Statistics

The binomial setting

Twelve seeds go into a tray and you count how many sprout. When does a count like that follow one standard pattern, so that a single formula can give every one of its probabilities?

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Compound eventsGrades 7 to 11 Independent events and the multiplication ruleMath 7

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The geometric distributionStatistics

Where you are 4 in The binomial distribution

Probability4

Random variables11

The binomial distribution4

Mean and standard deviation of a binomial Shape of a binomial and the normal approximation The binomial probability formula

The binomial setting

When you reach for it

Bolormaa, Ganbaatar and Oyuna run the seed trays at a community garden. Bolormaa fills a tray with 12 seeds of one kind of bean, one seed to a cell. From the seed company’s germination tests she takes each seed to sprout with probability 0.85, each seed on its own (made-up data, as is every number in this lesson). The question she cares about is a count: how many of the 12 will sprout?

Call that count X. It can be any whole number from 0 to 12, and it is a random variable: its value is not known until the tray has had its week. Finding its probabilities from scratch means listing outcomes of a compound event with 12 stages, which is hopeless by hand: each seed has two outcomes, so a tray has 2^{12} = 4{,}096 possible patterns of sprouts and failures.

Many counts like this share one structure, and when they do, a single formula gives every probability (the next lesson). This lesson is the test that comes first: checking whether a count has that structure, called the binomial setting. Reach for it whenever a question asks “how many of these will...” and before any binomial formula is used. Day 1 checks the four conditions on settings that pass; Day 2 studies settings where one fails.

The procedure

A chance process is a binomial setting when it is a run of repeated trials and all four of these hold.

Binary outcomes. Each trial ends in one of two outcomes, called success and failure. A seed sprouts or it does not.
Independent trials. Knowing how some trials came out does not change the probabilities for the others. Bolormaa’s seeds sit in separate cells and do not affect one another.
Number of trials fixed. The number of trials, n, is set before the process starts. The tray holds n = 12 seeds whatever happens.
Same chance of success. Every trial has the same probability of success, p. Each seed sprouts with p = 0.85.

The random variable that goes with the setting is X, the number of successes in the n trials. A random variable that counts successes in a binomial setting is a binomial random variable with parameters n and p; its possible values are 0, 1, 2, \ldots, n. Bolormaa’s X is binomial with n = 12 and p = 0.85.

Each condition is there because the binomial formula needs it, and seeing why is the best guide to checking it.

Why two outcomes. The variable counts one kind of outcome. If a trial could end three ways, a count of one of them is still possible, but only after the other two are lumped together as “failure”. So the first job is always to say, in words, what a success is. A success is simply the outcome being counted; it need not be a good thing. A seed that fails to sprout can be the success.

Why independence. The probability of a whole pattern, such as “the first seed sprouts, the second fails, the rest sprout”, comes from the multiplication rule for independent events: when one event does not change the chance of another, the chance that both happen is the product of their chances. That product is allowed only if the trials are independent. When they are not, the pattern’s probability is something else and the binomial formula gives a wrong answer.

Why a fixed number of trials. The formula answers “how many successes out of n”. If the number of trials is not set beforehand, for instance if planting stops at the first sprout, then there is no n to put in it, and the random variable is counting something else.

Why the same chance. With the same p on every trial, any pattern with k successes and n - k failures has probability made of k factors of p and n - k factors of 1 - p, in some order. Since the order of the factors does not change a product, every such pattern has the same probability. That is what lets the next lesson count patterns instead of listing them. If p changes from trial to trial, two patterns with the same count can have different probabilities.

Figure 1One possible outcome for Bolormaa’s tray. Each cell is one trial, sprout (S) or fail (F), and the binomial count is the number of S cells, here X = 10 (made-up data).

Checking the conditions is a judgment about how the data arise, stated in context with a reason for each.

Worked examples

Example 1

Day 1: Bolormaa’s tray

Bolormaa plants the 12 seeds described above, each in its own cell, each sprouting with probability 0.85, independently. Let X be the number that sprout. Check that X is binomial, give n, p and the possible values of X, and find the probability that every seed sprouts.

Solution

Check the four conditions in context.

Binary: each seed either sprouts or fails to sprout within the week. A success is “the seed sprouts”.
Independent: the seeds are in separate cells, so the outcome of one seed does not change the chance for another. This is an assumption about the tray, and it is a reasonable one when the cells share no cause that could make seeds fail together (Example 7 shows such a cause).
Fixed number: n = 12, set when the tray was filled.
Same chance: every seed is the same kind under the same conditions, so p = 0.85 for each.

