## Poisson approximation to binomial distribution

Let $X$ be a binomially distributed random variable with number of trials $n$ and probability of success $p$.

The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1-p)$.

The general rule of thumb to use Poisson approximation to binomial distribution is that the sample size $n$ is sufficiently large and $p$ is sufficiently small such that $\lambda=np$ (finite).

For sufficiently large $n$ and small $p$, $X\sim P(\lambda)$.

The probability mass function of Poisson distribution with parameter $\lambda$ is
```
$$
\begin{equation*}
P(X=x)= \left\{
\begin{array}{ll}
\dfrac{e^{-\lambda}\lambda^x}{x!} , & \hbox{$x=0,1,2,\cdots; \lambda>0$;} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

## Example 1

Suppose 1% of all screw made by a machine are defective. We are interested in the probability that a batch of 225 screws has at most one defective screw. Compute

a. the exact answer;

b. the Poisson approximation.

### Solution

Let $X$ denote the number of defective screw produced by a machine. Let $p$ be the probability that a screw produced by a machine is defective.

Given that $n=225$ (large) and $p=0.01$ (small). $X\sim B(225, 0.01)$.

a. Using Binomial Distribution: The probability that a batch of 225 screws has at most 1 defective screw is

```
$$
\begin{aligned}
P(X\leq 1) & =\sum_{x=0}^{1} P(X=x)\\
& =P(X=0) + P(X=1) \\
& = 0.1042+0.2368\\
&= 0.3411
\end{aligned}
$$
```

b. Using Poisson Approximation: If $n$ is sufficiently large and $p$ is sufficiently large such that that $\lambda = n*p$ is finite, then we use Poisson approximation to binomial distribution.

Here $\lambda=n*p = 225*0.01= 2.25$ (finite). Thus $X\sim P(2.25)$ distribution.

The probability mass function of $X$ is

```
$$
\begin{aligned}
P(X=x) &= \frac{e^{-2.25}2.25^x}{x!}; x=0,1,2,\cdots
\end{aligned}
$$
```

The probability that a batch of 225 screws has at most 1 defective screw is

```
$$
\begin{aligned}
P(X\leq 1) &= P(X=0)+ P(X=1)\\
&= \frac{e^{-2.25}2.25^{0}}{0!}+\frac{e^{-2.25}2.25^{1}}{1!}\\
&= 0.1054+0.2371\\
&= 0.3425
\end{aligned}
$$
```

## Example 2

On the average, 1 in 800 computers crashes during a severe thunderstorm. A certain company had 4,000 working computers when the area was hit by a severe thunderstorm.

a. Compute the expected value and variance of the number of crashed computers.

b. Compute the probability that less than 10 computers crashed.

c. Compute the probability that exactly 10 computers crashed.

### Solution

Let $X$ be the number of crashed computers out of $4000$. Let `$p=1/800$`

be the probability that a computer crashed during severe thunderstorm. Thus `$X\sim B(4000, 1/800)$`

.

Here $n=4000$ (sufficiently large) and `$p=1/800$`

(sufficiently small) such that `$\lambda =n*p =4000*1/800= 5$`

is finite. Thus we use Poisson approximation to Binomial distribution.

That is $X\sim P(5)$ distribution.

The probability mass function of $X$ is

```
$$
\begin{aligned}
P(X=x) &= \frac{e^{-5}5^x}{x!}; x=0,1,2,\cdots
\end{aligned}
$$
```

a. The expected value of the number of crashed computers
```
$$
\begin{aligned}
E(X)&= n*p\\
&=4000* 1/800\\
&=5
\end{aligned}
$$
```

The variance of the number of crashed computers
```
$$
\begin{aligned}
V(X)&= n*p*(1-p)\\
&=4000* 1/800*(1-1/800)\\
&=4.99
\end{aligned}
$$
```

b. The probability that less than 10 computers crashed is

```
$$
\begin{aligned}
P(X<10) &= P(X\leq 9)\\
&= 0.9682\\
& \quad \quad (\because \text{Using Poisson Table})
\end{aligned}
$$
```

c. The probability that exactly 10 computers crashed is
```
$$
\begin{aligned}
P(X= 10) &= P(X=10)\\
&= \frac{e^{-5}5^{10}}{10!}\\
&= 0.0181
\end{aligned}
$$
```

Hope this article helps you understand how to use Poisson approximation to binomial distribution to solve numerical problems.