Poisson Distribution to determine number of defective tyres.

Question

In a certain factory producing cycle tyres, there is a small chance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective tyres in a consignment of 10,000 lots.

First we have to calculate $\lambda$ which is equal to $np$. My confusion is what should be the criteria for choosing $n$ ? In this case I have to select $n=$ 10,000 or 10 ?

Answer 1

The number of defectives in a lot of $10$ has mean $10/500$. It is binomially distributed, but since $10/500$ is small, the probability a lot has no defectives is well-approximated by the probability that a Poisson with parameter $\lambda=10/500$ takes on the value $0$.

Thus the probability $p$ of no defectives in a lot of $10$ is given approximately by $p=e^{-10/500}$.

We interpret approximate number of lots with no defectives as the mean number of lots with no defective. This is $10000p$.

Remark: We dragged in the Poisson only because the problem seems to ask us to do so. However, finding the mean number of defective lots in $10000$ does not require the Poisson, and the Poisson approximation does not even simplify the computation.

Answer 2

I think it will be clearer, if a bit longer, to write out in some detail about what you're approximating with what. That's because there are really two $n$s and two $p$s in the model, but you're only going to use one of each in your Poisson approximation.

First, we have the probability distribution for the number of defective tires in a given lot of 10. This is a binomial random variable with $n=10,p=(1/500)$. The probability that this is zero is the probability that a given lot of tires has no defective tires in it. This is exactly $(499/500)^{10}$.

Next, we have the probability distribution for the number of lots in a consignment of 10000 lots which contain no defective tires. Now the probability of a lot to have no defective tires is $(499/500)^{10}$ (we already computed it), so this is a binomial random variable with $n=10000$ and $p=(499/500)^{10}$.

Now you approximate that binomial variable with a Poisson variable with those values of $n$ and $p$.