An average of three calls arrive every $5$ min. Assuming a Poisson arrival rate, compute the probabilities of the following events:
(a) exactly four calls will arrive during a $5$ minute interval.
$$P(x=4) = \frac{e^{-0.6} (.6)^4}{4!}$$ Is that right?
(b) $5$ min will pass without a call.
And this one is just $P(X=0)$? I can do this if I know I did part A correctly.
Thanks for any helpful feedback.
Best Answer
No, you didn't do part a) correctly. If you consider some interval $[s,t+s)$, then the distribution of the number of arrivals follows a Poisson distribution with parameter $\lambda [(t+s)-s] = \lambda t$. They tell us that the rate is $\lambda = \frac{3\text{ calls}}{5 \text{ mins}}$.
Hence, if you call this $X_{(0,5)}$, then $$P(X_{(0,5)} =4) = e^{-\lambda5}\frac{(\lambda 5)^{4}}{4!} = e^{-3}\frac{3^4}{4!}.$$