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| + | [[Category:random variables]] | ||
| + | [[Category:probability]] | ||
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| + | [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] | ||
| + | </font size> | ||
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| + | <font size= 4> | ||
| + | Communication, Networking, Signal and Image Processing (CS) | ||
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| + | Question 1: Probability and Random Processes | ||
| + | </font size> | ||
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| + | August 2001 | ||
| + | </center> | ||
| + | ---- | ||
| + | ---- | ||
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'''1. (10 Points)''' | '''1. (10 Points)''' | ||
Consider the following random experiment: A fair coin is repeatedly tossed until the same outcome (H or T) appears twice in a row. | Consider the following random experiment: A fair coin is repeatedly tossed until the same outcome (H or T) appears twice in a row. | ||
| − | (a) | + | (a) What is the probability that this experiment terminates on or before the seventh coin toss? |
| − | What is the probability that this experiment terminates | + | |
| + | |||
| + | '''(b)''' What is the probability that this experiment terminates with an even number of coin tosses? | ||
| + | ---- | ||
| + | ==Share and discuss your solutions below.== | ||
| + | ---- | ||
| + | ==Solution 1== | ||
| + | (a) | ||
Let N be the number of toss until the same outcome appears twice in a row. | Let N be the number of toss until the same outcome appears twice in a row. | ||
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<math class="inline">P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}.</math> | <math class="inline">P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}.</math> | ||
| − | + | (b) | |
| − | + | ||
| − | + | ||
<math class="inline">P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}.</math> | <math class="inline">P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}.</math> | ||
| + | ---- | ||
Latest revision as of 16:40, 13 March 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
August 2001
1. (10 Points)
Consider the following random experiment: A fair coin is repeatedly tossed until the same outcome (H or T) appears twice in a row.
(a) What is the probability that this experiment terminates on or before the seventh coin toss?
(b) What is the probability that this experiment terminates with an even number of coin tosses?
Solution 1
(a)
Let N be the number of toss until the same outcome appears twice in a row.
| $ N $th | $ \left(N - 1\right) $th | $ \left(N - 2\right) $th | $ \left(N - 3\right) $th | $ \cdots $ |
|---|---|---|---|---|
| H | H | T | H | $ \cdots $ |
| T | T | H | T | $ \cdots $ |
$ P\left(\left\{ N=n\right\} \right)=\frac{2}{2^{n}}=\frac{1}{2^{n-1}}\text{ for }n\geq2. $
$ P\left(\left\{ N\leq7\right\} \right)=\sum_{k=2}^{7}\frac{1}{2^{k-1}}=\sum_{k=1}^{6}\left(\frac{1}{2}\right)^{k}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{6}\right)}{1-\frac{1}{2}}=1-\frac{1}{64}=\frac{63}{64}. $
(b)
$ P\left(\left\{ N\text{ is even}\right\} \right)=\sum_{k=1}^{\infty}\frac{1}{2^{2k-1}}=2\sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^{k}=2\cdot\frac{\frac{1}{4}}{1-\frac{1}{4}}=2\cdot\frac{1}{3}=\frac{2}{3}. $
