Revision as of 09:02, 27 June 2012 by Mboutin (Talk | contribs)


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.


Share and discuss your solutions below.


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}}. $


Solution 2

Write it here.


Back to QE CS question 1, August 2000

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett