(New page: ==Question from ECE QE CS Q1 August 2000== ---- ==Share and discuss your solutions below.== ---- =Solution 1 (retrived from [[ECE600_QE_2000_August|her...) |
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| − | + | [[Category:ECE]] | |
| + | [[Category:QE]] | ||
| + | [[Category:CNSIP]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:random variables]] | ||
| + | [[Category:probability]] | ||
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| + | <center> | ||
| + | <font size= 4> | ||
| + | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
| + | </font size> | ||
| + | |||
| + | <font size= 4> | ||
| + | Communication, Networking, Signal and Image Processing (CS) | ||
| + | |||
| + | Question 1: Probability and Random Processes | ||
| + | </font size> | ||
| + | |||
| + | August 2000 | ||
| + | </center> | ||
| + | ---- | ||
| + | ---- | ||
| + | =Part 3= | ||
| + | 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. | ||
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==Share and discuss your solutions below.== | ==Share and discuss your solutions below.== | ||
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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> | ||
| + | <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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Latest revision as of 10:34, 13 September 2013
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2000
Contents
Part 3
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.
