \frac{800}{560} \approx 1.4286 $$ii. Standard deviation:
$$ \sigma = \sqrt{npq} = \sqrt{(100)\left(\frac{8}{560}\right)\left(\frac{552}{560}\right)} \approx 1.1867 $$Answer: \(X \sim B\left(100, \frac{8}{560}\right)\), with \(P(X = 2) = 0.2466\), \(P(X \le 6) = 0.9994\), \(P(X > 3) = 0.0557\), \(\mu \approx 1.4286\), and \(\sigma \approx 1.1867\).
During a certain NBA season, a player for the Los Angeles Clippers had the highest field goal completion rate in the league. This player scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by this player during the season. Let \(X\) be the number of shots that scored points.
a. What is the probability distribution for \(X\)?
b. Using the formulas, calculate the (i) mean and (ii) standard deviation of \(X\).
c. Find the probability that this player scored with 60 of these shots.
d. Find the probability that this player scored with more than 50 of these shots.
Solution
a. With \(n = 80\), \(p = 0.613\), and \(q = 0.387\),
$$ X \sim B(80, 0.613) $$b. i. Mean:
$$ \mu = np = 80(0.613) = 49.04 $$ii. Standard deviation:
$$ \sigma = \sqrt{npq} = \sqrt{80(0.613)(0.387)} \approx 4.3564 $$c. An exact count, so use the pdf:
$$ P(X = 60) = \text{binompdf}(80, 0.613, 60) = 0.0036 $$d. "More than 50" is the complement of "at most 50":
$$ P(X > 50) = 1 - P(X \le 50) = 1 - 0.6282 = 0.3718 $$Answer: \(X \sim B(80, 0.613)\), \(\mu = 49.04\), \(\sigma \approx 4.36\), \(P(X = 60) = 0.0036\), and \(P(X > 50) = 0.3718\). Scoring exactly 60 is rare because 60 sits about 2.5 standard deviations above the mean of 49.04.
The lifetime risk of developing cancer is about one in 67 (1.5%). Suppose we randomly sample 200 people. Let \(X\) be the number of people who will develop cancer.
a. What is the probability distribution for \(X\)?
b. Using the formulas, calculate the (i) mean and (ii) standard deviation of \(X\).
c. Find the probability that at most eight people develop cancer.
d. Is it more likely that five or six people will develop cancer? Justify your answer numerically.
Solution
a. With \(n = 200\) and \(p = 0.015\),
$$ X \sim B(200, 0.015) $$b. i. Mean:
$$ \mu = np = 200(0.015) = 3 $$ii. Standard deviation:
$$ \sigma = \sqrt{npq} = \sqrt{200(0.015)(0.985)} = 1.719 $$c. "At most eight" is cumulative:
$$ P(X \le 8) = 0.9965 $$d. Compare the two exact probabilities directly:
$$ P(X = 5) = 0.1011 \qquad P(X = 6) = 0.0500 $$Since \(0.1011 > 0.0500\), five is more likely than six. That fits the mean of 3: both values are above the centre, and 6 is further out than 5, so it carries less probability.
Answer: \(X \sim B(200, 0.015)\), \(\mu = 3\), \(\sigma = 1.719\), \(P(X \le 8) = 0.9965\), and five people is more likely than six because \(0.1011 > 0.0500\).
Comparing two outcomes is two clicks, not two formulas
Part d asks which of five or six is more likely, and that is the Exact tab of Calculators → Binomial run twice. Set \(n = 200\) and \(p = 0.015\) once, then enter \(k = 5\) (0.1011) and \(k = 6\) (0.0500). The shaded bar moves one step right and visibly shrinks, which is the answer before you have read either number: past the mean of 3, every further step out carries less probability than the one before it.
4.3.7 When an Experiment Is Not Binomial
The three conditions are a test, not a formality, and the one that fails most often in practice is independence. Sampling without replacement changes the composition of what is left, so the probability of a success shifts from trial to trial.
If the thing you draw goes back before the next draw, \(p\) holds steady and the experiment stays binomial. If it does not go back, \(p\) moves, and the binomial formula no longer applies.
