Binomial to Poisson Approximation

A2 · S2 · 11 min

When the number of trials is large and the probability of success is small, binomial probabilities are tedious to calculate but almost identical to Poisson probabilities with the same mean. The syllabus expects you to recognise when this approximation is appropriate, to justify it, and to use it. On Paper 6 it usually appears as "use a suitable approximation", and choosing the right approximation (Poisson or normal) is part of the mark.

Why the approximation works

The Poisson distribution was built as the limit of binomial distributions in which $n$ grows and $p$ shrinks while the mean $np$ stays fixed. So a binomial with a large $n$ and a small $p$ is already "most of the way" to its Poisson limit.

Compare the two sets of moments:

	$B(n, p)$	$\text{Po}(np)$
Mean	$np$	$np$
Variance	$np(1 - p)$	$np$

The means agree exactly. The variances differ by the factor $1 - p$ , which is close to $1$ when $p$ is small. Here is how close the probabilities are for $B(100, 0.03)$ and $\text{Po}(3)$ :

$r$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$B(100, 0.03)$	$0.0476$	$0.1471$	$0.2252$	$0.2275$	$0.1706$	$0.1013$	$0.0496$	$0.0206$
$\text{Po}(3)$	$0.0498$	$0.1494$	$0.2240$	$0.2240$	$0.1680$	$0.1008$	$0.0504$	$0.0216$

Agreement to about two decimal places, with a far simpler formula.

The conditions

Key result

If $X \sim B(n, p)$ with $n$ large and $p$ small, then approximately

X \sim \text{Po}(np).

The syllabus conditions are

n > 50 \quad\text{and}\quad np < 5 \qquad \text{(approximately)}.

Both conditions matter. A large $n$ alone is not enough: $B(200, 0.4)$ has $np = 80$ and is much better approximated by a normal distribution. A small $p$ alone is not enough either: $B(10, 0.05)$ is easy to compute exactly, and $n$ is too small for the approximation to be close.

Choosing between the Poisson and normal approximations

Situation	Approximation	Conditions
$n$ large, $p$ small	$\text{Po}(np)$	$n > 50$ , $np < 5$
$n$ large, $p$ not near $0$ or $1$	$N(np, npq)$ with continuity correction	$np > 5$ and $nq > 5$

The normal approximation to the binomial is from S1. The deciding number is $np$ : below $5$ , use the Poisson; above $5$ (with $nq > 5$ too), use the normal.

When $p$ is close to 1

If almost every trial is a success, the number of failures is the rare event. Define $Y = n - X$ , the number of failures. Then $Y \sim B(n, q)$ with $q = 1 - p$ small, and $Y$ can be approximated by $\text{Po}(nq)$ . Translate the question about successes into a question about failures before you approximate.

Using a Poisson approximation

Define $X$ and state the exact distribution: $X \sim B(n, p)$ .
Check the conditions: $n > 50$ and $np < 5$ . If $p$ is near $1$ , switch to counting failures.
State the approximating distribution: $X \sim \text{Po}(np)$ approximately, giving the numerical value of $np$ .
Calculate the required probability with the Poisson formula. No continuity correction: both distributions are discrete, on the same whole numbers.
Give the answer to 3 significant figures.

A routine approximation

On average $2\%$ of the light bulbs produced by a machine are faulty. A random sample of $150$ bulbs is taken.

(a) Use a suitable approximation to find the probability that the sample contains at most $2$ faulty bulbs.

(b) Justify your approximation.

Solution

(a) Let $X$ be the number of faulty bulbs. Then $X \sim B(150, 0.02)$ .

$np = 150 \times 0.02 = 3$ , so approximately $X \sim \text{Po}(3)$ .

P(X \le 2) \approx e^{-3}\left(1 + 3 + \frac{3^2}{2}\right) = 8.5e^{-3} = 0.423

(b) $n = 150 > 50$ and $np = 3 < 5$ .

For comparison, the exact binomial probability is $0.421$ , so the approximation is good.

Counting failures

$98\%$ of the seeds in a large batch germinate. A gardener plants $200$ seeds. Use a suitable approximation to find the probability that more than $195$ seeds germinate.

Solution

The number germinating is $B(200, 0.98)$ , but $p = 0.98$ is not small. Count the failures instead.

Let $Y$ be the number of seeds that do not germinate: $Y \sim B(200, 0.02)$ .

$n = 200 > 50$ and $nq = 4 < 5$ , so approximately $Y \sim \text{Po}(4)$ .

