## Question from ECE QE CS Q1 August 2000

Inquiries arrive at a recorded message device according to a Poisson process of rate 15 inquiries per minute. Find the probability that in a 1-minute period, 3 inquiries arrive during the first 10 seconds and 2 inquiries arrive during the last 15 seconds.

# Solution 1 (retrived from here)

$\lambda=\frac{15}{60\text{ sec}}=\frac{1}{4}\text{ sec}^{-1}.$

$P\left(\left\{ N\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left(\left(\lambda\left(t_{2}-t_{1}\right)\right)^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}\right)}{k!}.$

$P\left(\left\{ N\left(0,10\right)=3\right\} \cap\left\{ N\left(45,60\right)=2\right\} \right)=P\left(\left\{ N\left(0,10\right)=3\right\} \right)P\left(\left\{ N\left(45,60\right)=2\right\} \right) $$=\frac{\left(\frac{1}{4}\times10\right)^{3}e^{-\frac{1}{4}\times10}}{3!}\times\frac{\left(\frac{1}{4}\times15\right)^{2}e^{-\frac{1}{4}\times15}}{2!}$$ =\frac{1}{12}\cdot\left(\frac{5}{2}\right)^{3}\left(\frac{15}{4}\right)^{2}e^{-\frac{25}{4}}.$

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