The reason a dependent experiment breaks the formula is that the multiplication \(p^{x} q^{\,n-x}\) assumed each trial carried the same probability regardless of what came before. Once that assumption is gone, you need probabilities that are conditioned on the earlier outcomes instead.
Suppose you want the chance that \(A\) happens, but you already know \(B\) happened. Take the chance that both land together and scale it up by the chance of what you were told:
In a binomial experiment this formula would be doing no work at all, because independence means \(P(A \mid B) = P(A)\): knowing that \(B\) happened tells you nothing about \(A\). The moment that stops being true, the conditional probability differs from the unconditional one, and that gap is exactly what disqualifies the experiment.
A lacrosse team coached by Sione Tuipulotu is selecting a captain. He puts the names of all the seniors into a hat, and the first three that are drawn will be the captains. The names are not replaced once they are drawn (one person cannot be two captains). You want to see if the captains all play the same position. State whether this is binomial or not and state why.
Solution
Step 1 — Check for two outcomes. Each name drawn either plays the position of interest or does not, so this condition holds.
Step 2 — Check for a fixed number of trials. Exactly three names are drawn, so \(n = 3\) and this condition holds too.
Step 3 — Check independence. The names are not replaced. Once a senior is drawn, the pool shrinks by one, and the mix of positions left in the hat changes. The probability of drawing a player at the position of interest is therefore different on the second and third draws than it was on the first, and it depends on who was already drawn.
Answer: This is not binomial. Drawing without replacement makes the trials dependent, so the probability of a success does not stay the same from trial to trial.
The following example illustrates a problem that is not binomial. It violates the condition of independence. ABC College has a student advisory committee made up of ten 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.
Solution
Step 1 — Count the trials. There are two draws, so it looks at first like \(n = 2\) trials with two outcomes each (student or staff member).
Step 2 — Test independence. The committee has \(10 + 6 = 16\) members. The probability of drawing a student on the first draw is
$$ \frac{6}{16} $$Now look at the second draw. The probability of a student on the second draw is \(\frac{5}{15}\) when the first draw selected a student, because one student and one name are now gone. But it is \(\frac{6}{15}\) when the first draw selected a staff member.
Step 3 — Draw the conclusion. The probability of drawing a student's name changes depending on what happened first, so it is not the same for every trial and the outcome of trial one does help predict trial two. This violates the condition of independence.
Answer: This is not a binomial experiment. Because the two names are drawn without replacement, \(p\) changes between trials and condition 3 fails.
Problem Set 4.3
The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time first-year students from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly pick eight first-time, full-time first-year students from the survey. You are interested in the number that believes that same-sex couples should have the right to legal marital status. Use this information for the next eight exercises.
Problem 1. In words, define the random variable \(X\).
Solution
The survey question has exactly two replies of interest, and we are counting how many of the eight picked students give the "yes" reply.
Answer: \(X\) = the number of the eight first-year students picked who reply that they believe same-sex couples should have the right to legal marital status.
Problem 2. \(X \sim \) _____(_____, _____)
Solution
Step 1 — Identify \(n\). Eight students are picked, so \(n = 8\).
Step 2 — Identify \(p\). 71.3% of students replied yes, so \(p = 0.713\).
Answer: \(X \sim B(8, 0.713)\)
Problem 3. What values does the random variable \(X\) take on?
Solution
\(X\) counts successes out of eight trials, so the smallest possible count is 0 (nobody replies yes) and the largest is 8 (everybody does).
