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Question Number 123100 by pticantor last updated on 23/Nov/20
let P_k =(C/(k×4^(k ) )) 1) determine C so that the family (p_k ,k∈N^∗ ) define a measure of probability 2) we draw a number in N^∗ according to the probability P determined the probability that the number drawn is a multiple of 4 please i need your help
$$\boldsymbol{{let}}\:\boldsymbol{{P}}_{\boldsymbol{{k}}} =\frac{\boldsymbol{{C}}}{\boldsymbol{{k}}×\mathrm{4}^{\boldsymbol{{k}}\:} } \\ $$$$\left.\mathrm{1}\right)\:\boldsymbol{{determine}}\:\boldsymbol{{C}}\:\boldsymbol{{so}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{family}} \\ $$$$\left(\boldsymbol{{p}}_{\boldsymbol{{k}}} ,\boldsymbol{{k}}\in\mathbb{N}^{\ast} \right)\:\boldsymbol{{define}}\:\boldsymbol{{a}}\:\boldsymbol{{measure}}\:\boldsymbol{{of}}\:\boldsymbol{{probability}} \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{we}}\:\boldsymbol{{draw}}\:\boldsymbol{{a}}\:\boldsymbol{{number}}\:\boldsymbol{{in}}\:\mathbb{N}^{\ast} \:\boldsymbol{{according}}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\: \\ $$$$\boldsymbol{{probability}}\:\boldsymbol{{P}}\:\boldsymbol{{determined}}\:\boldsymbol{{the}}\:\boldsymbol{{probability}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{number}} \\ $$$$\boldsymbol{{drawn}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{multiple}}\:\boldsymbol{{of}}\:\mathrm{4} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\boldsymbol{{please}}\:\boldsymbol{{i}}\:\boldsymbol{{need}}\:\boldsymbol{{your}}\:\boldsymbol{{help}} \\ $$
Answered by Olaf last updated on 23/Nov/20
1) Σ_(k=1) ^∞ P_k = 1 CΣ_(k=1) ^∞ (1/(k4^k )) = 1 (1/(1−x)) = Σ_(k=0) ^∞ x^k −ln∣1−x∣ = Σ_(k=0) ^∞ (x^(k+1) /(k+1)) = Σ_(k=1) ^∞ (x^k /k) if ∣x∣ < 1 If x = (1/4), −ln(3/4) = Σ_(k=1) ^∞ (1/(k4^k )) ⇒ −C.ln(3/4) = 1 ⇒ C = (1/(ln(4/3))) 2) P_k = (1/(k4^k ln(4/3))) Let k = 4j, j∈N^∗ Σ_(j=1) ^∞ P_k = (1/(ln(4/3)))Σ_(j=1) ^∞ (1/(4j4^(4j) )) = (1/(4.ln(4/3)))Σ_(j=1) ^∞ (1/(j256^j )) Σ_(j=1) ^∞ P_k = (1/(4.ln(4/3)))Σ_(j=1) ^∞ (1/(j256^j )) = −((ln∣1−(1/(256))∣)/(4.ln(4/3))) Σ_(j=1) ^∞ P_k = ((ln((256)/(255)))/(4.ln(4/3))) (I′m not so sure of my result)
$$\left.\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{P}_{{k}} \:=\:\mathrm{1} \\ $$$$\mathrm{C}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}\mathrm{4}^{{k}} }\:=\:\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{k}} \\ $$$$−\mathrm{ln}\mid\mathrm{1}−{x}\mid\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}}\:\mathrm{if}\:\mid{x}\mid\:<\:\mathrm{1} \\ $$$$\mathrm{If}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}},\:−\mathrm{ln}\frac{\mathrm{3}}{\mathrm{4}}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}\mathrm{4}^{{k}} } \\ $$$$\Rightarrow\:−\mathrm{C}.\mathrm{ln}\frac{\mathrm{3}}{\mathrm{4}}\:=\:\mathrm{1}\:\Rightarrow\:\mathrm{C}\:=\:\frac{\mathrm{1}}{\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{P}_{{k}} \:=\:\frac{\mathrm{1}}{{k}\mathrm{4}^{{k}} \mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\mathrm{Let}\:{k}\:=\:\mathrm{4}{j},\:{j}\in\mathbb{N}^{\ast} \\ $$$$ \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{P}_{{k}} \:=\:\frac{\mathrm{1}}{\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}}\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}{j}\mathrm{4}^{\mathrm{4}{j}} }\:=\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}}\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{j}\mathrm{256}^{{j}} } \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{P}_{{k}} \:=\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}}\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{j}\mathrm{256}^{{j}} }\:=\:−\frac{\mathrm{ln}\mid\mathrm{1}−\frac{\mathrm{1}}{\mathrm{256}}\mid}{\mathrm{4}.\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{P}_{{k}} \:=\:\frac{\mathrm{ln}\frac{\mathrm{256}}{\mathrm{255}}}{\mathrm{4}.\mathrm{ln}\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\left({I}'{m}\:{not}\:{so}\:{sure}\:{of}\:{my}\:{result}\right) \\ $$

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