Introduction to Statistics · Chapter 4 · Random Variables
A fixed number of yes/no trials, run under identical conditions — one formula answers every probability question they can produce.
bookSHelf · Introduction to Statistics · §4.3 · a self-paced section
Learning objectives — by the end of this section you will be able to
§4.3.1 — three conditions, checked every time
Definition 4.3.1 — Binomial Experiment
Definition 4.3.1 — a fixed row of two-outcome trials, all run at the same p.
§4.3.1 — success is a label, not a judgment
"Success" does not mean "the good outcome." It means the outcome you decided to count. Studying who withdraws from a course? Withdrawing is the success — because withdrawing is what you're counting.
Pick the label first, keep it fixed for the whole problem, and p is the probability of the thing you labelled. Diego guesses correctly on true–false questions with p=0.6; every question keeps that same p, which is condition 3 doing its job.
§4.3.1 — one trial, n identical copies
Definition 4.3.2 — Bernoulli Trial
Any experiment with characteristics 2 and 3 where n=1 is a Bernoulli trial, named for Jacob Bernoulli. A binomial experiment counts successes over one or more Bernoulli trials.
Definition 4.3.2 — one two-outcome trial; n copies rebuild the binomial.
A binomial setup turns up constantly: ten coin flips, 50 yes/no answers, 32 multiple-choice guesses — the same simple action repeated, only two ways each repetition can come out.
Context Pause — counting successes, not outcomes
Which trials, or how many?
A binomial question never asks which trials succeeded. It asks how many did. Whether three heads land on flips 1, 4, 7 or on flips 2, 3, 9, they all land on the same value x=3.
Your turn — pick the label first
Try It Now 4.3.1 — fruit in school lunches
48% of schools in the state offer fruit in their lunches every day (52% do not). What would a "success" be here?
Answer: A success is a school that offers fruit every day: p=0.48, q=1−0.48=0.52. X counts, out of the n schools sampled, how many offer fruit every day.
Worked example — the same experiment, two valid labels
Example 4.3.1 — Naming the Success
At ABC College the withdrawal rate from an elementary physics course is 30% per term. Let X be the number of students who withdraw from the randomly selected class.
Check: two outcomes (withdraw / stay), independent students, fixed class size — all three conditions hold.
Answer: Binomial with p=0.30, q=0.70; X counts withdrawals. Calling staying enrolled the success instead would flip to p=0.70 with X counting who stayed — both labellings are correct, just different bookkeeping.
§4.3.2 — the binomial probability formula
Definition 4.3.3 — Binomial Probability Distribution Function
If X counts successes in n independent trials, each with P(success)=p, q=1−p:
P(X=x)=(xn)pxqn−x,x=0,1,…,nwhere (xn)=x!(n−x)!n! counts which x of the n trials succeed.
Definition 4.3.3 — every pattern with the same count carries the same probability; the coefficient counts the patterns.
§4.3.2 — reading the formula as a sentence
px is the probability that x particular trials all succeed. qn−x is the probability the remaining n−x all fail. Multiplying them is legal exactly because independence (condition 3) says so.
That product covers one specific pattern of successes and failures — so (xn) multiplies it by the number of patterns that produce the same count. A count of arrangements, times the probability of any one arrangement.
Insight Note — why the coefficient is there
Three patterns, one count
Getting 2 heads in 3 flips can happen as HHT, HTH, or THH. Each pattern has the same probability, and there are (23)=3 of them. The coefficient just stops you from counting one pattern when three would do.
Your turn — state the question, don't evaluate yet
Try It Now 4.3.2 — a dolphin catching fish
A dolphin catches a fish with probability 35% per attempt. Out of 20 attempts, state the probability it succeeds exactly 12 times.
Answer: n=20, p=0.35, q=0.65, so
P(X=12)=(1220)(0.35)12(0.65)8≈0.0136Worked example — win/lose game, 20 independent plays
Example 4.3.2 — Writing the Function for a Game
You win 55% of the time, lose 45%, each game independent. Playing 20 games, write the function for winning exactly 15.
Setup: p=0.55, q=0.45, n=20; want P(X=15).
P(X=15)=(1520)(0.55)15(0.45)5≈0.0365§4.3.3 — from formula to picture
Writing every value of P(X=x) at once gives the full probability density function. Graph it, and a question like "more than three" becomes a question about which bars to add up.
Your turn — translate before you compute
Try It Now 4.3.3 — a fair die, ten rolls
A fair six-sided die is rolled ten times. State the probability of rolling a one more than three times.
Answer: p=61, q=65, n=10; "more than three" is strict, so P(X>3), where X∼B(10,61), evaluating to about 0.0697.
Worked example — an altered coin, five flips
Example 4.3.3 — More Than Three Heads
A coin altered to p=0.25 (so q=0.75) is flipped n=5 times. Find P(X>3).
