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Question Number 123100 by pticantor last updated on 23/Nov/20

$$\mathit{l}\mathit{e}\mathit{t}{\mathit{P}}_{\mathit{k}}=\frac{\mathit{C}}{\mathit{k}\times 4^{\mathit{k}}}$$$$1)\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathit{C}\mathit{s}\mathit{o}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{t}\mathit{h}\mathit{e}\mathit{f}\mathit{a}\mathit{m}\mathit{i}\mathit{l}\mathit{y}$$$$({\mathit{p}}_{\mathit{k}},\mathit{k}\in {\mathbb{N}}^{\ast })\mathit{d}\mathit{e}\mathit{f}\mathit{i}\mathit{n}\mathit{e}\mathit{a}\mathit{m}\mathit{e}\mathit{a}\mathit{s}\mathit{u}\mathit{r}\mathit{e}\mathit{o}\mathit{f}\mathit{p}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}$$$$2)\mathit{w}\mathit{e}\mathit{d}\mathit{r}\mathit{a}\mathit{w}\mathit{a}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{i}\mathit{n}{\mathbb{N}}^{\ast }\mathit{a}\mathit{c}\mathit{c}\mathit{o}\mathit{r}\mathit{d}\mathit{i}\mathit{n}\mathit{g}\mathit{t}\mathit{o}\mathit{t}\mathit{h}\mathit{e}$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\mathit{P}\mathit{d}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathit{d}\mathit{t}\mathit{h}\mathit{e}\mathit{p}\mathit{r}\mathit{o}\mathit{b}\mathit{a}\mathit{b}\mathit{i}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}$$$$\mathit{d}\mathit{r}\mathit{a}\mathit{w}\mathit{n}\mathit{i}\mathit{s}\mathit{a}\mathit{m}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{p}\mathit{l}\mathit{e}\mathit{o}\mathit{f}4$$$$$$$$\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}\mathit{i}\mathit{n}\mathit{e}\mathit{e}\mathit{d}\mathit{y}\mathit{o}\mathit{u}\mathit{r}\mathit{h}\mathit{e}\mathit{l}\mathit{p}$$

Answered by Olaf last updated on 23/Nov/20

$$1)$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}{\mathrm{P}}_k=1$$$$\mathrm{C}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k{4}^k}=1$$$$\frac{1}{1-x}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{x}^k$$$$-\ln \mid 1-x\mid =\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{x^{k+1}}{k+1}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^k}{k}\mathrm{if}\mid x\mid <1$$$$\mathrm{If}x=\frac{1}{4},-\ln \frac{3}{4}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k{4}^k}$$$$\Rightarrow -\mathrm{C}.\ln \frac{3}{4}=1\Rightarrow \mathrm{C}=\frac{1}{\ln \frac{4}{3}}$$$$2)$$$${\mathrm{P}}_k=\frac{1}{k{4}^k\ln \frac{4}{3}}$$$$\mathrm{Let}k=4j,j\in {\mathbb{N}}^{\ast }$$$$$$$$\underset{j=1}{\sum ^{\mathrm{\infty }}}{\mathrm{P}}_k=\frac{1}{\ln \frac{4}{3}}\underset{j=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4j{4}^{4j}}=\frac{1}{4.\ln \frac{4}{3}}\underset{j=1}{\sum ^{\mathrm{\infty }}}\frac{1}{j{256}^j}$$$$\underset{j=1}{\sum ^{\mathrm{\infty }}}{\mathrm{P}}_k=\frac{1}{4.\ln \frac{4}{3}}\underset{j=1}{\sum ^{\mathrm{\infty }}}\frac{1}{j{256}^j}=-\frac{\ln \mid 1-\frac{1}{256}\mid }{4.\ln \frac{4}{3}}$$$$\underset{j=1}{\sum ^{\mathrm{\infty }}}{\mathrm{P}}_k=\frac{\ln \frac{256}{255}}{4.\ln \frac{4}{3}}$$$$\left(I^{\prime }mnotsosureofmyresult\right)$$

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