Ques: The number of patients arriving at a regional hospital has a Poisson
distribution with mean 20 per hour. Assume that 10% of the patients were involved in
an accident.
-
Find the probability that two or more patients arriving at the hospital over a 2-hour
period were involved in an accident. -
What if the average number of patients is 15 the first hour and 25 the second hour?
My understanding so far-
Let $X$ be the no. of patients that came to the hospital every hour.
Given: $E[X]=20$ or $X\sim Poisson(20)$
Now, assuming that 10% of the ones that arrive have met with an accident, I can create a new random variable named $Y$ which is the number of people arriving at the hospital who have met with an accident.
So, $E[Y]=2$ or $Y\sim Poisson(2)$.
$P(Y\geq 2 )= 1 – P(Y=1) – Pr(Y=0) $ = $1-e^{-2}\frac{2^0}{0!} – e^{-2}\frac{2^1}{1!}$
$= 1-3e^{-2}$
Now, for two hours, I can just multiply the above derived probability twice i.e. $(1-3e^{-2})^{2}$
Am I correct So far? Thanks a lot for any help!
Best Answer
You need to consider the Poisson process, $X(t)$ is the number of patients who arrived by time $t$, and $Y(t)$ is the number of patients with accidents who arrived by time $t$. The number of arrivals for the Poisson process with rate $\lambda$ follows Poisson distribution with parameter $\lambda t$. For the $Y-$process $\lambda=20\times 0.1=2$ as you computed. Hence the number of patients who arrived in two hours is Poisson with parameter $2\lambda=4$. From here you can easily derive the probability.