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Question Number 174737 by MathsFan last updated on 09/Aug/22

$$\mathrm{A}\mathrm{graduating}\mathrm{student}\mathrm{keeps}\mathrm{applying}$$$$\mathrm{for}\mathrm{a}\mathrm{job}\mathrm{until}\mathrm{she}\mathrm{gets}\mathrm{an}\mathrm{offer}.\mathrm{The}$$$$\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{offer}\mathrm{at}\mathrm{any}$$$$\mathrm{trial}\mathrm{is}0.35.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{expected}$$$$\mathrm{number}\mathrm{of}\mathrm{applications}?$$

Answered by aleks041103 last updated on 10/Aug/22

$$ifshedidNtrialsthenshefailedN-1timesandsucceded1time.$$$$Therefore,theprobabilityofgettinganoffer$$$$ontheN-thtrialis$$$$p_N={(1-0.35)}^{N-1}0.35=0.35{(0.65)}^{N-1}$$$$checkifthisisindeedaprobabilitydistribution$$$$\underset{i=1}{\sum ^{\mathrm{\infty }}}{p}_i=0.35\underset{i=0}{\sum ^{\mathrm{\infty }}}{(0.65)}^i=0.35\frac{1}{1-0.65}=1$$$$\Rightarrow itis!$$$$\Rightarrow ⟨N⟩=\underset{i=1}{\sum ^{\mathrm{\infty }}}{ip}_i=0.35\underset{i=1}{\sum ^{\mathrm{\infty }}}{ix}^{i-1}=$$$$=0.35\underset{i=1}{\sum ^{\mathrm{\infty }}}{\left(x^i\right)}^{\prime }=0.35{\left(\underset{i=1}{\sum ^{\mathrm{\infty }}}{x}^i\right)}^{\prime }=$$$$=0.35{\left(\frac{x}{1-x}\right)}^{\prime }=\frac{0.35}{{(1-x)}^2}=\frac{0.35}{0.{35}^2}=\frac{1}{0.35}$$$$\Rightarrow ⟨N⟩=\frac{100}{35}=\frac{20}{7}\approx 3$$

Commented by MathsFan last updated on 10/Aug/22

$$\mathrm{thanks}\mathrm{a}\mathrm{lot}$$

Commented by MathsFan last updated on 10/Aug/22

$$\mathrm{but}\mathrm{which}\mathrm{method}\mathrm{is}\mathrm{this}$$

Commented by aleks041103 last updated on 10/Aug/22

$$wdym?$$$$thisjustfindingtheexpectedvalue$$$$ofaquantitywithagivenprobability$$$$distribution$$

Commented by Tawa11 last updated on 10/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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