S2.1 Binomial and Poisson distributions
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A2 content (Statistics 2, specification sections 1.1, 1.2 and 1.3); a scientific calculator and the cumulative tables are all you need.
What this page covers
8 points
- The conditions for a binomial model \(B(n,p)\), stated in the context of a question
- Binomial probabilities from the probability function and from the cumulative tables, including the switch for \(p\gt0.5\)
- Turning \(P(X=r)\), \(P(X\gt r)\) and \(P(a\le X\lt b)\) into cumulative probabilities
- The conditions for a Poisson model \(Po(\lambda)\), and scaling \(\lambda\) to the length of the interval
- Adding independent Poisson variables
- Means and variances (quoted, not derived), and using "mean ≈ variance" to judge whether a Poisson model is sensible
- Approximating \(B(n,p)\) by \(Po(np)\) when \(n\) is large and \(p\) is small
- Chaining the two models, and "least \(n\)" questions solved with logarithms
In one minute
6 points
- The binomial distribution counts successes in a fixed number of independent trials, the Poisson distribution counts random events in an interval of time or space, and almost every question is the same three moves: name the distribution with its parameters, turn the event into cumulative probabilities, then use the tables or the formula.
- Binomial \(X\sim B(n,p)\): fixed \(n\); success or failure; independent trials; constant \(p\). \(P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\). Mean \(np\), variance \(np(1-p)\).
- Poisson \(X\sim Po(\lambda)\): events occur singly, independently, at random and at a constant mean rate. \(P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}\). Mean = variance = \(\lambda\).
- Tables give \(P(X\le x)\): \(P(X\ge r)=1-P(X\le r-1)\) and \(P(X=r)=P(X\le r)-P(X\le r-1)\).
- Rates scale: 3 per hour gives \(Po(1.5)\) for 30 minutes and \(Po(9)\) for 3 hours. Independent Poissons add: \(Po(\lambda)+Po(\mu)=Po(\lambda+\mu)\).
- Approximation: \(n\) large, \(p\) small \(\Rightarrow B(n,p)\approx Po(np)\).
Contents
13 sections
- When is a binomial model suitable?
- How do you use the cumulative binomial tables?
- When is a Poisson model suitable?
- How do Poisson variables combine?
- What are the mean and variance, and how do they help you choose a model?
- When can a Poisson distribution approximate a binomial?
- How do these ideas combine in a full question?
- Exam tips for S2.1
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
When is a binomial model suitable?
A binomial situation is a repeated experiment where you only care about one outcome. Flip a seed into a tray and ask "does it germinate?"; take an email from an inbox and ask "is it spam?". Repeat a fixed number of times and count the yeses.
Learn the four conditions as a list, and then say them in the context of the question:
- a fixed number of trials, \(n\);
- each trial has two outcomes, success or failure;
- the trials are independent;
- the probability of success, \(p\), is the same for every trial.
\(X\) can take the values \(0,1,2,\ldots,n\) and no others. Write the model in full every time: "Let \(X\) be the number of …; \(X\sim B(8,0.35)\)".
Use the formula for a single value, or for a sum of two or three values when the parameters are not in the tables. The bar chart below shows a whole binomial distribution, every bar computed from the formula.
A climber completes a particular route on any one attempt with probability 0.35, independently of other attempts. She makes 8 attempts. (a) Find the probability that she completes the route exactly 3 times. (b) Find the probability that she completes it at least twice. (c) State, in context, two assumptions needed for your model. Give probabilities to 4 significant figures.
Define the model: \(X\) = number of completed attempts; \(X\sim B(8,0.35)\).
(a) \(P(X=3)=\binom{8}{3}(0.35)^3(0.65)^5=56\times0.042875\times0.11603=\) 0.2786
(b) "At least 2" is everything except 0 and 1: \[P(X\ge2)=1-(0.65)^8-8(0.35)(0.65)^7=1-0.03186-0.13726=\] 0.8309
(c) The attempts are independent of each other; the probability of completing the route is 0.35 on every attempt.
