A traveler's tale of counting, chance, and the quiet mathematics of "exactly how many out of how many." Walk the road with Nova the apprentice, and learn the Binomial Distribution one gate at a time.
A situation is Binomial only when all four conditions below are true at once. Think of it as a checklist — if any single item fails, the situation is not binomial, even if the other three are perfect.
Every trial has only two possible results: "success" or "failure." No third option, no partial credit.
One trial's result must never change the odds of the next trial. What happens to traveler #1 cannot influence traveler #2.
The total number of trials, n, is decided before you start counting. You never stop early or keep going based on results.
The probability of success, p, stays exactly the same on every single trial, from the first to the last.
A fair coin is flipped 8 times in a row. Nova counts the number of heads.
Five cards are drawn from a 52-card deck without replacement. Nova counts how many are aces.
Nova rolls a die again and again until the first 6 appears, then counts how many rolls it took.
A royal census worker randomly surveys 50 people from a kingdom of two million, asking "Do you enjoy the harvest festival?" (70% of the whole kingdom would say yes.)
A teacher gives a pop quiz to the same 10 students every day for a week, and each day the questions are different so no student's score affects another day's score. She counts, across the week, how many days each individual student passes (assuming each day that student has the same 80% chance of passing, independent of other days). Is counting one student's number of "pass" days across n = 5 days binomial?
Piece 1 — Arrangements: nCk counts how many different orders the k successes could land in among n trials (e.g. success-success-fail vs fail-success-success are different orders, same total).
Piece 2 — Success chance: pk is the chance that all k successes happen, multiplied together.
Piece 3 — Failure chance: (1 − p)n−k is the chance that the remaining (n−k) trials all fail.
Multiply the three pieces, and you get the exact probability of landing precisely k successes.
where n = number of trials, k = number of successes wanted, p = probability of success on one trial
A fair coin (p = 0.5) is flipped n = 5 times. Find the probability of getting exactly k = 3 heads.
A castle archer hits her target with probability p = 0.7 on each shot. She fires n = 6 arrows. Find the probability that exactly k = 4 hit.
A 10-question multiple-choice quiz has 4 options per question, so a random guess succeeds with p = 0.25. Find the probability of exactly k = 2 correct guesses out of n = 10.
A knight hits a dragon with p = 0.6 per arrow, firing n = 4 arrows. Find the probability of at least 3 hits — this needs two k-values added together.
A fair coin is flipped n = 4 times. Which expression correctly gives P(exactly 2 heads)?
Mean (μ) is the expected number of successes if you repeated the whole experiment many, many times and averaged the results — the "typical" number of successes you'd expect. It is simply n × p.
Standard deviation (σ) measures how much the actual number of successes tends to swing above or below that mean, trial-run to trial-run. A small σ means results cluster tightly near μ; a large σ means results scatter more widely.
A farmer plants n = 20 seeds, each with a p = 0.3 chance of sprouting. Find the mean and standard deviation of the number that sprout.
A fair coin is flipped n = 100 times. Find the mean and standard deviation of the number of heads.
Recall the archer with p = 0.7, firing n = 6 arrows. Find her mean and standard deviation of hits.
A dice game has n = 50 trials with p = 0.4 chance of success each. What is the mean number of successes?
binompdf(n, p, k) — "probability density function" — instantly gives P(X = k), the exact same number you'd get from the formula in 3.2, without doing any arithmetic by hand.
binomcdf(n, p, k) — "cumulative density function" — gives P(X ≤ k), the sum of every probability from X = 0 all the way up through X = k.
On a TI-84: press 2nd → VARS (DISTR) → scroll to binompdf or binomcdf → enter n, p, x in that order.
Confirm Case 1 from Chapter 3.2 using the calculator: 5 coin flips, find P(exactly 3 heads).
The archer (n = 6, p = 0.7) wants to know: what's the chance she lands 4 hits or fewer?
A free-throw shooter makes p = 0.8 of her shots, attempting n = 10. Find the probability of at least 8 makes.
You want P(X ≤ 6) for a binomial with n = 15, p = 0.4. Which spell do you cast?
Binary outcomes · Independent trials · Number of trials fixed (n) · Same probability (p) each time. All four must hold for a binomial setting.
Gives the exact probability of precisely k successes out of n trials.
μ is the expected number of successes; σ measures typical spread around μ.
Use 1 − binomcdf(n,p,k) for P(X>k), the complement trick.