So X is binomial with n = 12 and p = 0.85, and its possible values are 0, 1, 2, \ldots, 12. Figure 1 shows one possible outcome, X = 10.

“Every seed sprouts” is the single pattern of 12 successes. By independence its probability is the product of the twelve chances:

P(X = 12) = 0.85^{12} \approx 0.1422

Even with seeds this reliable, a full tray of sprouts happens only about 14 times in 100.

Example 2

Day 1: counting failures instead

Ganbaatar looks at the same tray and counts the seeds that do not sprout; call that count Y. Is Y binomial? If so, give n and p. What is Y when X = 10? Find the probability that seeds 1 and 2 fail and the other ten sprout, and compare it with the probability that seeds 11 and 12 fail and the first ten sprout.

Solution

Nothing about the tray has changed, only which outcome is called a success. Now a success is “the seed fails to sprout”, which has probability 1 - 0.85 = 0.15 on every seed. The outcomes are still two, the seeds still independent, and there are still n = 12 of them. So Y is binomial with n = 12 and p = 0.15.

Each seed is counted in exactly one of X and Y, so X + Y = 12. When X = 10, Y = 12 - 10 = 2: ten sprouts and two failures.

For the pattern “seeds 1 and 2 fail, seeds 3 to 12 sprout”, independence lets the probabilities multiply, two factors of 0.15 and ten of 0.85:

0.15^2 \times 0.85^{10} \approx 0.0044

The pattern “seeds 11 and 12 fail, seeds 1 to 10 sprout” has the same ten factors of 0.85 and two of 0.15, in a different order, so its probability is the same, about 0.0044. This is the same-chance condition at work: within a binomial setting, only the number of successes affects a pattern’s probability, not where they fall.

Example 3

Day 1: making the outcomes binary

Oyuna sorts each seedling at the end of the week into one of three groups: strong, weak, or did not sprout, with probabilities 0.60, 0.25 and 0.15 for each seed, independently. She plants 12 seeds. For each variable, say whether it is binomial, and if so give n and p.

(a) S, the number of strong seedlings. (b) T, the number of seeds that sprouted at all. (c) H, the total height of all 12 seedlings in centimeters.

Solution

(a) Each seed has three outcomes, but the variable counts only one of them. Call “strong” a success and lump “weak” and “did not sprout” together as failure. Then each trial is binary with p = 0.60, the seeds are independent, and n = 12. So S is binomial with n = 12 and p = 0.60.

(b) Here a success is “strong or weak”. The two groups cannot both happen to one seed, so their probabilities add: p = 0.60 + 0.25 = 0.85. The other conditions hold as before, so T is binomial with n = 12 and p = 0.85, the same model as Bolormaa’s X.

(c) H is not a count of successes. A seedling’s height can be any of a whole range of values, and the variable adds measurements instead of counting trials of one kind. No choice of “success” turns a total of heights into a number of successes, so H is not binomial.

Example 4

Day 2: drawing from a small packet

Bolormaa has a small packet of 10 seeds: 8 are coated with a seed treatment and 2 are plain. Without looking, she draws 4 seeds one at a time, without putting any back, and lets C be the number of coated seeds drawn. (a) Is C binomial? (b) Compare with drawing 4 seeds from a sack of 2{,}000 seeds, 1{,}600 of them coated.

Solution

(a) Each draw is binary (coated or plain) and there are n = 4 draws. The trouble is the other two conditions. The first draw is coated with probability \tfrac{8}{10} = 0.8. The second depends on the first, as the right-hand tree of Figure 2 shows:

if the first seed was coated, 7 of the 9 left are coated, so the chance is \tfrac{7}{9} \approx 0.778;
if the first seed was plain, 8 of the 9 left are coated, so the chance is \tfrac{8}{9} \approx 0.889.

Knowing the first draw changes the chance for the second, so the draws are not independent, and the chance of a coated seed is not the same on every draw. C is not binomial. Its exact probabilities come from counting draws without replacement, the subject of Probability with counting formulas.

Figure 2In the tray the chance on the second seed is the same on every branch. In the small packet it depends on the first draw, so the draws are neither independent nor at a constant chance (made-up data).