Answer: \(x = 0, 1, 2, 3, 4, 5, 6, 7, 8\)
Problem 4. Construct the probability distribution function (PDF). Fill in Table 4.3.1.
| \(x\) | \(P(x)\) |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 | |
| 7 | |
| 8 |
Solution
Step 1 — Use the binomial formula for each value.
$$ P(X = x) = \binom{8}{x}(0.713)^{x}(0.287)^{8-x} $$Step 2 — Evaluate at every \(x\) from 0 to 8.
| \(x\) | \(P(x)\) |
|---|---|
| 0 | 0.0000 |
| 1 | 0.0009 |
| 2 | 0.0080 |
| 3 | 0.0395 |
| 4 | 0.1227 |
| 5 | 0.2439 |
| 6 | 0.3030 |
| 7 | 0.2151 |
| 8 | 0.0668 |
Answer: The table above. As a check, the nine probabilities sum to 1, and the tallest bar sits at \(x = 6\), just above the mean of 5.7.
Problem 5. On average \((\mu)\), how many would you expect to answer yes?
Solution
Step 1 — Use the binomial mean formula.
$$ \mu = np = 8(0.713) = 5.704 $$Answer: About 5.7 of the eight students. That fits the table in Problem 4.3.4, where the probability piles up around \(x = 6\).
Problem 6. What is the standard deviation \((\sigma)\)?
Solution
Step 1 — Find \(q\).
$$ q = 1 - 0.713 = 0.287 $$Step 2 — Apply the standard deviation formula.
$$ \sigma = \sqrt{npq} = \sqrt{8(0.713)(0.287)} = \sqrt{1.6371} = 1.2795 $$Answer: \(\sigma \approx 1.28\) students.
Problem 7. What is the probability that at most five of the first-year students reply "yes"?
Solution
Step 1 — Translate "at most five." At most means less than or equal to, so we want \(P(X \le 5)\).
Step 2 — Evaluate the cumulative probability with binomcdf(8, 0.713, 5), which adds the rows for \(x = 0\) through \(x = 5\) in the table from Problem 4.3.4.
Answer: \(P(X \le 5) = 0.4151\), a little under a 42% chance. Note this is below 50%, which makes sense because the mean of 5.7 sits above 5.
Problem 8. What is the probability that at least two of the first-year students reply "yes"?
Solution
Step 1 — Translate "at least two." At least two means \(P(X \ge 2)\).
Step 2 — Use the complement. Going the direct way would mean adding seven terms. The complement of "at least 2" is "at most 1," which is only two terms:
$$ P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X = 0) + P(X = 1)] $$Step 3 — Evaluate. From the table in Problem 4.3.4, \(P(X = 0) \approx 0.0000\) and \(P(X = 1) = 0.0009\), so
$$ P(X \ge 2) = 1 - 0.0009 = 0.9990 $$Answer: \(P(X \ge 2) = 0.9990\). With 71.3% of students replying yes, getting fewer than two yes replies out of eight would be very surprising.
Problem 9. 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.
Solution
Step 1 — Set up the binomial. Two babies per 1,000 are born deaf in a healthy baby nursery, so \(p = \frac{2}{1000} = 0.002\). We surveyed \(n = 1{,}000\) babies, and \(X\) counts how many were born deaf:
$$ X \sim B(1000, 0.002) $$Step 2 — Translate "exactly two." An exact count needs the pdf, so we want \(P(X = 2)\).
Step 3 — Evaluate.
$$ P(X = 2) = \binom{1000}{2}(0.002)^{2}(0.998)^{998} = 0.2709 $$Answer: \(P(X = 2) = 0.2709\). The mean here is \(\mu = np = 2\), so two is the single most likely outcome, and it still only happens about 27% of the time.
Use the following information to answer the next four exercises. Recently, a nurse named Lena Vang commented that when a patient calls the medical advice line she staffs claiming to have the flu, the chance that they truly have the flu (and not just a nasty cold) is only about 4%. Of the next 25 patients calling in claiming to have the flu, we are interested in how many actually have the flu.
Problem 10. Define the random variable and list its possible values.
Solution
We are counting, among the next 25 callers who claim to have the flu, how many actually do.
Answer: \(X\) = the number of patients calling in and claiming to have the flu who actually have the flu, and \(x = 0, 1, 2, \ldots, 25\).
Problem 11. State the distribution of \(X\).
Solution
Step 1 — Identify \(n\). The next 25 callers, so \(n = 25\).