"Trials are independent" on its own is generic. Name the trials: attempts. In reality tiredness or practice could change \(p\) — which is exactly why it is an assumption.
How do you use the cumulative binomial tables?
The tables give \(P(X\le x)\) for chosen \(n\) and for \(p=0.05,0.10,\ldots,0.50\), to 4 decimal places. Because \(X\) is a whole number, every inequality can be rewritten with "\(\le\)": \(X\lt7\) is \(X\le6\), and \(X\gt5\) is the complement of \(X\le5\). That one idea handles every event.
| You want | Write it as | Example, \(X\sim B(20,0.15)\) |
|---|---|---|
| \(P(X\le r)\) | read directly | \(P(X\le4)=0.8298\) |
| \(P(X\lt r)\) | \(P(X\le r-1)\) | \(P(X\lt5)=P(X\le4)=0.8298\) |
| \(P(X=r)\) | \(P(X\le r)-P(X\le r-1)\) | \(P(X=4)=0.8298-0.6477=0.1821\) |
| \(P(X\gt r)\) | \(1-P(X\le r)\) | \(P(X\gt5)=1-0.9327=0.0673\) |
| \(P(X\ge r)\) | \(1-P(X\le r-1)\) | \(P(X\ge6)=1-0.9327=0.0673\) |
| \(P(a\le X\le b)\) | \(P(X\le b)-P(X\le a-1)\) | \(P(2\le X\le6)=0.9781-0.1756=0.8025\) |
Of the emails arriving at an office inbox, 15% are spam, independently of each other. A random sample of 20 emails is taken. Find the probability that the number of spam emails is (a) at most 4, (b) exactly 4, (c) more than 5, (d) at least 2 but fewer than 7.
Model: \(X\sim B(20,0.15)\).
(a) \(P(X\le4)=\) 0.8298 (b) \(P(X=4)=0.8298-0.6477=\) 0.1821
(c) "More than 5" starts at 6: \(P(X\gt5)=1-P(X\le5)=1-0.9327=\) 0.0673
(d) List the values: 2, 3, 4, 5, 6. \(P(2\le X\le6)=P(X\le6)-P(X\le1)=0.9781-0.1756=\) 0.8025
What if p is bigger than 0.5?
The tables stop at \(p=0.5\). If \(X\sim B(n,p)\) counts successes, then \(Y=n-X\) counts failures and \(Y\sim B(n,1-p)\). Translate the event: "\(X\ge12\) out of 15" is the same as "\(Y\le3\)". Large values of \(X\) are small values of \(Y\).
A ferry leaves on time on 70% of crossings, independently. (a) For 15 crossings, find the probability that at least 12 leave on time. (b) For 15 crossings, find the smallest value of \(k\) such that the probability of at most \(k\) late departures is at least 0.9. (c) Find the least number of crossings for which the probability of at least one late departure exceeds 0.95.
(a) Switch to late departures: \(L\) = number late, \(L\sim B(15,0.3)\). "At least 12 on time" = "at most 3 late", so \(P(L\le3)=\) 0.2969
(b) Scan the \(B(15,0.3)\) column for the first value \(\ge0.9\): \(P(L\le6)=0.8689\lt0.9\) and \(P(L\le7)=0.9500\ge0.9\), so \(k=7\)
(c) \(P(\text{at least one late})=1-P(\text{none late})=1-0.7^n\). \[1-0.7^n\gt0.95\Rightarrow0.7^n\lt0.05\Rightarrow n\gt\frac{\ln0.05}{\ln0.7}=8.40\] so \(n=9\) (check: \(1-0.7^8=0.9424\), \(1-0.7^9=0.9596\)). Dividing by \(\ln0.7\), which is negative, reverses the inequality — the check line confirms the direction.
