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=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | =Solution 1 (retrived from [[ECE600_QE_2000_August|here]])= | ||
| + | <math class="inline">\lambda=\frac{15}{60\text{ sec}}=\frac{1}{4}\text{ sec}^{-1}.</math> | ||
| + | <math class="inline">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!}.</math> | ||
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| + | <math class="inline">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)</math><math class="inline">=\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!}</math><math class="inline">=\frac{1}{12}\cdot\left(\frac{5}{2}\right)^{3}\left(\frac{15}{4}\right)^{2}e^{-\frac{25}{4}}.</math> | ||
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Revision as of 08:56, 27 June 2012
Contents
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}}. $
Solution 2
Write it here.