Only x=4,5 exceed 3, so P(X>3)=P(X=4)+P(X=5):
P(X=4)=(45)(.25)4(.75)1=.0146, P(X=5)=.0010Answer: P(X>3)=0.0156 — about a 1.6% chance.
Figure 4.3.1 — the pdf for X∼B(5,0.25): skewed right, mean 1.25. The two bars past 3 are the shortest on the graph.
§4.3.4 — before any arithmetic
Most of the work happens before the calculation. Decide four things: 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 right and the calculation is mechanical.
Context Pause — the phrase tells you the inequality
Translate the English literally
"At least 40" means X≥40. "At most 12" means X≤12. "More than 3" means X>3. Mistranslating one of these is the single most common way a correct formula produces a wrong answer.
Your turn — check the three conditions
Try It Now 4.3.4 — the driver's exam
65% of people pass the driver's exam on the first try. A group of 50 test-takers is randomly selected. Give two reasons this is binomial.
Answer: (1) Two outcomes per trial — pass (p=0.65) or not (q=0.35). (2) A fixed n=50. (Independence holds too: random selection makes one person's result not affect another's.)
Worked example — fill in the blanks, then state the question
Example 4.3.4 — Homework on Time
70% of statistics students do homework on time; a class of 50 is randomly selected. What is the probability that at least 40 do?
Answer: The probability question is P(X≥40), with X∼B(50,0.70).
§4.3.5 — packing a whole setup into a symbol
Definition 4.3.4 — Binomial Notation
X∼B(n,p)read "X is a random variable with a binomial distribution." n is the number of trials, p the probability of success per trial, q=1−p, and X takes values x=0,1,…,n.
§4.3.5 — eleven characters, everything you need
It means "is distributed as." Writing X∼B(20,0.41) tells a reader every single thing needed 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.
Your turn — a cumulative probability
Try It Now 4.3.5 — community volunteers
32% of students volunteer outside school. Of 30 selected at random, find the probability that at most 14 volunteer.
Answer: X∼B(30,0.32). "At most 14" is cumulative:
P(X≤14)=0.9695Worked example — a cumulative probability, plus the mean
Example 4.3.5 — Workers with a High School Diploma
41% of adult workers have a diploma but no further education. Of 20 randomly selected, find P(X≤12) and the expected count.
X∼B(20,0.41):
P(X≤12)=0.9738μ=np=(20)(0.41)=8.2Answer: 0.9738; expect about 8.2 workers out of 20.
Figure 4.3.2 — X∼B(20,0.41), centred near 8.2, most of its area at or below 12.
Try it in rāSHio — one panel, four tabs
No graphing calculator needed
Calculators → Binomial gives four tabs — Exact for P(X=k), At Most for P(X≤k), At Least for P(X≥k), and Range for P(a≤X≤b). Fill in Trials (n) and Success probability (p) once; they stay put as you switch tabs. For Example 4.3.5: n=20, p=0.41, At Most, enter 12 — the panel shades every bar at or below 12 and reports 0.9738.
§4.3.5 — the panel, in motion
Figure 4.3.3: One panel, four tabs — Exact, At Most, At Least, Range — printing the complement beside the shaded region. Shown with n=10, p=0.5 as an illustration, not this section's numbers.
The panel also prints the complement: the At Most tab answers P(X>12) at the same time as P(X≤12) — 0.0262 sits right beside 0.9738 — so you rarely need to write 1−P(X≤k) yourself.
§4.3.6 — the mean and standard deviation
Run 20 trials at 41% success each, and you expect 20×0.41=8.2 successes — the mean formula should feel obvious once said aloud.
For most distributions you would sum x⋅P(x) across every value; the binomial gets its mean in a single multiplication.
§4.3.6 — mean, variance, standard deviation
Definition 4.3.5 — Mean and Standard Deviation of a Binomial
For X∼B(n,p) with q=1−p:
μ=npσ2=npqσ=npqDefinition 4.3.5 — the whole x⋅P(x) column collapses into a single product, np.
Insight Note — why the mean can be a fraction
A long-run average, not a headcount
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.
Your turn — exact, cumulative, complement, mean, sd
Try It Now 4.3.6 — saving over spending
60% of American adults prefer saving to spending. Of 50 randomly sampled, find several probabilities plus the mean and standard deviation.
X∼B(50,0.60):
P(X=25)=0.0405P(X≤20)=0.0034P(X>30)=0.4465 μ=np=30σ=npq=12≈3.4641Worked example — a fraction for p, straight from the data
Example 4.3.6 — Signature Artists in a Catalog
Of 560 catalog pages, 8 feature signature artists. Sampling 100 pages, let X count pages with signature artists.
X∼B(100,5608):
P(X=2)=0.2466P(X≤6)=0.9994P(X>3)=0.0557Answer: μ≈1.4286, σ≈1.1867.
