Let $N_t$ denote the number of accidents upto time $t$. Assume that $ N_t$ is Poisson process with intensity 2. Given that there are exactly 5 accidents in the period $[20,30]$ what is the probability that there is exactly one accident during the time period $[15,25]$
Solution
$[15,25]_1 \longrightarrow $ event of exactly one accident in $[15,25]$ time period
$[25,30]_4 \longrightarrow $ event of exactly 4 accidents in $[25,30]$ time period
$[20,30]_5 \longrightarrow $ event of exactly 5 accidents in $[20,30]$ time period
So we need to calculate $ P([15,25]_1)$ given $[20,30]_5$. So using Bays theorem the probability is
\begin{align} \frac{P \left([20,30]_5 \cap [15,25]_1\right)}{P([20,30]_5)} & = \frac{P([15,25]_1 \cap [25,30]_4)}{P([20,30]_5)}\\ & = \frac{P([15,25]_1) \cap P([25,30]_4)}{P([20,30]_5)}\\ & = \frac{\frac{5^1e^{-5}}{1!}\frac{2.5^4e^{-2.5}}{4!}}{\frac{5^5e^{-5}}{5!}} \end{align}
I am using the fact that intensity is 2 (i.e an accident every 2 minutes), we have that $ \lambda =0.5$ (accidents per minute).
This calculation leads to the final answer $\frac{5}{16}e^{-2.5} $. But the answer provided is $\frac{15}{32}e^{-10}$ where did I go wrong?
It appears you have not taken into account the time elapsed in each interval. Specifically, if the intensity is $\lambda = 2$ per unit of time, then $$\Pr[N_t = n] = e^{-\lambda t} \frac{(\lambda t)^n}{n!}, \quad n = 0, 1, 2, \ldots.$$
The other problem is that you need to take into account the fact that there are two cases to consider: either the event that occurs in $(15,25]$ occurs in $(20,25]$, or it occurs in $(15,20]$. In the first case, then we have $0$ events in $(15,20]$, $1$ event in $(20,25]$, and $4$ events in $(25,30]$. In the second case, we have $1$ event in $(15,20]$, $0$ events in $(20,25]$, and $5$ events in $(25,30]$. To obtain the total probability of the desired event, you need to consider these two mutually exclusive cases.