Writing \(P(X\ge6)=1-P(X\le6)\). The complement of "\(X\ge6\)" is "\(X\le5\)", so \(P(X\ge6)=1-P(X\le5)\). List the values on each side whenever you are unsure.
When is a Poisson model suitable?
Some counts have no fixed number of trials: calls to a help desk in an hour, flaws in a metre of cloth, clicks of a Geiger counter in a minute. There is no "\(n\)", just events scattered at random through time or space.
State the conditions in context, for example "calls arrive one at a time", "calls arrive independently of one another", "calls arrive at a constant average rate". \(X\) can take the values \(0,1,2,\ldots\) with no upper limit.
Scaling. The parameter is proportional to the size of the interval. A rate of 3 per 10 minutes gives \(Po(0.6)\) for 2 minutes and \(Po(9)\) for 30 minutes. This is the single most important habit on the topic.
Calls arrive at a help desk at a mean rate of 3 every 10 minutes. (a) State two conditions, in context, for a Poisson model of the number of calls. (b) Find the probability of exactly 5 calls in a 10-minute period. (c) Find the probability of more than 8 calls in half an hour. (d) Find the probability of no calls in a 2-minute period.
(a) Calls arrive independently of each other; calls arrive at a constant mean rate (and one at a time).
(b) \(X\sim Po(3)\): \(P(X=5)=e^{-3}\dfrac{3^5}{5!}=\) 0.1008
(c) 30 minutes is three 10-minute periods, so \(Y\sim Po(9)\): \(P(Y\gt8)=1-P(Y\le8)=1-0.4557=\) 0.5443
(d) 2 minutes is 0.2 of a period, so \(W\sim Po(0.6)\): \(P(W=0)=e^{-0.6}=\) 0.5488
Before any calculation, write the scaled distribution for the interval in the question: "in 30 minutes, \(Y\sim Po(9)\)". Using the original \(\lambda\) for a different interval is the most expensive slip on this topic, because every later part depends on it.
How do Poisson variables combine?
Two independent streams of random events merge into one stream whose rate is the sum of the rates. Scaling a rate is the same idea: three independent 10-minute periods, each \(Po(3)\), add to \(Po(9)\). Independence is essential and must be stated. The difference \(X-Y\) is not Poisson (it can be negative).
A bakery receives online orders at a mean rate of 2.5 per hour and phone orders at a mean rate of 1.5 per hour, independently, each following a Poisson distribution. (a) Find the probability that the bakery receives at least 6 orders in total in an hour. (b) Find the probability that, in an hour, it receives exactly 2 online orders and exactly 1 phone order. (c) Find the probability of exactly 10 orders in total in a 2-hour period.
(a) Add the rates: \(T\) = total orders in an hour, \(T\sim Po(2.5+1.5)=Po(4)\). \(P(T\ge6)=1-P(T\le5)=1-0.7851=\) 0.2149
(b) Independent, so multiply: \(e^{-2.5}\dfrac{2.5^2}{2!}\times e^{-1.5}\dfrac{1.5^1}{1!}=0.2565\times0.3347=\) 0.08585
(c) Scale the total rate to 2 hours, \(Po(8)\): \(P=e^{-8}\dfrac{8^{10}}{10!}=\) 0.09926
In (b) the total is 3 orders, but \(P(T=3)\) would be wrong: it includes 3 online and 0 phone, 1 online and 2 phone, and so on.
What are the mean and variance, and how do they help you choose a model?
| Distribution | Mean \(E(X)\) | Variance \(\text{Var}(X)\) | Standard deviation |
|---|---|---|---|
| \(B(n,p)\) | \(np\) | \(np(1-p)\) | \(\sqrt{np(1-p)}\) |
| \(Po(\lambda)\) | \(\lambda\) | \(\lambda\) | \(\sqrt{\lambda}\) |
These are quoted results: S2 does not ask you to derive them. They are more useful than they look:
- For a binomial, the variance is always less than the mean, since \(1-p\lt1\).
