The binomial distribution: modelling a real-world situation

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This section starts the A Level unit on the binomial and Poisson distributions, which is examined on the Statistics 2 paper (WST02/01).

Across the three most recent S2 papers, roughly a quarter of the marks came from the binomial and Poisson distributions: setting up a model, criticising it in context, reading tables and approximating one distribution by the other.

S2 assumes the whole of S1, and this unit leans on two pieces of it: discrete random variables (a probability distribution whose probabilities add to 1, with a mean and a variance) and the binomial coefficients $\binom{n}{r}$ from the binomial expansion in your pure units.

How the marks work

M1 rewards a correct method applied to this question's numbers; a slip in arithmetic can still earn it. dM1 depends on the M mark before it, so it is lost if that one was.

A1 is for a correct answer and is only available after its M mark. A1* sits on a result printed in the question: every step must be visible, including the line just before the printed value. cso means "correct solution only".

B1 is an independent mark: a model with its parameters, a table value to 4 d.p., an assumption in context.

ft (follow through) lets a later mark stand when you use your own wrong earlier answer correctly; awrt means "anything which rounds to"; isw means correct work is not cancelled by later extra work. A misread that does not make the question easier costs two A or B marks.

S2 test rules (taught in the hypothesis-testing unit, and built on the tables here): hypotheses in symbols such as $p$ or $\lambda$ , attached to $\mathrm{H}_0$ and $\mathrm{H}_1$ ; a critical region written as values of the variable, never as a probability; a plain decision, then a conclusion in context; nothing that contradicts it.

S2 habits: table values to 4 d.p. as printed, or calculator values to the same accuracy; final answers to 3 s.f. unless the stem asks for something else.

The idea: counting successes in repeated trials

A basketball player takes 10 free throws.

A packet holds 12 seeds.

A quiz has 8 multiple-choice questions and a student guesses every one.

In each case the same thing is done a fixed number of times, each attempt either "works" or "does not", and we want to know how many times it works.

The count can be $0, 1, 2, \ldots$ up to the number of attempts, and we want a probability for each value.

The binomial distribution is the model for that count.

It is built from four conditions, and the whole skill of "modelling a real-world situation" is checking those four conditions for the situation in front of you:

1. There is a fixed number of trials, $n$ .

2. Each trial has two possible outcomes, called success and failure.

3. The trials are independent: the outcome of one does not change the probability for another.

4. The probability of success, $p$ , is the same on every trial.

If $X$ is the number of successes, we write $X \sim \mathrm{B}(n, p)$ , and the two numbers in the brackets are the parameters.

Naming the distribution means naming both: "binomial" on its own is not a model, because it does not say which binomial.

Why the formula is what it is

Take $n = 4$ trials with $p = 0.3$ , and ask for $\mathrm{P}(X = 2)$ .

One way to get exactly two successes is SSFF, with probability $0.3 \times 0.3 \times 0.7 \times 0.7 = 0.3^{2} \times 0.7^{2}$ , using independence to multiply.

But SFSF, SFFS, FSSF, FSFS and FFSS also give two successes, and each of them has the same probability $0.3^{2} \times 0.7^{2}$ : the order changes, the factors do not.

There are $\binom{4}{2} = 6$ such arrangements, because we are choosing which 2 of the 4 places hold the S. So

\mathrm{P}(X = 2) = 6 \times 0.3^{2} \times 0.7^{2} = 0.2646.

The same argument for $n$ trials and $r$ successes gives the probability function

\mathrm{P}(X = r) = \binom{n}{r} p^{r} (1 - p)^{n - r}, \qquad r = 0, 1, 2, \ldots, n.

Each factor has a job: $\binom{n}{r}$ counts the arrangements, $p^{r}$ is the chance of the $r$ successes, and $(1 - p)^{n - r}$ is the chance of the $n - r$ failures.

The probabilities do add to 1, and the binomial expansion from your pure units shows why: $\sum_{r} \binom{n}{r} p^{r} (1 - p)^{n - r} = \big(p + (1 - p)\big)^{n} = 1^{n} = 1$ .

That is where the name "binomial" comes from.

Learn it or look it up? The binomial probability function, its mean $np$ and its variance $np(1 - p)$ are printed in the Mathematical Formulae and Statistical Tables booklet, together with cumulative binomial tables.

What the booklet cannot give you is the judgement about whether the model fits.

What the shape tells you

The chart below shows three binomial distributions with $n = 10$ .

When $p = 0.5$ the distribution is symmetrical about 5.

When $p = 0.2$ most of the probability is on small values and there is a tail to the right (positive skew).

When $p = 0.8$ the picture is the mirror image, because counting successes with $p = 0.8$ is the same as counting failures with $p = 0.2$ .

That mirror is the reason the booklet's binomial tables only go up to $p = 0.5$ , which you will use in the section on cumulative probabilities.

Bar chart of P(X = r) for r = 0 to 10 for three binomial distributions with n = 10. B(10, 0.2) peaks at r = 2 (about 0.30) and is skewed with a long tail to the right; B(10, 0.5) is symmetrical about r = 5 (about 0.25); B(10, 0.8) is the mirror image of B(10, 0.2), peaking at r = 8.

Commenting critically: does the model fit?

You are expected to "comment critically" on whether a model is appropriate.

That means testing each condition against the context, and saying in the context's own words what could go wrong.

The usual failures are:

Not independent. Choosing from a small group without replacement changes the probabilities after each pick: 5 cards from a pack of 52 is not binomial for "number of hearts", since each heart removed makes the next less likely. Weather on consecutive days, or a player whose confidence rises after a success, also breaks independence.

$p$ not constant. Bulbs taken from two machines with different defect rates; a learner whose success rate improves with practice; seeds stored in different conditions.

No fixed $n$ . "The number of throws needed to score" has no fixed number of trials, so it is not binomial at all.

More than two outcomes that matter. This one is usually fixed by grouping: "a six" and "not a six".

When a question asks for an assumption, write one condition in the context's words: "each seed germinates independently of the others", or "the probability that a customer pays by card is the same for every customer".

A bare list of the four conditions scores nothing, and neither does a Poisson condition such as "events occur at a constant rate", which belongs to a different model.

Sampling from a large population is the exception to the without-replacement worry: if 3 bulbs are taken from a batch of 10 000 of which 2% are faulty, removing one faulty bulb changes the proportion so little that $\mathrm{B}(3, 0.02)$ is an excellent model.

How it connects

The binomial is the first named distribution in the course, and much of S2 grows out of it.

The section on the Poisson distribution shows what happens when $n$ is large and $p$ is small.

Later in S2 a binomial with large $n$ is approximated by a Normal distribution, and hypothesis tests use binomial tables to test a claim about $p$ : "is this coin biased?", "has the defect rate fallen?".

In genetics, the number of offspring with a trait is binomial; in quality control, the number of faulty items in a sample is the classic case; in data science, a "conversion rate" on a website is a binomial $p$ .

Key terms

Binomial distribution: $X \sim \mathrm{B}(n, p)$ counts the successes in $n$ independent trials, each with the same probability of success $p$ ; $\mathrm{P}(X = r) = \binom{n}{r} p^{r}(1 - p)^{n - r}$ for $r = 0, 1, \ldots, n$ (in the booklet).

Parameters: the numbers that fix one distribution in a family; for the binomial, $n$ and $p$ .

Trial: one repetition, with outcome success or failure.

Independent trials: the outcome of any trial does not change the probability of success on any other.

Worked example 1: stating a model and using the probability function

A gardener knows that 85% of the seeds of a certain variety germinate.

She plants a packet of 12 of these seeds.

The random variable $G$ is the number of seeds in the packet that germinate.

(a) State a suitable distribution to model $G$ . (1)

(b) State one assumption needed for this model, in context. (1)

(d) Find the probability that all 12 of the seeds germinate. (2)

1. (a) $G \sim \mathrm{B}(12, 0.85)$ B1 (both parameters)

2. (b) Each seed germinates independently of the other seeds (or: every seed has the same probability, 0.85, of germinating) B1

3. (c) $\mathrm{P}(G = 10) = \binom{12}{10}(0.85)^{10}(0.15)^{2}$ M1

4. $= 66 \times 0.196874\ldots \times 0.0225 = 0.29236\ldots = 0.292$ (3 s.f.) A1

5. (d) $\mathrm{P}(G = 12) = (0.85)^{12}$ M1

6. $= 0.142$ (3 s.f.) A1

Answer: (a) $\mathrm{B}(12, 0.85)$ (c) $0.292$ (d) $0.142$ .

Even with a high germination rate, a full packet germinating happens only about 1 time in 7, which is why seed packets carry a few spares.

Worked example 2: commenting critically on a binomial model

For each random variable, state whether a binomial distribution is a suitable model.

If it is, state the parameters; if it is not, give a reason in context.