More than $195$ germinate means $196$ to $200$ germinate, so $0$ to $4$ fail:

P(Y \le 4) \approx e^{-4}\left(1 + 4 + \frac{4^2}{2} + \frac{4^3}{6} + \frac{4^4}{24}\right) = e^{-4}(1 + 4 + 8 + 10.667 + 10.667) = 0.629

Finding the sample size

The probability that a randomly chosen person has a particular rare blood group is $0.004$ . Use a suitable approximation to find the least number of people who must be tested so that the probability of finding at least one person with this blood group is greater than $0.95$ .

Solution

Let $X$ be the number with the blood group among $n$ people: $X \sim B(n, 0.004)$ . With $n$ large and $p$ small, approximately $X \sim \text{Po}(0.004n)$ .

P(X \ge 1) = 1 - e^{-0.004n} > 0.95

e^{-0.004n} < 0.05 \;\Rightarrow\; -0.004n < \ln 0.05 \;\Rightarrow\; n > \frac{\ln 20}{0.004} = 748.9

The least number is $749$ .

Check the approximation is reasonable: $n = 749 > 50$ and $np = 3.0 < 5$ . (The exact binomial calculation, $1 - 0.996^n > 0.95$ , gives $748$ . The approximation is close but not identical, which is why the question specifies the method.)

An exam-style multi-part question

$0.8\%$ of the eggs from a farm have a double yolk. The eggs are sold in boxes of $250$ .

(a) Use a suitable approximation to find the probability that a box contains more than $3$ double-yolked eggs.

(b) Explain why a normal approximation would not be appropriate here.

(c) A shop buys $6$ boxes. Find the probability that at most one of these boxes contains more than $3$ double-yolked eggs.

Solution

(a) Let $X$ be the number of double-yolked eggs in a box. $X \sim B(250, 0.008)$ .

$n = 250 > 50$ and $np = 2 < 5$ , so approximately $X \sim \text{Po}(2)$ .

P(X > 3) = 1 - P(X \le 3) \approx 1 - e^{-2}\left(1 + 2 + 2 + \frac{4}{3}\right) = 1 - 0.85712 = 0.14288

So $P(X > 3) = 0.143$ (3 s.f.).

(b) For a normal approximation we need $np > 5$ , but $np = 2$ . The distribution is very skewed, not bell-shaped.

P(N \le 1) = 0.85712^6 + 6(0.14288)(0.85712)^5 = 0.39651 + 0.39658 = 0.793

Note that $N$ is not approximated: $n = 6$ is small, so the exact binomial is used.

Common mistakes

Applying a continuity correction. The continuity correction is only for approximating a discrete distribution by a continuous one. Binomial to Poisson is discrete to discrete: no correction.
Using the variance $npq$ as $\lambda$ . The Poisson parameter is the mean, $np$ .
Approximating when $p$ is near 1 without switching. $B(200, 0.98)$ has $np = 196$ , nowhere near $5$ . Count failures.
Forgetting to rewrite the event after switching. "More than 195 successes" becomes "at most 4 failures". Check with an extreme case: 200 successes is 0 failures.
Using the Poisson when $np$ is large. If $np > 5$ and $nq > 5$ , the normal is the suitable approximation.
Justifying with only one condition. "Because $n$ is large" is incomplete. Give both $n > 50$ and $np < 5$ , with numbers.

Exam tip

"Use a suitable approximation" means you must choose and name it. Write " $X \sim B(150, 0.02)$ , approximated by $\text{Po}(3)$ ".
"Justify" or "explain why your approximation is valid": quote the numbers, " $n = 150 > 50$ and $np = 3 < 5$ ". Words alone (" $n$ is large and $p$ is small") are often accepted, but numbers are safer.
If you calculate the exact binomial probability when an approximation was asked for, you may lose the method marks even though the answer is close. Follow the instruction.
If a question does not ask for an approximation and $n$ is small enough to compute exactly (for example $n = 12$ ), use the exact binomial.
Watch for the switch back: once you have found a probability with the Poisson, a follow-up "how many of these boxes" question is an exact binomial with small $n$ .

Summary

$B(n, p) \approx \text{Po}(np)$ when $n$ is large and $p$ is small: $n > 50$ and $np < 5$ .
Means agree exactly; variances $npq$ and $np$ are close because $q \approx 1$ .
No continuity correction: both distributions are discrete.
If $p$ is close to $1$ , count failures: $n - X \sim B(n, q) \approx \text{Po}(nq)$ .
Use the normal approximation instead when $np > 5$ and $nq > 5$ .
Justify with the numbers; follow "use a suitable approximation" literally.