- For a Poisson, mean = variance. So if a sample of counts has a mean and a variance that are close, a Poisson model is plausible; if the variance is much bigger or smaller than the mean, it is not.
- Given a binomial mean and variance, divide: \(\dfrac{np(1-p)}{np}=1-p\) gives \(p\), then \(n=\text{mean}\div p\).
(a) \(X\sim B(n,p)\) has mean 6 and variance 4.2. Find \(n\) and \(p\), and hence find \(P(X\gt6)\). (b) The numbers of surface flaws on 100 ceramic tiles are recorded.
| Flaws, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| Tiles, \(f\) | 16 | 30 | 26 | 16 | 8 | 3 | 1 |
Find the mean and variance of the number of flaws per tile and explain whether a Poisson model is reasonable.
(a) \(1-p=\dfrac{4.2}{6}=0.7\Rightarrow p=0.3\); \(n=\dfrac{6}{0.3}=20\). Then \(P(X\gt6)=1-P(X\le6)=1-0.6080=\) 0.3920
(b) \(\sum f=100\), \(\sum fx=183\), \(\sum fx^2=517\). Mean \(=\dfrac{183}{100}=1.83\); variance \(=\dfrac{517}{100}-1.83^2=1.8211\).
Mean ≈ variance, which is a property of the Poisson distribution, so \(Po(1.83)\) is a reasonable model. With the model, the expected number of tiles with no flaws is \(100e^{-1.83}=16.0\), against 16 observed.
When can a Poisson distribution approximate a binomial?
Keep the mean the same: \(\lambda=np\). It works because the binomial variance \(np(1-p)\approx np\) when \(p\) is small — mean and variance nearly equal, just like a Poisson. A common rule of thumb is \(n\gt50\) and \(np\lt5\); it is still fair a little beyond this, as long as \(p\) is small. If \(p\) is close to 1, approximate the number of failures, \(B(n,1-p)\), instead.
2% of the components made by a machine are faulty, independently. A batch of 150 components is checked. (a) Explain why a Poisson distribution can be used to approximate the number of faulty components, and state its parameter. (b) Use it to find the probability of at most 2 faulty components. (c) Use it to find the probability of more than 5 faulty components.
(a) \(X\sim B(150,0.02)\); \(n=150\) is large and \(p=0.02\) is small, so \(X\approx Po(150\times0.02)=Po(3)\)
(b) Poisson table, \(\lambda=3\): \(P(X\le2)\approx\) 0.4232 (the exact binomial value is 0.4209)
(c) \(P(X\gt5)\approx1-P(X\le5)=1-0.9161=\) 0.0839 (exact: 0.08188)
\(B(150,0.02)\) is not in any table, and the exact sum would need six binomial terms; the approximation turns it into one table look-up.
Using the Poisson approximation just because \(n\) is large. With \(B(60,0.4)\), \(n\) is large but \(p\) is not small: the mean is 24 and the variance 14.4, far from equal, so a Poisson model with \(\lambda=24\) would be far too spread out.
How do these ideas combine in a full question?
Longer questions often chain the two distributions: a Poisson probability becomes the \(p\) of a binomial ("how many of the 12 months have no breakdowns?"). Keep that \(p\) unrounded in your calculator and round only the final answer. "Least number of …" questions lead to an inequality with a power or an exponential — solve with logarithms, then check the integers either side.
Breakdowns of a lift occur at random at a mean rate of 1.5 per month. (a) Find the probability of exactly 2 breakdowns in a month. (b) Find the probability of more than 4 breakdowns in a 3-month period. (c) Find the probability that in a year (12 months) there are exactly 3 months with no breakdowns. (d) Find the least whole number of months for which the probability of at least one breakdown exceeds 0.99.