Your turn — a high success rate, a large n
Try It Now 4.3.7 — an NBA shooter's season
A player scores with 61.3% of shots. In a random sample of 80 shots, find the mean, sd, and two exact/tail probabilities.
X∼B(80,0.613): μ=49.04, σ≈4.36
P(X=60)=0.0036P(X>50)=1−P(X≤50)=0.3718Worked example — which single value is more likely?
Example 4.3.7 — Comparing Two Nearby Outcomes
The lifetime risk of cancer is about 1.5%. Sampling 200 people, is five or six people developing cancer more likely?
X∼B(200,0.015): μ=3, σ=1.719, P(X≤8)=0.9965
P(X=5)=0.1011P(X=6)=0.0500Answer: Five is more likely — both sit above the mean of 3, and six is further out, so it carries less probability.
Try it in rāSHio — comparing two outcomes is two clicks
Not two formulas
Set n=200, p=0.015 once in the Exact tab, then enter k=5 (0.1011) and k=6 (0.0500). The shaded bar moves one step right and visibly shrinks — the answer before you've read either number.
§4.3.7 — the condition that fails most often
The three conditions are a test, not a formality — and the one that fails most in practice is independence. Sampling without replacement changes what is left, so p shifts trial to trial.
The multiplication pxqn−x assumed every trial carried the same probability regardless of what came before. Once that's gone, you need probabilities conditioned on earlier outcomes instead.
§4.3.7 — what replaces independence
The chance of A, already knowing B
P(A∣B)=P(B)P(A and B)In a binomial experiment this formula does no work at all — independence means P(A∣B)=P(A). The moment that stops being true, the experiment is disqualified.
Context Pause — replacement is the tell
Does it go back before the next draw?
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.
Your turn — binomial, or not?
Try It Now 4.3.8 — choosing lacrosse captains
Three captains are drawn from a hat of seniors' names, not replaced. Is checking whether all three play the same position binomial?
Answer: Not binomial. Names are not replaced, so the pool — and the probability of drawing that position — changes after each draw. Independence fails.
Worked example — p changes between draws
Example 4.3.8 — A Committee Drawn Without Replacement
A committee of 10 staff and 6 students draws a chairperson and recorder, without replacement. Are both being students independent trials?
First draw: P(student)=166. Second draw: 155 if the first was a student, but 156 if not.
Answer: Not binomial. The second draw's probability depends on the first, so independence (condition 3) fails.
Key Terminology — the vocabulary this section put to work
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∼B(n,p).
Bernoulli trial — a single trial with two outcomes; a binomial experiment with n=1.
success / failure — the outcome counted (probability p), not necessarily the desirable one; and the outcome not counted (probability q=1−p).
n, p, q — the number of trials; the probability of success per trial; the probability of failure per trial (q=1−p).
independence — one trial's outcome does not change any other trial's probabilities; the condition sampling without replacement violates.
Appendix — two-term complement
Try It Now 4.A1 — a smoking-related lung condition
A 13% chance of a severe lung condition, across four smoking friends. How likely is it that two or more develop one?
X∼B(4,0.13): P(X≥2)=1−P(X=0)−P(X=1)=1−0.5729−0.3424
Answer: About 0.0847, roughly 8.5%.
Appendix — is independence a fair assumption?
Try It Now 4.A2 — a placekicker's attempts
A placekicker takes many field-goal attempts over a season. Is treating his kicks as independent a fair assumption?
Answer: A reasonable first model, but an approximation — fatigue, confidence after a miss, weather, and a defense that adjusts can all slightly couple later attempts to earlier ones.
Appendix — beyond the binomial: waiting for a fixed count
Try It Now 4.A3 — a streak of field goals
Converting 80% of kicks independently, how likely is the fourth success by the fifth attempt? (The count of successes is fixed, not the number of trials — negative binomial, not binomial.)
Fourth on attempt 4: (0.80)4=0.4096. Fourth on attempt 5: (34)(0.80)3(0.20)(0.80)=0.3277.
Answer: 0.4096+0.3277=0.7373, about 74%.
The headline result
Three numbers describe any binomial experiment completely.
n and p fix the whole distribution; every probability comes from one formula, P(X=x)=(xn)pxqn−x.
And the same two parameters give the mean and spread directly: μ=np, σ=npq — no summing required.
† All of it depends on independence holding. Sample without replacement and p drifts trial to trial — check for replacement before reaching for this formula.
§4.3 — conclusions
Check the three conditions (fixed n, two outcomes, independence), name the success, and translate the English into an inequality. One formula then answers exact, at-most, at-least, and range questions — by hand or with rāSHio's Calculators → Binomial panel.
Replacement is the tell. If what you draw doesn't go back, p shifts between trials, independence fails, and you need conditional probability instead of the binomial formula.
Next: §4.4 — a discrete distribution built from a playing-card experiment, where you build the pdf table yourself.