(a) $A$ , the number of sixes in 30 rolls of a fair die. (1)

(b) $B$ , the number of hearts when 5 cards are dealt from a well-shuffled standard pack of 52 cards. (1)

(d) $D$ , the number of rolls of a fair die needed to get the first six. (1)

1. (a) Suitable: $A \sim \mathrm{B}\!\left(30, \frac{1}{6}\right)$ ; 30 independent rolls with the same chance of a six B1

2. (b) Not suitable: the cards are dealt without replacement, so the probability of a heart changes after each card (for example $\frac{13}{52}$ for the first card, then $\frac{12}{51}$ if the first was a heart) B1

3. (c) Not suitable: rain on one day makes rain the next day more likely, so the days are not independent, and the chance of rain is unlikely to be the same on every day B1

4. (d) Not suitable: there is no fixed number of trials; $D$ counts trials, not successes B1

Answer: only (a) is binomial.

Each "not suitable" answer names the condition that fails and says why in the setting of the question, which is what "comment critically" means.

Worked example 3: a printed probability, then complements

A quiz has 8 questions.

Each question has 5 options, only one of which is correct.

Tom guesses the answer to every question, independently.

The random variable $X$ is the number of questions he answers correctly.

(a) Show that the probability that Tom answers exactly 2 questions correctly is 0.2936 to 4 decimal places. (2)

(b) Find the probability that Tom answers at least one question correctly. (2)

To pass the quiz a student must answer at least 7 questions correctly.

1. (a) $X \sim \mathrm{B}(8, 0.2)$ , so $\mathrm{P}(X = 2) = \binom{8}{2}(0.2)^{2}(0.8)^{6}$ M1

2. $= 28 \times 0.04 \times 0.262144 = 0.293601\ldots = 0.2936$ (4 d.p.) A1*

3. (b) $\mathrm{P}(X \geqslant 1) = 1 - \mathrm{P}(X = 0) = 1 - (0.8)^{8}$ M1

4. $= 1 - 0.167772\ldots = 0.832$ (3 s.f.) A1

5. (c) $\mathrm{P}(X \geqslant 7) = \mathrm{P}(X = 7) + \mathrm{P}(X = 8)$ M1

6. $= 8(0.2)^{7}(0.8) + (0.2)^{8}$ M1

7. $= 0.00008192 + 0.00000256 = 0.0000845$ (3 s.f.) A1

Answer: (a) shown (b) $0.832$ (c) $0.0000845$ (or $8.45 \times 10^{-5}$ ).

In (a) the unrounded $0.293601\ldots$ is the line before the printed $0.2936$ ; writing only "0.2936" would not show it.

Part (c) says guessing almost never passes: fewer than 1 student in 10 000.

Worked example 4: a probability from an earlier result, then a binomial count

The delivery time, $T$ minutes, of an online grocery order has upper quartile $q_3$ .

A random sample of 6 orders is taken, and the times are independent.

The random variable $Y$ is the number of these orders whose delivery time is more than $q_3$ .

(a) Explain why $Y \sim \mathrm{B}(6, 0.25)$ . (1)

(b) Find $\mathrm{P}(Y = 2)$ . (2)

1. (a) By the definition of the upper quartile, $\mathrm{P}(T > q_3) = 0.25$ for each order, and the 6 orders are independent, so the count above $q_3$ is $\mathrm{B}(6, 0.25)$ B1

2. (b) $\mathrm{P}(Y = 2) = \binom{6}{2}(0.25)^{2}(0.75)^{4} = 15 \times 0.0625 \times 0.31640625$ M1

3. $= 0.297$ (3 s.f.) A1

4. (c) At least 5 below $q_3$ means at most 1 above: $\mathrm{P}(Y \leqslant 1) = \mathrm{P}(Y = 0) + \mathrm{P}(Y = 1)$ M1

5. $= (0.75)^{6} + 6(0.25)(0.75)^{5}$ M1

6. $= 0.177978\ldots + 0.355957\ldots = 0.534$ (3 s.f.) A1

Answer: (b) $0.297$ (c) $0.534$ .

In (c) you could instead count orders below $q_3$ with $\mathrm{B}(6, 0.75)$ and find $\mathrm{P}(\geqslant 5)$ ; the answer is the same, but the reports found that the $p = 0.25$ route, counting the rarer event, led to fewer errors.

Exam tip

"State a suitable model" earns its mark only with the parameters: write $X \sim \mathrm{B}(12, 0.85)$ , not "binomial".

An assumption must be one condition written about the people or objects in the question ("each customer chooses independently"), never the bare word "independent" or a list of all four conditions.

When you write the probability function with numbers in, check that the two powers add to $n$ : $\binom{12}{10}(0.85)^{10}(0.15)^{2}$ has $10 + 2 = 12$ .

