Problem source

Four students, one of whom is a mathematician, take turns at washing up over a long period of time. The number of plates broken by any student in this time obeys a Poisson distribution, the probability of any given student breaking $n$ plates being  $\e^{-\lambda} \lambda^n/n!$ for some fixed constant $\lambda$, independent of the number of breakages by other students. 
Given that five plates are broken, find the probability that three or more were broken by the mathematician.

Solution source

Let $X$ be the number of plates broken by the mathematician and $Y$ by the other student. Then $X \sim Po(\lambda), Y \sim Po(3\lambda)$ and $X+Y \sim Po(4\lambda)$

\begin{align*}
&& \mathbb{P}(X = k | X+Y = n) &= \frac{\mathbb{P}(X = k, Y = n-k)}{\mathbb{P}(X+Y=n)} \\ 
&&&= \frac{e^{-\lambda} \lambda^k/k! \cdot e^{-3\lambda} (4\lambda)^{n-k}/(n-k)!}{e^{-4\lambda}(4\lambda)^n/n!} \\
&&&= \binom{n}{k} \left ( \frac{1}{4} \right)^k \left ( \frac{3}{5} \right)^{n-k}
\end{align*}

Therefore $X | X+Y = n \sim Binomial(n, \tfrac14)$

\begin{align*}
\mathbb{P}(X \geq 3 | X + Y = n) &= \binom{5}{3} \frac{3^2}{4^5} + \binom{5}{4} \frac{3}{4^5} + \binom{5}{5} \frac{1}{4^5} \\
&= \frac{1}{4^5} \left ( 90+ 15 + 1 \right) \\
&= \frac{106}{4^5} = \frac{53}{512} \approx \frac1{10}
\end{align*}