Poisson Coin tossing on to a page question

Question

I am having difficulties with a question that goes as follows:

"The mean no. of misprints on a page of area 400 $cm^2$ is 10, A silver dollar of area 20 $cm^2$ is tossed on the page at random. Find the approximate probability that it covers at least one misprint"

I believe that one is supposed to use poisson variable but I cannot seem to reconcile the probability together. Could someone please explain please ?

Thank you.

Answer 1

We are told that there are $10$ errors per $400\text{ cm}^2$. Then the rate of errors $\lambda$ is $$\lambda = \frac{10\text{ errors}}{400\text{ cm}^2}.$$ Then the number of errors per unit area $N(A)$ follows a Poisson distribution with rate $\lambda|A|$, $$N\sim \text{Poisson}(\lambda|A|)$$ where $|A| = 20 \text{ cm}^2$, the 'size' of $A$.

Hence, if $N(A)$ is the number of errors that the coin covers, then \begin{align*} P(N(A)\geq 1) &= 1-P(N(A) = 0) \\ &= 1- e^{-\lambda|A|}\frac{(\lambda|A|)^0}{0!} \\ &= 1-\exp\left\{-\frac{10}{400}\cdot 20\right\} \\ &= 0.3934693. \end{align*}

Answer 2

The average number of misprints in a page of 20 $cm^2$ is 0.5. Hence our Poisson distribution will be of the form $P(x=n)=((0.5)^n/n!)e^{-0.5}$ The probability of getting at least one misprint is the complement of getting 0 misprints, hence $1-((0.5)^0/0!)e^{-0.5}=1-e^{-1/2}$ Note that $\lambda$= average in a Poisson distribution $(\lambda^n/n!)e^{-\lambda}$.