Leaving out the $(1 - p)^{n - r}$ factor counts the successes but forgets the failures.

Common mistakes

"Binomial" with no parameters. It is tempting because the word feels like the answer, but the mark is for the model, and $\mathrm{B}(n, p)$ is a model only when $n$ and $p$ are given.

Assumptions out of context, or from the wrong model. "Trials are independent" scores nothing; "events occur singly at a constant rate" is a Poisson condition and is wrong here. Fix: name the objects, "each seed germinates independently".

Without replacement from a small group. The S2 reports on sampling from a small bag note candidates missing that the draws were made without replacement, so the probabilities change after each draw. The same fault breaks a binomial model: if $p$ changes after each pick, the model is not binomial. If the population is large, say so: the change is negligible.

Over-rounding. A single binomial value given to 2 s.f., or with no method seen, can lose both marks. Write the expression with the numbers in, then give 3 s.f.

Check your understanding

1. A box holds 12 light bulbs, 3 of which are faulty. Four bulbs are taken from the box at random, without replacement, and $F$ is the number of faulty bulbs taken.

Explain why $\mathrm{B}(4, 0.25)$ is not a suitable model for $F$ , and suggest how the selection could be changed so that a binomial model would be suitable.

(2)

Total: [2]

2. A netball player scores with 35% of her shots from a long distance. In one match she takes 9 long-distance shots. The random variable $X$ is the number of these shots with which she scores.

(a) State a suitable model for $X$ . (1)

(b) State one assumption you have made, in context. (1)

(d) Find the probability that she scores with at least 8 of the shots. (3)

Total: [7]

3. The random variable $X \sim \mathrm{B}(7, 0.4)$ . Show that $\mathrm{P}(X = 3) = 0.2903$ to 4 decimal places. (2)

Total: [2]

4. The lifetime of a type of battery has lower quartile $q_1$ . Five batteries are chosen at random.

(a) Find the probability that exactly one of the batteries has a lifetime less than $q_1$ . (2)

(b) Find the probability that at least 4 of the batteries have a lifetime more than $q_1$ . (3)

Total: [5]

5. Explain why the number of heads in 10 tosses of a coin can be modelled by a binomial distribution, but the number of tosses needed to obtain the first head cannot. (2)

Total: [2]

6. The random variable $Y \sim \mathrm{B}(4, p)$ , where $p > 0$ . Given that $\mathrm{P}(Y = 0) = 0.0256$ ,

(a) find the value of $p$ (2)

(b) find $\mathrm{P}(Y = 3)$ . (2)

Total: [4]

Answers

1. The bulbs are taken without replacement from only 12, so the probability that a bulb is faulty changes after each draw (for example $\frac{3}{12}$ , then $\frac{2}{11}$ after a faulty one): the trials are not independent B1. Replace each bulb before the next is taken (or take the bulbs from a very large batch with 25% faulty) B1

2. (a) $X \sim \mathrm{B}(9, 0.35)$ B1 (b) Each shot is scored independently of the others, or the probability of scoring is 0.35 for every shot B1 (c) $\binom{9}{3}(0.35)^{3}(0.65)^{6}$ M1 $= 0.272$ (3 s.f.) A1 (d) $\mathrm{P}(X = 8) + \mathrm{P}(X = 9)$ M1; $9(0.35)^{8}(0.65) + (0.35)^{9}$ M1; $= 0.00140$ (3 s.f.) A1

3. $\binom{7}{3}(0.4)^{3}(0.6)^{4} = 35 \times 0.064 \times 0.1296$ M1; $= 0.290304 = 0.2903$ (4 d.p.) A1*

4. (a) $L \sim \mathrm{B}(5, 0.25)$ , the number below $q_1$ : $\binom{5}{1}(0.25)(0.75)^{4}$ M1; $= 0.396$ (3 s.f.) A1 (b) at least 4 above means at most 1 below: $\mathrm{P}(L \leqslant 1)$ M1; $(0.75)^{5} + 5(0.25)(0.75)^{4}$ M1; $= 0.633$ (3 s.f.) A1

5. For heads in 10 tosses, the number of trials is fixed at 10, the tosses are independent and the probability of a head is the same each time, with two outcomes B1. For the first head, the number of tosses is what is being counted, so there is no fixed number of trials B1

6. (a) $(1 - p)^{4} = 0.0256$ M1; $1 - p = 0.4$ , so $p = 0.6$ A1 (b) $\binom{4}{3}(0.6)^{3}(0.4)$ M1; $= 0.3456$ A1

The binomial distribution: modelling a real-world situation

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