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Anyone have any guidance on this? Thanks --[[User:Haddada|Haddada]] 20:46, 21 October 2010 (UTC) | Anyone have any guidance on this? Thanks --[[User:Haddada|Haddada]] 20:46, 21 October 2010 (UTC) | ||
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| + | * Here is part of my solution. For question a, suppose N is the rv. that equals to the number of balls you need and p is the prob. that a ball is put into ith bag | ||
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| + | Then p=1/n | ||
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| + | <math>P(N=k)=(1-p)^kp=(1-\frac{1}{n})^k\frac{1}{n}</math> | ||
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| + | <math> | ||
| + | \begin{align} | ||
| + | E[N]=\sum_{k=0}^{\infty}kP(N=k)&=\sum_{k=0}^{\infty}k(1-\frac{1}{n})^k\frac{1}{n} \\ | ||
| + | &=(1-\frac{1}{n})\frac{1}{n}+2(1-\frac{1}{n})^2\frac{1}{n}+3(1-\frac{1}{n})^3\frac{1}{n}+... \\ | ||
| + | &=(1-\frac{1}{n})\frac{1}{n}+(1-\frac{1}{n})^2\frac{1}{n}+(1-\frac{1}{n})^3\frac{1}{n}+... \\ | ||
| + | &+(1-\frac{1}{n})^2\frac{1}{n}+(1-\frac{1}{n})^3\frac{1}{n}+... \\ | ||
| + | &+(1-\frac{1}{n})^3\frac{1}{n}+... \\ | ||
| + | &=(1-\frac{1}{n})+(1-\frac{1}{n})^2+(1-\frac{1}{n})^3+... \\ | ||
| + | &=n-1 | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | For question b, I was stucked. | ||
| + | |||
| + | --[[User:zhao148|Zhao]] 23:50, 21 October 2010 (UTC) | ||
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Revision as of 17:48, 21 October 2010
Homework 5 Discussion
Please post your questions or comments here
- How to solve question 2?
Anyone have any guidance on this? Thanks --Haddada 20:46, 21 October 2010 (UTC)
- Here is part of my solution. For question a, suppose N is the rv. that equals to the number of balls you need and p is the prob. that a ball is put into ith bag
Then p=1/n
$ P(N=k)=(1-p)^kp=(1-\frac{1}{n})^k\frac{1}{n} $
$ \begin{align} E[N]=\sum_{k=0}^{\infty}kP(N=k)&=\sum_{k=0}^{\infty}k(1-\frac{1}{n})^k\frac{1}{n} \\ &=(1-\frac{1}{n})\frac{1}{n}+2(1-\frac{1}{n})^2\frac{1}{n}+3(1-\frac{1}{n})^3\frac{1}{n}+... \\ &=(1-\frac{1}{n})\frac{1}{n}+(1-\frac{1}{n})^2\frac{1}{n}+(1-\frac{1}{n})^3\frac{1}{n}+... \\ &+(1-\frac{1}{n})^2\frac{1}{n}+(1-\frac{1}{n})^3\frac{1}{n}+... \\ &+(1-\frac{1}{n})^3\frac{1}{n}+... \\ &=(1-\frac{1}{n})+(1-\frac{1}{n})^2+(1-\frac{1}{n})^3+... \\ &=n-1 \end{align} $
For question b, I was stucked.
--Zhao 23:50, 21 October 2010 (UTC)