Step 2 — Identify \(p\). About 4% of such callers truly have the flu, so \(p = 0.04\).
Answer: \(X \sim B(25, 0.04)\)
Problem 12. Find the probability that at least four of the 25 patients actually have the flu.
Solution
Step 1 — Translate "at least four." At least four means \(P(X \ge 4)\).
Step 2 — Use the complement so that we add four terms instead of twenty-two:
$$ P(X \ge 4) = 1 - P(X \le 3) $$Step 3 — Evaluate with binomcdf(25, 0.04, 3):
Answer: \(P(X \ge 4) = 0.0165\), under a 2% chance.
Problem 13. On average, for every 25 patients calling in, how many do you expect to have the flu?
Solution
Step 1 — Use the binomial mean formula.
$$ \mu = np = 25(0.04) = 1 $$Answer: One patient. Out of every 25 callers convinced they have the flu, on average just one actually does, which is why Problem 4.3.12 gave such a small probability for four or more.
Problem 14. People visiting a hot dog stand often purchase more than one hot dog. The probability distribution for purchases of number of hot dogs per customer at a hot dog stand is given in Table 4.3.2.
| \(x\) | \(P(x)\) |
|---|---|
| 0 | 0.03 |
| 1 | 0.50 |
| 2 | 0.24 |
| 3 | |
| 4 | 0.07 |
| 5 | 0.04 |
a) Describe the random variable \(X\) in words.
b) Find the probability that a customer purchases three hot dogs.
c) Find the probability that a customer purchases at least four hot dogs.
d) Find the probability that a customer purchases at most two hot dogs.
Solution
a. \(X\) = the number of hot dogs purchased by one customer at the hot dog stand.
b. Every probability distribution's probabilities must add to 1, so the missing entry is whatever is left over:
$$ P(X = 3) = 1 - (0.03 + 0.50 + 0.24 + 0.07 + 0.04) = 1 - 0.88 = 0.12 $$c. "At least four" means 4 or 5:
$$ P(X \ge 4) = 0.07 + 0.04 = 0.11 $$d. "At most two" means 0, 1, or 2:
$$ P(X \le 2) = 0.03 + 0.50 + 0.24 = 0.77 $$Answer: (a) the number of hot dogs one customer buys; (b) 0.12; (c) 0.11; (d) 0.77.
Problem 15. Kaitlyn Brennan, 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% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.
a) In words, define the random variable \(X\).
b) List the values that \(X\) may take on.
c) Give the distribution of \(X\).
d) How many of the 12 students do we expect to attend the festivities?
e) Find the probability that at most four students will attend.
f) Find the probability that more than two students will attend.
Solution
a. \(X\) = the number of the 12 surveyed students who will attend the Tet festivities.
b. \(x = 0, 1, 2, \ldots, 12\)
c. With \(n = 12\) and \(p = 0.18\):
$$ X \sim B(12, 0.18) $$d. The expected count is the mean:
$$ \mu = np = 12(0.18) = 2.16 $$e. "At most four" is cumulative, so use binomcdf(12, 0.18, 4):
f. "More than two" is the complement of "at most two":
$$ P(X > 2) = 1 - P(X \le 2) = 1 - 0.6298 = 0.3702 $$Answer: (a) the number attending; (b) 0 through 12; (c) \(X \sim B(12, 0.18)\); (d) 2.16 students; (e) 0.9511; (f) 0.3702.
The probability that the San Jose Sharks 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. Let \(X\) be the number of games won in that upcoming month. Use this information for the next three exercises.
Problem 16. The expected number of wins for that upcoming month is:
a) 1.67
b) \(\frac{12}{382}\)
c) 1043
d) 4.43
Solution
Step 1 — Set up the binomial. There are \(n = 12\) games and the win probability is \(p = 0.3694\).
Step 2 — Apply the mean formula.
$$ \mu = np = 12(0.3694) = 4.4328 $$Answer: d. 4.43
Problem 17. What is the probability that the San Jose Sharks win six games in that upcoming month?
a) 0.1476
b) 0.2336
c) 0.7664
d) 0.8903
Solution
Step 1 — Translate "win six games." This is an exact count, so we need the pdf at \(x = 6\).
