I would like to ask the (iii) part of the following question:
My solution: Consier the time period of lenght $t$, starting from when they both start fishing. The probability that Adam catches one or more fishes is given by $$1-e^{\frac{-\lambda t}{T}},$$ as the number of fish cought in a time interval $T$ follows a poisson distribution. Similarly, the probability that Eve catches no fishes in the time period of $t$ is $$e^{\frac{-\mu t}{T}}.$$ Therefore, the probability that Adam catches a fish first, in the time period $t$, is $$e^{\frac{- \mu t}{T}}\times(1-e^{\frac{-\lambda t}{T}})$$. We need now consider all cases, which are given by $0\leq t \leq T$, done by intigration. Therefore, the desired probability is $$=\int_{0}^{T} e^{\frac{- \mu t}{T}}\times(1-e^{\frac{-\lambda t}{T}}) dt=\frac{-T}{\mu}e^\mu + \frac{T}{\mu + \lambda}e^{-(\mu + \lambda)}+\frac{-T}{\mu}-\frac{T}{\lambda + \mu}.$$
However, this is wrong. According to the markscheme, one just has to note that this is the same as setting $r=1$ (in part (i)) and $k=1$ (in part (ii)). This give the answer as $$\frac{\lambda}{\lambda + \mu}.$$
My question: Where have I gone wrong with my methodology?