(b) From the sack the first draw is coated with probability \tfrac{1600}{2000} = 0.8. After a coated seed, the chance for the second is

\frac{1599}{1999} \approx 0.7999

and after a plain seed it is \tfrac{1600}{1999} \approx 0.8004. The draws are still not exactly independent, but the chance barely moves, because 4 seeds are only \tfrac{4}{2000} = 0.002, or 0.2\%, of the sack. Treating C as binomial with n = 4 and p = 0.8 is a close approximation. The rule of thumb for when this is safe, a sample of less than a tenth of the population, is the The ten percent condition.

Example 5

Day 2: planting until the first sprout

Ganbaatar has a packet of the same bean seeds, each sprouting with probability 0.85 independently. He plants one seed a day and stops on the first day a seed sprouts. Let D be the number of seeds he plants. Is D binomial? Compare with planting exactly 5 seeds and counting the sprouts.

Solution

Three conditions hold: each seed sprouts or fails, the seeds are independent, and each has the same chance 0.85. The fourth fails. The number of trials is not fixed in advance; it is decided by the outcomes, since planting stops at the first success. And D does not count successes at all: it always ends with exactly one sprout. It counts trials. D could be 1, 2, 3, or any larger whole number, with no upper limit, whereas a binomial count can never exceed its n. So D is not binomial.

A count of trials up to the first success in repeated independent trials with the same p is a geometric setting, which has its own formula in The geometric distribution.

If Ganbaatar instead plants exactly 5 seeds and counts how many sprout, all four conditions hold and the count is binomial with n = 5 and p = 0.85, with possible values 0 to 5. The seeds are the same; the question asked of them decides the model.

Example 6

Day 2: when the soil dries

Oyuna’s tray of 12 seeds sits partly under a vent. The 6 cells away from the vent keep moist, and a seed there sprouts with probability 0.85. The 6 cells under the vent dry out, and a seed there sprouts with probability 0.60. The seeds are independent. Let X be the number of the 12 that sprout. Is X binomial? Find the probability that all 12 sprout.

Solution

The outcomes are binary, n = 12 is fixed, and the seeds are independent, but the chance of success is not the same: it is 0.85 for half the seeds and 0.60 for the other half. So X is not binomial. A pattern with ten sprouts in which both failures are under the vent has a different probability from one in which both failures are in the moist cells, and that breaks the counting that the binomial formula depends on.

Independence still holds, so the multiplication rule still gives the chance of a single pattern:

0.85^6 \times 0.60^6 \approx 0.0176

Each half of the tray on its own is binomial: the sprouts in the moist half follow n = 6, p = 0.85, and those under the vent n = 6, p = 0.60. Their total is not binomial with any single p.

Example 7

Day 2: a shared cause

All of Ganbaatar’s 12 cells sit on one watering timer. On a given week the timer works with probability 0.9. If it works, each seed sprouts with probability 0.9, independently of the others; if it fails, the soil stays dry and no seed sprouts. Let X be the number of sprouts. Find the probability that a given seed sprouts, then the probability that two given seeds both sprout, and decide whether X is binomial.

Solution

A given seed sprouts only if the timer works and then the seed sprouts. By the multiplication rule along that path, and since the other path gives no sprouts at all,

P(\text{a seed sprouts}) = 0.9 \times 0.9 = 0.81

for every seed. So the outcomes are binary, n = 12 is fixed, and the chance is the same, p = 0.81, on every seed.

For two given seeds to sprout, the timer must work and then both seeds must sprout:

P(\text{both sprout}) = 0.9 \times 0.9 \times 0.9 = 0.729

If the seeds were independent, this would be 0.81 \times 0.81 = 0.6561. It is larger: learning that one seed sprouted tells you the timer worked, which raises the chance for the other seed from 0.81 to 0.9. The seeds are not independent, because they share a cause. So X is not binomial, even though three of the four conditions hold.

Boundaries

The binomial setting is a model. Whether it fits is judged from how the data arise: separate cells, one kind of seed, a count fixed in advance. It cannot be read off the numbers.
Binary means each trial has two outcomes for the purpose of the count. More outcomes are fine if they are lumped into success and failure (Example 3), but a total of measurements, such as heights, is not a count of successes and is never binomial.
Independence fails when trials share a cause (one timer, one heat mat, seedlings crowding each other) or when drawing without replacement from a small group (Example 4). Drawing a small part, less than a tenth, of a large group is close enough to independent; see The ten percent condition.
Fixed n fails when the process stops because of its outcomes. Counting trials until the first success is the geometric setting, not the binomial one.
Same p fails when conditions differ between trials (Example 6). Each group with its own chance may be binomial, but their combined count is not.
Any one condition failing is enough to rule the binomial model out. Example 7 meets three conditions and still fails.
A single trial is the smallest case: n = 1, and the count is 1 with probability p and 0 with probability 1 - p.