Step 2 — Evaluate with binompdf(12, 0.3694, 6):
Answer: a. 0.1476. Six wins is above the mean of 4.43, so it is possible but not the most likely result.
Problem 18. What is the probability that the San Jose Sharks win at least five games in that upcoming month?
a) 0.3694
b) 0.5266
c) 0.4734
d) 0.2305
Solution
Step 1 — Translate "at least five." At least five means \(P(X \ge 5)\).
Step 2 — Use the complement.
$$ P(X \ge 5) = 1 - P(X \le 4) $$Step 3 — Evaluate with binomcdf(12, 0.3694, 4):
Answer: c. 0.4734. Notice option b (0.5266) is the cumulative probability itself, which is the trap if you forget to subtract from 1.
Problem 19. A student takes a ten-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% of the questions correct.
Solution
Step 1 — Set up the binomial. There are \(n = 10\) true-false questions, and randomly guessing gives \(p = 0.5\) on each:
$$ X \sim B(10, 0.5) $$Step 2 — Translate the passing grade. At least 70% of 10 questions is at least 7 questions, so we want \(P(X \ge 7)\).
Step 3 — Use the complement.
$$ P(X \ge 7) = 1 - P(X \le 6) = 1 - 0.8281 = 0.1719 $$Answer: \(P(X \ge 7) = 0.1719\). Guessing your way to a passing grade happens about 17% of the time on a ten-question true-false quiz, which is why true-false quizzes are usually longer than ten questions.
Problem 20. 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% of the questions correctly.
Solution
Step 1 — Set up the binomial. There are \(n = 32\) questions, each with three choices, so a random guess is correct with probability \(p = \frac{1}{3}\):
$$ X \sim B\left(32, \frac{1}{3}\right) $$Step 2 — Translate the threshold. 75% of 32 is 24, and "more than 75%" is strictly greater, so we want \(P(X > 24)\).
Step 3 — Use the complement. The event "more than 24" is the complement of "less than or equal to 24":
$$ P(X > 24) = 1 - P(X \le 24) = 1 - \text{binomcdf}\left(32, \frac{1}{3}, 24\right) $$Step 4 — Evaluate.
$$ P(X > 24) \approx 0.0000003 $$Answer: \(P(X > 24) \approx 0\). The probability of guessing more than 75% of the 32 questions correctly is very small and practically zero. The mean is only \(\mu = np \approx 10.7\), so 24 correct answers sits far out in the tail.
Problem 21. Six different colored dice are rolled. Of interest is the number of dice that show a one.
a) In words, define the random variable \(X\).
b) List the values that \(X\) may take on.
c) Give the distribution of \(X\).
d) On average, how many dice would you expect to show a one?
e) Find the probability that all six dice show a one.
f) Is it more likely that three or that four dice will show a one? Justify your answer numerically.
Solution
a. \(X\) = the number of the six dice that show a one.
b. \(x = 0, 1, 2, 3, 4, 5, 6\)
c. Each die is a separate trial with a \(\frac{1}{6}\) chance of showing a one, and the dice being different colors does not change any probability, it just makes the six trials easy to tell apart:
$$ X \sim B\left(6, \frac{1}{6}\right) $$d. The expected count is the mean:
$$ \mu = np = 6\left(\frac{1}{6}\right) = 1 $$e. All six showing a one means \(x = 6\):
$$ P(X = 6) = \binom{6}{6}\left(\frac{1}{6}\right)^{6} = \frac{1}{46{,}656} \approx 0.0000214 $$f. Compare the two exact probabilities:
$$ P(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{3} = 0.0536 $$ $$ P(X = 4) = \binom{6}{4}\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)^{2} = 0.0080 $$Since \(0.0536 > 0.0080\), three dice showing a one is more likely than four.