Practice questions

Question

State, with a reason, which approximation (if any) is suitable for each distribution: (a) $B(60, 0.03)$ , (b) $B(40, 0.05)$ , (c) $B(200, 0.45)$ , (d) $B(500, 0.996)$ .
$X \sim B(80, 0.025)$ . Use a suitable approximation to find (a) $P(X = 3)$ , (b) $P(X \le 1)$ .
A rare condition affects $1$ in $2000$ people. Use a suitable approximation to find the probability that, in a town of $6000$ people, more than $4$ people have the condition.
$99\%$ of components pass a quality test. A batch of $300$ components is tested. Use a suitable approximation to find the probability that at least $297$ pass.
The probability that an item is defective is $0.01$ . Use a suitable approximation to find the least number of items that must be inspected for the probability of finding at least one defective to be at least $0.9$ .
$X \sim B(60, 0.05)$ . Calculate $P(X = 2)$ exactly and using a Poisson approximation. Find the percentage error in the approximation.
The probability that a sample of $100$ items from a large batch contains no defective items is $0.135$ . Use a Poisson approximation to estimate the proportion of defective items in the batch, and hence estimate the probability that a sample of $250$ items contains at most $2$ defectives.
The probability that a hen's egg has a double yolk is $0.004$ . Eggs are packed in crates of $360$ . (a) Use a suitable approximation to find the probability that a crate contains at least $3$ double-yolked eggs, and justify the approximation. (b) A shop sells $10$ crates in a week. Find the probability that more than one of these crates contains at least $3$ double-yolked eggs. (c) Using a Poisson approximation, find the largest number of eggs that can be packed in a box if the probability that the box contains no double-yolked egg is to be greater than $0.95$ .

Answers

(a) Poisson, $\text{Po}(1.8)$ : $n = 60 > 50$ and $np = 1.8 < 5$ . (b) None needed; $n = 40$ is not large, so calculate exactly (Poisson would be borderline at best). (c) Normal, $N(90, 49.5)$ : $np = 90 > 5$ and $nq = 110 > 5$ . (d) Count failures: $B(500, 0.004)$ for failures, approximated by $\text{Po}(2)$ since $n > 50$ and $nq = 2 < 5$ .
$np = 2$ , so $X \approx \text{Po}(2)$ . (a) $P(X = 3) = e^{-2}\dfrac{8}{6} = 0.180$ . (b) $P(X \le 1) = e^{-2}(1 + 2) = 3e^{-2} = 0.406$ .
$X \sim B(6000, 0.0005) \approx \text{Po}(3)$ . $P(X > 4) = 1 - e^{-3}\left(1 + 3 + 4.5 + 4.5 + 3.375\right) = 1 - 16.375e^{-3} = 1 - 0.8153 = 0.185$ .
Failures $Y \sim B(300, 0.01) \approx \text{Po}(3)$ . At least $297$ pass means at most $3$ fail. $P(Y \le 3) = e^{-3}(1 + 3 + 4.5 + 4.5) = 13e^{-3} = 0.647$ .
$X \approx \text{Po}(0.01n)$ . $1 - e^{-0.01n} \ge 0.9 \Rightarrow e^{-0.01n} \le 0.1 \Rightarrow n \ge 100\ln 10 = 230.3$ . Least $n = 231$ .
Exact: $\binom{60}{2}(0.05)^2(0.95)^{58} = 0.22588$ . Approximation $\text{Po}(3)$ : $e^{-3}\dfrac{9}{2} = 0.22404$ . Percentage error $= \dfrac{0.22588 - 0.22404}{0.22588} \times 100 = 0.81\%$ .
$e^{-100p} = 0.135 \Rightarrow 100p = -\ln 0.135 = 2.0025 \Rightarrow p = 0.0200$ (3 s.f.). For $250$ items, $\lambda = 250 \times 0.020025 = 5.006$ , so $X \approx \text{Po}(5.006)$ . $P(X \le 2) = e^{-5.006}\left(1 + 5.006 + \dfrac{5.006^2}{2}\right) = 0.124$ . (Using $\lambda = 5$ gives $0.125$ , also acceptable.)
(a) $X \sim B(360, 0.004)$ ; $n = 360 > 50$ , $np = 1.44 < 5$ , so $X \approx \text{Po}(1.44)$ . $P(X \ge 3) = 1 - e^{-1.44}\left(1 + 1.44 + \dfrac{1.44^2}{2}\right) = 1 - e^{-1.44}(3.4768) = 1 - 0.82375 = 0.176$ . (b) $N \sim B(10, 0.17625)$ . $P(N > 1) = 1 - 0.82375^{10} - 10(0.17625)(0.82375)^9 = 1 - 0.14386 - 0.30782 = 0.548$ . (c) For $n$ eggs, $P(\text{none}) \approx e^{-0.004n} > 0.95 \Rightarrow n < \dfrac{-\ln 0.95}{0.004} = 12.82$ . Largest number $= 12$ .

Why the approximation works

The conditions

Choosing between the Poisson and normal approximations

When ppp is close to 1

Practice questions

Console

When $p$ is close to 1