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The four binomial conditions and the formula they license, the mean and spread of a binomial count, the geometric setting and its mean, and the complement trick for 'at least one'.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can recognise a binomial setting by its four conditions and a geometric one by what it leaves out, compute exact probabilities and means for both, and use the complement when a question asks for 'at least one'. You will also have seen what happens to the distribution when the number of trials or the success rate is changed — including the fact that the spread grows like the square root of $n$, which is what every confidence interval later in the course rests on.
You can compute an expectation and a variance for a random variable given its distribution, and you know that independent variances add. Two settings turn up so often that their distributions have been worked out once and for all, and recognising them is most of the work.
Trial: one repetition, with two possible outcomes.
Binomial setting: a fixed number of independent trials, each with the same probability of success; $X$ counts the successes.
Geometric setting: independent trials with the same probability, run until the first success; $X$ counts the trials.
$\binom{n}{k}$: the number of ways $k$ successes can be arranged among $n$ trials.
Complement: the opposite event. $P(\text{at least one}) = 1 - P(\text{none})$.
Picture a machine that runs $n$ trials, each succeeding with probability $p$, and reports how many succeeded. It has two dials, and watching what each does to the output is the fastest route to understanding both models.
Turn up $n$, hold $p$. The distribution slides to the right and widens — the mean $np$ grows in proportion to $n$, but the standard deviation $\sqrt{np(1-p)}$ grows only like $\sqrt{n}$. So the count becomes more variable while the proportion becomes less so. That single asymmetry is why a national poll of 1,000 people works.
Turn up $p$, hold $n$. The distribution slides right and, past $p = 0.5$, narrows again: it is widest in the middle, where each trial is least predictable, and narrow at either end, where almost every trial goes the same way.
Set $p$ very small and $n$ very large, keeping $np$ fixed. The distribution settles into a fixed shape, piled up near zero with a long right tail — rare events in many opportunities.
Change the question from 'how many' to 'how long'. Now $n$ is not a dial at all: you run until the first success, and the count of trials is what varies. Its mean is $1/p$, and its distribution is always the same downhill shape, highest at one trial, whatever $p$ is. If the number of trials is not fixed in advance, no binomial formula can apply — there is nothing to put in place of $n$.
The binomial setting has four conditions, and all four must hold: a fixed number $n$ of trials, two outcomes each time, the same probability $p$ every time, and independence. Then $X$, the number of successes, satisfies
$$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}, \qquad \mu = np, \qquad \sigma = \sqrt{np(1-p)}$$
The formula reads left to right as a sentence: the probability of one particular arrangement is $p^k(1-p)^{n-k}$, there are $\binom{n}{k}$ arrangements, and because they are mutually exclusive they add.
The geometric setting keeps the last three conditions and drops the first: the trials run until the first success, so the number of trials is what varies. Then $P(X = k) = (1-p)^{k-1}p$ and $\mu = \dfrac{1}{p}$ — if one attempt in five works, expect five attempts.
Two habits make these usable. Check the conditions before the formula: a formula applied to the wrong setting returns a number and no warning. And for anything phrased as at least, use the complement — $P(\text{at least one}) = 1 - P(\text{none})$ — which turns a sum of many terms into a single one.
Another way: picture
Picture the binomial as $n$ boxes in a row, each ticked or not. The probability of one filled-in row is $p^k(1-p)^{n-k}$, and $\binom{n}{k}$ counts how many different rows have exactly $k$ ticks. The geometric is the same row of boxes with no right-hand end: you keep adding boxes until one is ticked.
Another way: steps
| Change | Mean $np$ | Standard deviation $\sqrt{np(1-p)}$ | Shape |
|---|---|---|---|
| $n$ doubled | doubles | grows by $\sqrt{2}$ | wider, more symmetric |
| $p$ doubled (below $0.5$) | doubles | grows | slides right |
| $p$ near $0$ or $1$ | — | small | skewed, piled at one end |
| $p = 0.5$ | $n/2$ | largest | symmetric |
The second column is the one that matters later: the spread grows like $\sqrt{n}$ while the count grows like $n$, so the proportion of successes settles down as the sample grows. Every confidence interval in this course is a consequence of that row.
"Any count of successes is binomial." Drawing five cards from a deck and counting the aces is not: the draws are not independent and $p$ changes after each one. The four conditions are a test, not a description.
"$P(X = k)$ is just $p^k(1-p)^{n-k}$." That is the probability of one particular order. There are $\binom{n}{k}$ orders, and leaving the coefficient out makes the answer far too small — usually by a factor of several hundred.
"At least one means add them all up." It means $1$ minus the probability of none, which is a single term. The long way is not wrong, only long enough that it usually goes wrong.
Fixed $n = 10$, two outcomes, constant $p$, independent: binomial.
Check first.
Mean $= 10 \times 0.7 = 7$ made throws.
$np$.
$P(X = 8) = \binom{10}{8}(0.7)^8(0.3)^2 = 45 \times 0.0576 \times 0.09 \approx 0.233$.
Coefficient times arrangement.
Adding $P(X=1) + P(X=2) + P(X=3) + P(X=4)$ would work.
The long way.
$P(\text{no six}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296}$.
One term.
$P(\text{at least one}) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518$.
Just over half.
A trial is repeated $4$ times, independently, and succeeds with probability $\frac{1}{2}$ each time. What is the expected number of successes?
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In $8$ independent trials, each succeeding with probability $\frac{1}{3}$, find $P(X = 4)$.
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A trial succeeds with probability $\frac{1}{4}$ each time, independently. How many trials are expected before the first success, counting the successful one?
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In $6$ independent trials, each succeeding with probability $\frac{1}{5}$, find the probability of **at least one** success.
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Which of these is a binomial setting?
Put the steps of computing $P(X = k)$ in the order they are done.
Number the steps in order (write the number in the box):
A machine jams on $\frac{1}{5}$ of its cycles, independently. An engineer will keep running it until it jams, then investigate. How many cycles should they expect to run?
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Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A binomial machine is set to $n = 40$ trials with a success rate of $25\%$, so its mean is $10$ successes. Using the controls, set it so that the mean is $40$ successes.
This task has no paper form; do it on a device.
You can identify and use the binomial and geometric models. Without looking: which of the four conditions does a geometric setting break, and why does $p^k(1-p)^{n-k}$ on its own give an answer that is far too small?