Answer: (a) the number of dice showing a one; (b) 0 through 6; (c) \(X \sim B\left(6, \frac{1}{6}\right)\); (d) 1 die; (e) \(\frac{1}{46{,}656} \approx 0.0000214\); (f) three is more likely, because \(0.0536 > 0.0080\).
Key Terms
binomial experiment — a statistical experiment with a fixed number of independent trials, exactly two outcomes per trial, and a constant probability of success.
binomial probability distribution — the distribution of \(X\), the number of successes in \(n\) independent trials, written \(X \sim B(n, p)\).
Bernoulli trial — a single trial with two outcomes; a binomial experiment with \(n = 1\).
success — the outcome being counted in a binomial experiment, whose probability is \(p\); not necessarily the desirable outcome.
failure — the outcome not being counted, whose probability is \(q = 1 - p\).
\(n\) — the number of trials in a binomial experiment.
\(p\) — the probability of a success on any one trial.
\(q\) — the probability of a failure on any one trial, equal to \(1 - p\).
independence — the property that the outcome of one trial does not change the probabilities for any other trial; the condition that sampling without replacement violates.
Appendix — Additional Examples and Practice
Suppose a smoker carries a 13% chance of developing a severe lung condition at some point in life. Four of your friends smoke. How likely is it that two or more of them end up with such a condition?
Solution
Step 1 — State the distribution. Let \(X\) be the number of the four friends who develop a severe lung condition, with \(n = 4\), \(p = 0.13\), and \(q = 0.87\), so \(X \sim B(4, 0.13)\).
Step 2 — Use the complement. "At least 2" is everything except 0 and 1, so
$$ P(X \ge 2) = 1 - P(X = 0) - P(X = 1) $$Step 3 — Evaluate the two terms.
$$ P(X = 0) = \binom{4}{0}(0.13)^{0}(0.87)^{4} = 0.5729 $$ $$ P(X = 1) = \binom{4}{1}(0.13)^{1}(0.87)^{3} = 4(0.13)(0.6585) = 0.3424 $$Step 4 — Subtract.
$$ P(X \ge 2) = 1 - 0.5729 - 0.3424 = 0.0847 $$Answer: About \(0.0847\), or roughly an 8.5% chance.
Mateo Reyes is a placekicker who takes many field goal attempts over a season. A negative binomial model is only valid when the attempts do not influence one another. Is treating Mateo's kicks as independent a fair assumption to make?
Solution
It is a reasonable working assumption, but it is an assumption rather than a fact, and it is worth naming what could break it.
Independence would hold if Mateo's chance of making a kick were the same on every attempt regardless of how the earlier ones went. In practice several things could couple the attempts together: fatigue late in a game, a shift in confidence after a miss, changing wind or field conditions, or a defence that adjusts after seeing his first few attempts.
Answer: Yes, independence is reasonable as a first model, because his technique is largely the same from attempt to attempt. But it is an approximation. Fatigue, psychological effects after a miss, and changing weather all make the real attempts slightly dependent.
Take the attempts to be independent, with Mateo converting 80% of them. His husband is timing how long the streak takes. How likely is it that Mateo's fourth successful kick arrives no later than the fifth attempt?
Solution
Step 1 — Recognize the structure. We are waiting for the fourth success, and asking that it arrive on or before attempt 5. This is a negative binomial question, not a binomial one: the number of trials is not fixed in advance, the number of successes is.
Step 2 — Split into the two ways it can happen. The fourth field goal lands within 5 attempts if it happens on attempt 4 or on attempt 5.
Step 3 — Fourth goal on attempt 4. All four attempts must be successes:
$$ P = (0.80)^{4} = 0.4096 $$Step 4 — Fourth goal on attempt 5. Exactly 3 of the first 4 attempts succeed, and attempt 5 succeeds:
$$ P = \binom{4}{3}(0.80)^{3}(0.20)^{1} \times (0.80) = 4(0.512)(0.20)(0.80) = 0.3277 $$Step 5 — Add them.
$$ P(\text{fourth goal within 5 attempts}) = 0.4096 + 0.3277 = 0.7373 $$Answer: About \(0.7373\), or roughly a 74% chance.