Where this goes next

Once a count passes all four checks, The binomial probability formula gives the probability of exactly k successes from n, p and k alone, by counting the patterns with k successes and multiplying by the probability that every one of them shares. Mean and standard deviation of a binomial gives the count’s mean and standard deviation from n and p, and Shape of a binomial and the normal approximation shows when a normal curve can stand in for the whole distribution. Example 5’s contrast leads to The geometric distribution, where the question is how long the wait for the first success is. All of these belong to the The binomial distribution and its neighbors.

Exercises

All data are made up. Round probabilities to four decimal places.

Day 1: checking the four conditions

Bolormaa plants 20 lettuce seeds in a tray, one to a cell. Each sprouts with probability 0.7, independently. Let X be the number that sprout. Is X binomial? If so, give n, p, the possible values of X, and the probability that a given seed fails.
In the tray of exercise 1, let Y be the number of seeds that fail. Is Y binomial? If so, give n and p. If 15 seeds sprout, what is Y?
Ganbaatar fills 3 trays of 12 bean seeds each, every seed sprouting with probability 0.85 independently, and counts the sprouts across all three trays. Is the total binomial? If so, give n and p.
Each of Oyuna’s 15 seedlings ends the month tall, medium or short, with probabilities 0.2, 0.5 and 0.3, independently. For each count, say whether it is binomial and give n and p if it is: (a) the number of tall seedlings; (b) the number that are not short.
Ganbaatar weighs the bean pods from each of 12 plants and records the total weight in grams. Is the total binomial? Explain.
Oyuna plants 4 bean seeds, each sprouting with probability 0.85, independently. Find the probability that the first three sprout and the fourth fails, and explain why the pattern “the first fails and the other three sprout” has the same probability.
Bolormaa plants 5 lettuce seeds of the kind in exercise 1. Find the probability that all five sprout.
A seed sprouts with probability 0.85, and a seed that sprouts survives being moved to the garden bed with probability 0.9. Bolormaa plants 10 seeds and counts the plants that both sprout and survive the move. Each seed’s fate is independent of the others. Is the count binomial? If so, give n and p.

Day 2: when a condition fails

A packet holds 12 seeds, 9 coated and 3 plain. Bolormaa draws 3 without replacement and counts the coated ones. Find (a) the probability that the first seed is coated, (b) the probability that the second is coated if the first was coated, and (c) the probability that the second is coated if the first was plain. (d) Is the count binomial?
In exercise 9, Bolormaa instead puts each seed back and shakes the packet before the next draw. Is the count of coated seeds now binomial? If so, give n and p.
A sack holds 5{,}000 seeds, 4{,}500 of them coated. Oyuna draws 20 without replacement. (a) What fraction of the sack does she draw? (b) If the first seed drawn is coated, what is the probability that the second is coated? (c) Is a binomial model reasonable for the number of coated seeds? If so, give n and p.
Ganbaatar plants lettuce seeds one at a time and stops when the first one sprouts. Is the number of seeds he plants binomial? Which condition fails?
Oyuna plants bean seeds one at a time until 5 have sprouted and counts the seeds used. Is the count binomial? Which condition fails?
In a tray of 12 seeds, the 8 cells by the window sprout with probability 0.85 each and the 4 cells by the door, where it is cold, with probability 0.5 each, all independently. (a) Is the number of sprouts in the whole tray binomial? (b) Is the number of sprouts by the window binomial? (c) Find the probability that all 4 seeds by the door sprout.
Bolormaa plants one seed a day for 7 days in soil that is drying out, so the seed planted on day 1 sprouts with probability 0.85, and each later day’s chance is 0.05 lower than the day before. The seeds are independent. (a) Is the number of sprouts binomial? (b) Find the probability that all 7 sprout.
A tray of seeds sits on one heat mat. The mat works with probability 0.8; if it works, each seed sprouts with probability 0.9, and if it fails, with probability 0.4, independently in either case. Find (a) the probability that a given seed sprouts, (b) the probability that two given seeds both sprout, and (c) decide whether the number of sprouts is binomial.
Oyuna sows 2 seeds in each of 6 cells. In a cell where both sprout, the stronger seedling shades the other, which then dies. She counts the seedlings alive after two weeks among the 12 seeds. Explain why this count is not binomial.

