Mathematics 9709 · Probability & Statistics 2

Poisson ≈ binomial

This is the same limit that birthed the Poisson, read backwards. A binomial with a huge $n$ and a tiny $p$ is already “rare events over endless trials” — so it is a Poisson in a binomial costume, with $\lambda=np$ . You trade a two-number formula with brutal coefficients for one easy number.

B(n, p) ≈ Po(np) when n large, p small

Trigger: $n$ large and $p$ small (guide $n>50,\ np<5$ ). Set $\lambda=np$ and use the ordinary Poisson formula. State the justification in your working (“ $n$ large, $p$ small, so use $\text{Po}(np)$ ”) — it scores a mark.

Worked example

Start simple: convert, then one tail

$X\sim B(60,0.05)$ . Here $n=60>50$ and $np=3<5$ , so $X\approx\text{Po}(3)$ . Find $P(X\ge4)$ :

$P(X\ge4)=1-P(X\le3)=1-0.647=0.353$

Answer

\approx0.353

n large AND p small (np < 5) → Po(np), one parameter. State the justification; it scores a mark.

Worked example

Rare faults in a big box

Components are faulty with probability 0.4%; a box has 1000. Find $P(\text{more than 5 faulty})$ .

$X\sim B(1000,0.004)$ ; $n$ large, $p$ small, $np=4<5$ , so $X\approx\text{Po}(4)$ .
“more than 5” excludes 5: $P(X>5)=1-P(X\le5)=1-0.785$ .

Answer

P(X>5)\approx0.215

Worked example

How good is the approximation?

Glassware is faulty with probability 0.012; a batch of 375 gives $np=4.5$ , so $B(375,0.012)\approx\text{Po}(4.5)$ . The two agree to 2–3 decimals:

x	Binomial B(375, 0.012)	Poisson Po(4.5)
0	0.0108	0.0111
1	0.0492	0.0500
2	0.1118	0.1125

Answer

np=4.5\Rightarrow\text{Po}(4.5)\text{ tracks the binomial closely.}

The bigger $n$ and smaller $p$ (with $np$ fixed), the tighter the match.

Examiner note

Write the justification “

n

large,

p

small” (or

n>50

and

np<5

) explicitly; for

P(Y>2)

write

1-P(0)-P(1)-P(2)

with no term dropped. (9709/62/M/J/23 Q2) (9709/62/O/N/22 Q3)

Common mistake

Writing

\text{Po}(n,p)

— Poisson has one parameter,

\text{Po}(np)

. You need

n

large and

p

small (if

np

is large it's a normal-approx job instead), and you must name the approximating distribution.

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Poisson ≈ binomial

B(n, p) ≈ Po(np) when n large, p small

Worked example

Start simple: convert, then one tail

$X\sim B(60,0.05)$ . Here $n=60>50$ and $np=3<5$ , so $X\approx\text{Po}(3)$ . Find $P(X\ge4)$ :

$P(X\ge4)=1-P(X\le3)=1-0.647=0.353$

Answer

\approx0.353

n large AND p small (np < 5) → Po(np), one parameter. State the justification; it scores a mark.

Worked example

Rare faults in a big box

Components are faulty with probability 0.4%; a box has 1000. Find $P(\text{more than 5 faulty})$ .

$X\sim B(1000,0.004)$ ; $n$ large, $p$ small, $np=4<5$ , so $X\approx\text{Po}(4)$ .
“more than 5” excludes 5: $P(X>5)=1-P(X\le5)=1-0.785$ .

Answer

P(X>5)\approx0.215

Worked example

How good is the approximation?

Glassware is faulty with probability 0.012; a batch of 375 gives $np=4.5$ , so $B(375,0.012)\approx\text{Po}(4.5)$ . The two agree to 2–3 decimals:

x	Binomial B(375, 0.012)	Poisson Po(4.5)
0	0.0108	0.0111
1	0.0492	0.0500
2	0.1118	0.1125

Answer

np=4.5\Rightarrow\text{Po}(4.5)\text{ tracks the binomial closely.}

The bigger $n$ and smaller $p$ (with $np$ fixed), the tighter the match.

Examiner note

Write the justification “

n

large,

p

small” (or

n>50

and

np<5

) explicitly; for

P(Y>2)

write

1-P(0)-P(1)-P(2)

with no term dropped. (9709/62/M/J/23 Q2) (9709/62/O/N/22 Q3)

Common mistake

Writing

\text{Po}(n,p)

— Poisson has one parameter,

\text{Po}(np)

. You need

n

large and

p

small (if

np

is large it's a normal-approx job instead), and you must name the approximating distribution.