(a) \(X\sim Po(1.5)\): \(P(X=2)=e^{-1.5}\dfrac{1.5^2}{2!}=\) 0.2510
(b) Scale to 3 months, \(Y\sim Po(4.5)\): \(P(Y\gt4)=1-P(Y\le4)=1-0.5321=\) 0.4679
(c) The probability that a given month has no breakdowns is \(P(X=0)=e^{-1.5}=0.22313\ldots\). Months are independent, so \(M\) = number of months with no breakdowns, \(M\sim B(12,e^{-1.5})\): \[P(M=3)=\binom{12}{3}(e^{-1.5})^3(1-e^{-1.5})^9=\] 0.2519
(d) In \(m\) months the number of breakdowns is \(Po(1.5m)\): \(1-e^{-1.5m}\gt0.99\Rightarrow e^{-1.5m}\lt0.01\Rightarrow m\gt\dfrac{\ln100}{1.5}=3.07\), so 4 months (3 months gives 0.9889; 4 months gives 0.9975).
When one distribution feeds another, write both definitions on separate lines: "\(X\sim Po(1.5)\), breakdowns in a month" and "\(M\sim B(12,0.2231\ldots)\), months with none". It makes clear which variable each probability belongs to.
Exam tips for S2.1
Start every part with the random variable in words and its distribution with parameters: "\(X\) = number of spam emails, \(X\sim B(20,0.15)\)". It is often worth credit in itself, and it stops you using the wrong \(n\) or \(\lambda\) later.
Rewrite the event as \(P(X\le\ldots)\) or \(1-P(X\le\ldots)\) before opening the tables, and show that line. If \(p\gt0.5\), switch to the other outcome first.
"The trials are independent" is a definition, not an answer. Write "whether one seed germinates does not affect whether another does". Give exactly as many conditions as the question asks for.
Table values are quoted to 4 decimal places. Use the probability function when the parameters are not in the tables (for example \(p=0.35\) or \(\lambda=1.83\)), and give those answers to 4 significant figures unless told otherwise.
Solve the inequality with logarithms, round up to the next whole number, and show a check at \(n\) and \(n-1\). Remember that dividing by the log of a number less than 1 reverses the inequality.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(P(X\lt4)=P(X\le4)\). | \(X\) is discrete, so the value 4 itself makes a difference: for \(B(10,0.4)\) these are 0.3823 and 0.6331. | \(P(X\lt4)=P(X\le3)\). |
| \(P(X\ge4)=1-P(X\le4)\). | That subtracts the value 4, which should be included. | \(P(X\ge4)=1-P(X\le3)\). |
| A Poisson rate is the same whatever the interval. | Twice the time means twice as many events on average. | Scale \(\lambda\) in proportion to the length (or area) of the interval. |
| Any two Poisson variables add to a Poisson. | The result needs the variables to be independent. | Independent: \(X+Y\sim Po(\lambda+\mu)\); state the independence. |
| Counting trials until the first success is binomial. | A binomial needs a fixed number of trials decided in advance. | Check all four conditions, starting with a fixed \(n\). |
| The Poisson approximation works whenever \(n\) is large. | \(p\) must also be small, so that \(np(1-p)\approx np\). | Use \(Po(np)\) when \(n\) is large and \(p\) is small. |
How this topic connects
S1 discrete random variables
Discrete random variables, \(E(X)\) and \(\text{Var}(X)\) from a probability table, and the mean and variance of a frequency table (Worked example 6) all come from S1; the binomial and Poisson are special discrete distributions with ready-made formulae.
Pure mathematics (P1–P4)
\(\binom{n}{r}\) is the same coefficient as in the binomial expansion of \((1+x)^n\) in P2; logarithms solve the "least \(n\)" questions; the series for \(e^{\lambda}\) is why Poisson probabilities add up to 1. In IAL Physics, counts from a radioactive source in a fixed time are a textbook Poisson process, which is why repeated counts vary by about \(\sqrt{\text{mean}}\).