Reasoning

Ganbaatar says: “A seed either sprouts or it doesn’t, so the number of sprouts in any tray is binomial.” Explain why two outcomes per seed is not enough, using an example from this lesson.
Explain why, in a binomial setting, every pattern with the same number of successes has the same probability, and name the two conditions this uses.
In a binomial setting with n trials, let X count successes and Y count failures. Explain why Y is binomial too, with which n and p, and why X + Y = n.
Explain why a binomial count can never exceed n, and why that rules out “the number of seeds planted until the first sprout” as a binomial variable.

Answers

Yes: binary, independent, n = 20, same p = 0.7. X takes the values 0, 1, \ldots, 20. A given seed fails with probability 1 - 0.7 = 0.3.
Yes, with success “the seed fails”: n = 20, p = 0.3. If 15 sprout, Y = 20 - 15 = 5.
Yes. The 3 \times 12 = 36 seeds are 36 independent trials with the same chance, so the total is binomial with n = 36 and p = 0.85 (assuming the trays share no cause that makes seeds fail together).
(a) Binomial with success “tall”: n = 15, p = 0.2. (b) Binomial with success “tall or medium”: n = 15, p = 0.2 + 0.5 = 0.7.
No. A total weight adds measurements; it is not a count of trials that end in success, so the outcomes are not binary.
0.85^3 \times 0.15 \approx 0.0921. The other pattern has the same three factors of 0.85 and one of 0.15 in a different order, and the order of factors does not change a product.
0.7^5 \approx 0.1681, by the multiplication rule for the five independent seeds.
Yes. Each seed is a trial, a success is “sprouts and survives”, with p = 0.85 \times 0.9 = 0.765 for every seed; the seeds are independent and n = 10.
(a) \tfrac{9}{12} = 0.75. (b) \tfrac{8}{11} \approx 0.7273. (c) \tfrac{9}{11} \approx 0.8182. (d) No: the chance on the second draw depends on the first, so the draws are not independent and p is not the same.
Yes. Each draw is from the full packet, so every draw is coated with probability \tfrac{9}{12} = 0.75 whatever came before: n = 3, p = 0.75.
(a) \tfrac{20}{5000} = 0.004, or 0.4\%. (b) \tfrac{4499}{4999} \approx 0.9000. (c) Yes: the sample is far less than a tenth of the sack, so the draws are close to independent, with n = 20 and p = 0.9.
No. The number of trials is not fixed; it depends on when the first sprout appears. This is a geometric setting.
No. The number of trials is not fixed; planting stops when the fifth sprout appears, and the variable counts seeds planted, not successes in a set number of trials.
(a) No: the chance is 0.85 for some seeds and 0.5 for others. (b) Yes, n = 8, p = 0.85. (c) 0.5^4 = 0.0625.
(a) No: the chance changes from day to day. (b) The chances are 0.85, 0.80, \ldots, 0.55, and by independence their product is 0.85 \times 0.80 \times \cdots \times 0.55 \approx 0.0766.
(a) 0.8 \times 0.9 + 0.2 \times 0.4 = 0.72 + 0.08 = 0.80. (b) 0.8 \times 0.9^2 + 0.2 \times 0.4^2 = 0.648 + 0.032 = 0.68. (c) Not binomial: 0.80^2 = 0.64, not 0.68, so the seeds are not independent; they share the mat.
Seeds in the same cell are not independent: whether one seedling lives depends on whether its neighbor sprouted. Learning that one seedling in a cell is alive lowers the chance the other is alive; in fact the two can never both be alive.
Binary outcomes are one condition of four. Oyuna’s tray under the vent (Example 6) has two outcomes per seed but different chances, and Ganbaatar’s timer (Example 7) has two outcomes and one chance but dependent seeds. Neither count is binomial.
Independence makes a pattern’s probability the product of the trials’ chances. With the same p on every trial, a pattern with k successes is a product of k factors of p and n - k of 1 - p in some order, and reordering factors does not change a product.
Relabeling which outcome is the success changes no trial: they stay binary, independent and n in number, and the new success has chance 1 - p on every trial. So Y is binomial with n trials and probability 1 - p. Each trial is counted in exactly one of X and Y, so X + Y = n.
X counts successes among n trials, and there cannot be more successes than trials, so 0 \le X \le n. The number of seeds planted until the first sprout has no upper limit (every seed could fail for a long time), so no fixed n bounds it.