In simple terms
A friendly intro before the formal notes — no formulas yet.
Counting Successes in a Fixed Number of Tries
The binomial distribution answers one question: if I repeat the same yes/no experiment a fixed number of times, how likely is each possible number of 'successes'? It is the maths behind any pass/fail, hit/miss or defective/good situation that you repeat under identical conditions.
Imagine taking exactly 10 penalty kicks. Each kick is a goal ('success') or a miss ('failure'), your chance of scoring each one is the same, and no kick affects the next. The binomial distribution lets you compute the probability of exactly 7 goals, or at least 8 goals, out of the 10 — turning 'how many will go in?' into an exact number.
- 1
Check the four conditions: a fixed number of trials , exactly two outcomes per trial, a constant success probability , and independent trials.
- 2
Write the distribution as , naming what counts.
- 3
Translate the words into a probability statement: 'exactly' is , 'at most' is , 'at least ' is .
- 4
Use the GDC (binomPdf for a single value, binomCdf for a cumulative one) and quote the answer, usually to 3 significant figures, in context.
Explore the concept
Use the live diagram and synced steps — play it or tap a step card to walk through.
Key formulas
Tap any symbol to reveal exactly what it means and its units.
$E(X)=\mu=np,\qquad \text{Var}(X)=\sigma^2=np(1-p)=npq.$
Full topic notes
Formal explanation with the rigour you need for the exam.
Recognising a binomial situation
Before any calculation, you must be sure the situation is genuinely binomial. Four conditions must all hold, and they are worth checking explicitly because a single failed condition changes the model entirely. A handy memory aid is the acronym BINS.
The condition most often overlooked is constant with independence. Sampling without replacement — drawing counters from a bag and not putting them back — changes the probability from one draw to the next, so those trials are neither independent nor identically distributed and the situation is not binomial. Sampling with replacement, or an experiment repeated under identical conditions, keeps constant and is binomial.
Binary: each trial has exactly two outcomes, which we label 'success' and 'failure'. A coin toss is heads (success) or tails (failure).
Independent: the outcome of one trial has no effect on any other. Rolling a die repeatedly gives independent trials; drawing cards without replacement does not.
Number: the number of trials is fixed in advance. You decide to toss the coin 20 times, so .
Success: the probability of success is the same on every trial. The probability of failure is then .
Notation and the probability formula
When a random variable counts the number of successes and satisfies the four conditions, we write it in the shorthand . This states everything at once: a binomial distribution with trials and success probability . Naming the distribution this way is not decoration — in an exam it is the line that earns the first method mark, because it shows the examiner you have identified the correct model.
The probability of exactly successes in trials is where counts the number of ways to arrange successes among the trials.
Read the formula as three factors: counts the arrangements, is the probability of successes, and is the probability of the remaining failures. The two exponents must always add to ; if they do not, you have mixed up and . On Paper 2 you rarely evaluate this by hand — the GDC command binomPdf(n, p, x) does it — but understanding the structure is what lets you set problems up correctly.
Cumulative probabilities: 'at least' and 'at most'
Exam questions rarely stop at 'exactly'. They ask for ranges: 'at most 3', 'at least 10', 'fewer than 5', 'between 8 and 12'. These are cumulative probabilities, and your GDC's binomCdf function is built for them. The single most important habit is to translate the English into a probability statement before touching the calculator, because that statement is where the method mark lives and where careless slips are caught.
binomPdf(n, p, x) gives . Use for 'exactly '.
binomCdf(n, p, x) gives . Use for 'at most ', 'no more than ', ' or fewer'.
'At least ' (): , i.e. 1 - binomCdf(n, p, k-1).
'More than ' (): , i.e. 1 - binomCdf(n, p, k).
'Between and inclusive' (): .
The most common error in this whole topic is mishandling the boundary of an inequality. For 'at least 5' () you subtract the probability of '4 or fewer' (), NOT '5 or fewer'. Always write the statement on the page first: it earns the method mark and forces you to get the right.
Mean and variance
For any distribution we want a measure of centre (the mean) and of spread (the variance). For the binomial these come from two remarkably simple formulas that you should be able to write down instantly — they are frequently worth an easy method-and-answer pair of marks.
For : The standard deviation is .
The mean is the average number of successes over many repetitions. Toss a fair coin 100 times (, ) and you expect heads, with variance and standard deviation . Note that the mean need not be a whole number even though itself only takes whole values — it is a long-run average, not a single outcome.
