Suppose calls to Dryden Fire Department arrive according to a Poisson Process with rate $0.5$ per hour. Suppose the time $T$ needed to respond to a call, return to the station and be ready for the next call is uniformly distributed between $0.5 $ hour and $1$ hour. If a new call comes before Dryden is ready to respond, then Ithaca fire station is requested to help. Suppose the Dryden fire station is ready to respond now. Then, find the probability distribution of the number of calls they will handle before they have to request Ithaca to help.
Okay, so i gathered that letting $N(t)$ be the number of calls by time $t$, $N(t)\in Poisson(0.5t)$. Hence the interarrival time is $Exponential(0.5)$.
If $S$ is the interarrival time, then we "should be" interested in $P(S<T)$, which is the probability that the time between two consecutive calls is less than the time taken to respond to the previous call.
Now denoting the number of calls taken by Dryden before having to ask Ithaca for help, by $N$, we want to find $P(N=n)$ where $n\geq1$.
It seems that it has a Geometric distribution with $p=P(S<T)$ i.e. first success occurs when $S<T$.
However, I am getting $P(S<T)=(1-2e^{-1/4})^2$ which is not at all the answer. Answer is $P(N=n)=(1-p)^{n-1}p$ for $n\geq1$ where $p=e^{-1}-e^{-2}$. Also, is my reasoning right?
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $S\sim \exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.