Problem source

The probability that there are exactly $n$ misprints in an issue of a newspaper is $\mathrm{e}^{-\lambda}\lambda^{n}/n!$ where $\lambda$ is a positive constant. The probability that I spot a particular misprint is $p$, independent of what happens for other misprints, and $0 < p < 1.$
\begin{questionparts}
	\item If there are exactly $m+n$ misprints, what is the probability that I spot exactly $m$ of them?
\item Show that, if I spot exactly $m$ misprints, the probability that I have failed to spot exactly $n$ misprints is 
\[
\frac{(1-p)^{n}\lambda^{n}}{n!}\mathrm{e}^{-(1-p)\lambda}.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item $\binom{m+n}{m} p^m (1-p)^n$

\item $\,$
\begin{align*}
\mathbb{P}(\text{failed to spot }n\text{ misprints}|\text{spotted }m\text{ misprints}) &= \frac{\mathbb{P}(\text{failed to spot }n\text{ misprints and spotted }m\text{ misprints}) }{\mathbb{P}(\text{spotted }m\text{ misprints})} \\
&= \frac{\binom{m+n}{n}p^m(1-p)^n e^{-\lambda} \lambda^{m+n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}p^m(1-p)^k e^{-\lambda} \lambda^{m+k}/(n+k)!} \\
&= \frac{\binom{m+n}{n}(1-p)^n \lambda^{n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}(1-p)^k  \lambda^{k}/(n+k)!} \\
&= \frac{(1-p)^n \lambda^{n}/n!}{\sum_{k=0}^{\infty} (1-p)^k  \lambda^{k}/k!} \\
&= \frac{(1-p)^n\lambda^n}{n!} e^{-(1-p)\lambda}
\end{align*}

Alternatively, given the missed misprints and spotted misprints are independent, we can view them as both following $Po(p\lambda)$ and $Po((1-p)\lambda)$ and so we obtain exactly this result, without calculation.
\end{questionparts}