The rest of S2
Normal approximations to the binomial and Poisson, with a continuity correction (S2.3), are used when \(n\) or \(\lambda\) is too large for the tables; hypothesis tests about \(p\) and \(\lambda\) (S2.4) use the cumulative tables exactly as in this page's second section.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}\) | binomial probability of exactly \(r\) successes | Yes |
| \(E(X)=np\), \(\text{Var}(X)=np(1-p)\) | mean and variance of \(B(n,p)\) | Yes |
| \(P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}\) | Poisson probability of exactly \(r\) events | Yes |
| \(E(X)=\text{Var}(X)=\lambda\) | mean and variance of \(Po(\lambda)\); judging whether data fit a Poisson model | Yes |
| \(P(X\le x)\) from cumulative binomial and Poisson tables | probabilities for tabulated \(n\), \(p\) and \(\lambda\) | Yes |
| \(P(X=r)=P(X\le r)-P(X\le r-1)\); \(P(X\ge r)=1-P(X\le r-1)\) | turning any event into cumulative probabilities | No – derive it |
| \(X\sim B(n,p)\Rightarrow n-X\sim B(n,1-p)\) | using the tables when \(p\gt0.5\) | No – learn it |
| rate \(\lambda\) per unit \(\Rightarrow Po(t\lambda)\) for an interval of \(t\) units | scaling the Poisson parameter | No – learn it |
| \(X\sim Po(\lambda)\), \(Y\sim Po(\mu)\) independent \(\Rightarrow X+Y\sim Po(\lambda+\mu)\) | combining independent Poisson variables | No – learn it |
| \(B(n,p)\approx Po(np)\) for \(n\) large, \(p\) small | Poisson approximation to the binomial | No – learn it |
Check your understanding
1. \(X\sim B(25,0.2)\). Write down the mean and the variance of \(X\).
Mean \(np=25\times0.2=5\); variance \(np(1-p)=5\times0.8=4\).
2. Write \(P(4\lt X\le9)\) in terms of cumulative probabilities.
The values are 5, 6, 7, 8, 9, so \(P(4\lt X\le9)=P(X\le9)-P(X\le4)\).
3. Particles are detected at a mean rate of 2 per minute. Find the probability that none is detected in a 30-second interval.
In 30 seconds the count is \(Po(1)\), so \(P(0)=e^{-1}=0.3679\).
4. Why is \(Po(24)\) not a good approximation to \(B(40,0.6)\)?
\(p=0.6\) is not small: the binomial mean is 24 but its variance is only 9.6, whereas a Poisson has variance equal to its mean.
5. \(X\sim Po(3)\) and \(Y\sim Po(5)\) are independent. State the distribution of \(X+Y\) and its variance.
\(X+Y\sim Po(8)\), so its variance is 8 (equal to its mean).
6. A seed germinates with probability 0.9, independently. 10 seeds are planted. Find the probability that at least 9 germinate.
Switch to failures: \(F\sim B(10,0.1)\), and "at least 9 germinate" = "at most 1 fails". \(P(F\le1)=0.7361\) from the tables (or \(0.9^{10}+10(0.1)(0.9)^9=0.3487+0.3874=0.7361\)).
Specification checklist
- (S2 1.1) I can state, in context, the conditions under which a binomial model \(B(n,p)\) is suitable, and calculate binomial probabilities from the probability function.
- (S2 1.1) I can turn any event into cumulative probabilities and use the binomial tables, including by switching to \(n-X\) when \(p\gt0.5\).
- (S2 1.1) I can state, in context, the conditions for a Poisson model \(Po(\lambda)\), calculate Poisson probabilities by formula or from tables, and scale \(\lambda\) to the interval in the question.
- (S2 1.1) I can use the fact that the sum of independent Poisson variables is a Poisson variable whose parameter is the sum of the parameters.
- (S2 1.2) I can quote and use the mean and variance of the binomial and Poisson distributions, and use mean ≈ variance to judge whether a Poisson model fits data.
- (S2 1.3) I can explain when a Poisson distribution approximates a binomial (\(n\) large, \(p\) small) and use \(Po(np)\) to calculate approximate probabilities.