1 problem found
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: 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*}