19.1.2 Poisson Approximations of Binomials

Introduction

The Poisson distribution can be used to approximate a binomial distribution under specific conditions. This approximation simplifies calculations when the number of trials ( $n$ ) is large, and the probability of success ( $p$ ) is small. Instead of working directly with $B(n, p)$ , we use a Poisson distribution with mean $\lambda = np$

This note covers:

Conditions for the Poisson approximation.
How to use the Poisson approximation in calculations.
Worked examples with step-by-step solutions.

Binomial Distribution

If $X \sim B(n, p)$ , then:

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

where:

$n$ : Number of trials,
$p$ : Probability of success,
$x$ : Number of successes.

Poisson Approximation to Binomial

When $n$ is large and $p$ is small ( $np \leq 10$ ), the binomial distribution $B(n, p)$ can be approximated by the Poisson distribution $\text{Po}(\lambda)$ , where:

\lambda = np

For $Y \sim \text{Po}(\lambda)$ , the probability mass function is:

P(Y = x) = \frac{\lambda^x e^{-\lambda}}{x!}

where $\lambda$ is the mean and variance of the distribution.

Why Use the Poisson Approximation?

Simplifies Calculations: For large $n$ and small $p$ , computing $\binom{n}{x}$ and powers of $p$ becomes cumbersome. The Poisson formula is much simpler.
Real-Life Applications: The Poisson distribution is often used in contexts like rare events over time or space (e.g., accidents, calls to a call center).

Worked Examples

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Example 1: Using the Poisson Approximation

Problem

A factory produces lightbulbs, and each bulb has a $0.002$ probability of being defective. If $1000$ bulbs are tested, find the probability that:

Exactly 2 bulbs are defective.
At most 1 bulb is defective.

Step 1: Check Conditions for Poisson Approximation

For the binomial distribution $B(n, p)$

$n=1000$
$p=0.002$ Calculate $\lambda = np$

\lambda = 1000 \times 0.002 = 2

Since $n$ is large and $p$ is small, the Poisson approximation is appropriate: $Y \sim \text{Po}(2)$

Step 2: Find $P(Y = 2)$

Use the Poisson formula:

P(Y = x) = \frac{\lambda^x e^{-\lambda}}{x!}

Substitute $\lambda = 2$ and $x = 2$

P(Y = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4 \times e^{-2}}{2}

Using $e^{-2} \approx 0.1353$

P(Y = 2) = \frac{4 \times 0.1353}{2} = 0.2706

Step 3: Find $P(Y \leq 1)$

P(Y \leq 1) = P(Y = 0) + P(Y = 1)

Calculate $P(Y = 0)$

P(Y = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = \frac{1 \times e^{-2}}{1} = e^{-2} \approx 0.1353

Calculate $P(Y = 1)$

P(Y = 1) = \frac{\lambda^1 e^{-\lambda}}{1!} = \frac{2 \times e^{-2}}{1}

= 2 \times 0.1353 = 0.270

Add the probabilities:

P(Y \leq 1) = 0.1353 + 0.2706 = 0.4059

Final Answer:

$P(Y = 2) = :success[0.2706]$
$P(Y \leq 1) = :success[0.4059]$

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Example 2: Comparing Binomial and Poisson Probabilities

Problem

A raffle has $200$ tickets, and each ticket has a $0.01$ chance of being a winner. Find the probability of exactly 3 winning tickets:

Using the binomial distribution.
Using the Poisson approximation.

Part 1: Using the Binomial Distribution

For $X \sim B(200, 0.01)$ , the probability is:

P(X = 3) = \binom{200}{3} (0.01)^3 (1-0.01)^{197}

Step 1: Compute $\binom{200}{3}$

\binom{200}{3} = \frac{200 \times 199 \times 198}{3 \times 2 \times 1} = 1313400

Step 2: Compute powers of $0.01$ and $0.99$

(0.01)^3 = 0.000001, \quad (0.99)^{197} \approx 0.1348

Step 3: Calculate:

P(X = 3) = 1313400 \times 0.000001 \times 0.1348 = 0.1769

Part 2: Using the Poisson Approximation

Step 1: Check conditions:

n = 200, p = 0.01, \lambda = np = 200 \times 0.01 = 2

Poisson approximation is appropriate: $Y \sim \text{Po}(2)$

Step 2: Use the Poisson formula:

P(Y = 3) = \frac{\lambda^3 e^{-\lambda}}{3!}

Substitute $\lambda = 2$

P(Y = 3) = \frac{2^3 e^{-2}}{6} = \frac{8 \times 0.1353}{6} = 0.1804

Final Answer:

Binomial: $P(X = 3) = :success[0.1769]$
Poisson: $P(Y = 3) = :success[0.1804]$ The Poisson approximation is close to the binomial result.

Note Summary

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Common Mistakes

Using Poisson approximation when $n$ is small or $p$ is large: The approximation is only valid when $n$ is large and $p$ is small ( $np \leq 10$ ).
Forgetting to calculate $\lambda = np$ : Ensure the Poisson mean is correctly computed.
Ignoring the factorial in the Poisson formula: The denominator $x!$ is essential when calculating probabilities.
Assuming exact results: The Poisson approximation is an estimate and may slightly differ from the binomial result.

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Key Formulas

Binomial Distribution:

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

Poisson Approximation:

P(Y = x) = \frac{\lambda^x e^{-\lambda}}{x!}, \quad \lambda = np

Conditions for Approximation:

$n$ is large,
$p$ is small,
$\lambda = np \leq 10$

Poisson Approximations of Binomials Simplified Revision Notes for A-Level Edexcel Further Maths Further Statistics 1

19.1.2 Poisson Approximations of Binomials

Introduction

Binomial Distribution

Poisson Approximation to Binomial

Why Use the Poisson Approximation?

Worked Examples

Example 1: Using the Poisson Approximation

Example 2: Comparing Binomial and Poisson Probabilities

Note Summary

Common Mistakes

Key Formulas

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