Common mistakes examiners penalise
Applying the binomial model when a condition fails — sampling without replacement makes change from trial to trial, so the trials are not independent and the situation is not binomial. Always check all four BINS conditions first.
Swapping and in the formula — in , is the number of trials and the number of successes. The two exponents must add to ; if they don't, you've mixed them up.
Getting the boundary wrong for 'at least' — 'at least ' is , so subtract at , not . Writing the probability statement before using the GDC prevents this.
Adding cases for 'at least one' instead of using the complement — is one short line; summing every case invites arithmetic error and wastes time.
Confusing with — because is discrete, and . Shifting a strict inequality by one is a classic dropped mark.
Using where is needed — the mean is but the variance is ; the standard deviation is the square root of the variance, not of the mean.
Over-rounding mid-calculation — carry full calculator accuracy through the working and round only the final answer, normally to 3 significant figures with any units.
Model answer — marked the way our engine marks it
On Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) for the correct approach, or an accuracy mark (A) for the correct value that depends on that method being present. Follow-through (FT) means a wrong number early on need not cost you the marks that follow, provided your later steps are correct on your own figure. The engine also applies 'ignore subsequent working' (ISW) once a correct answer is seen, accepts equivalent forms, and accepts any answer that rounds correctly. But that protection only exists if the method is written down. Study how each mark below is earned by a specific line.
Where this leads
The binomial distribution is your first named probability model, and the habits it teaches carry forward. Recognising conditions before choosing a model, translating words into a precise probability statement, and using the complement for 'at least' all reappear in the normal distribution that follows, where the same GDC-and-statement discipline applies to continuous data. The mean and variance formulas foreshadow the more general expectation algebra of HL. Master the routine here — check BINS, write , state the probability, read it off, round sensibly — and the rest of the statistics topic becomes variations on a method you already own.
Worked examples
See the formulas applied — reveal one step at a time, like the exam.
A factory's light bulbs are defective with probability 0.08, independently of one another. An inspector selects 15 bulbs at random. Find the probability that exactly 2 are defective. [3]
- 1
Identify the distribution. A trial is checking one bulb: defective (success) or not (failure); bulbs, constant , independent. Let be the number of defective bulbs, so . [M1: correct model]
In a large school, 65% of students own a GDC. A teacher randomly selects a class of 25 students. (a) Find the probability that exactly 18 students own a GDC. (b) Find the probability that at least 20 students own a GDC. [5]
- 1
Identify the distribution. Let be the number of the 25 students who own a GDC. With and constant , . [M1: correct model]
A biased coin has probability 0.3 of landing heads. It is tossed 40 times and is the number of heads. (a) State the distribution of . (b) Find the mean and standard deviation of . (c) Find if instead the coin were tossed 80 times. [5]
- 1
(a) . [A1: distribution stated]
A fair die is rolled 10 times. Find the probability of exactly 3 sixes, and the probability of at least one six. [5]
- 1
Model answer — full working.
How it all connects
The big idea sits in the middle — tap a linked idea to explore the link.
Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.
Glossary
Try to recall each definition before you reveal it.
Quick check
Answer in your head first — then tap to check. No pressure.
Revision flashcards
Flip the card. Test yourself before the exam.
The four conditions for a binomial distribution
Binary: each trial has exactly two outcomes (success/failure). Independent: trials do not affect one another. Number: a fixed number of trials . Success: the probability of success is constant for every trial. (Acronym: BINS.)
Key takeaways
Review these before you close the topic — retrieval beats re-reading.
- ✓
Binary: each trial has exactly two outcomes, which we label 'success' and 'failure'. A coin toss is heads (success) or tails (failure).
- ✓
Independent: the outcome of one trial has no effect on any other. Rolling a die repeatedly gives independent trials; drawing cards without replacement does not.
- ✓
Number: the number of trials is fixed in advance. You decide to toss the coin 20 times, so .
- ✓
Success: the probability of success is the same on every trial. The probability of failure is then .
Practice — then mark it
The whole point: a real Cambridge question, marked mark-by-mark.
Get a Paper 2 binomial question marked: solve it with full working
Get a Paper 2 binomial question marked: solve it with full working
Extra simulations & links
PhET, GeoGebra and other curated tools — open in a new tab.
Frequently asked
Checkpoint
One marked question is worth ten re-reads — close the loop before you move on.
Reading it isn’t knowing it — prove it.
Before you move on: do Get a Paper 2 binomial question marked: solve it with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.