Question Number 174737 by MathsFan last updated on 09/Aug/22
$$\mathrm{A}\:\mathrm{graduating}\:\mathrm{student}\:\mathrm{keeps}\:\mathrm{applying} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{job}\:\mathrm{until}\:\mathrm{she}\:\mathrm{gets}\:\mathrm{an}\:\mathrm{offer}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{an}\:\mathrm{offer}\:\mathrm{at}\:\mathrm{any} \\ $$$$\mathrm{trial}\:\mathrm{is}\:\mathrm{0}.\mathrm{35}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{expected}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{applications}? \\ $$
Answered by aleks041103 last updated on 10/Aug/22
$${if}\:{she}\:{did}\:{N}\:{trials}\:{then}\:{she}\:{failed}\:{N}−\mathrm{1}\:{times}\:{and}\:{succeded}\:\mathrm{1}\:{time}. \\ $$$${Therefore},\:{the}\:{probability}\:{of}\:{getting}\:{an}\:{offer} \\ $$$${on}\:{the}\:{N}−{th}\:{trial}\:{is} \\ $$$${p}_{{N}} =\left(\mathrm{1}−\mathrm{0}.\mathrm{35}\right)^{{N}−\mathrm{1}} \mathrm{0}.\mathrm{35}=\mathrm{0}.\mathrm{35}\left(\mathrm{0}.\mathrm{65}\right)^{{N}−\mathrm{1}} \\ $$$${check}\:{if}\:{this}\:{is}\:{indeed}\:{a}\:{probability}\:{distribution} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{p}_{{i}} =\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{0}.\mathrm{65}\right)^{{i}} =\mathrm{0}.\mathrm{35}\frac{\mathrm{1}}{\mathrm{1}−\mathrm{0}.\mathrm{65}}=\mathrm{1} \\ $$$$\Rightarrow{it}\:{is}! \\ $$$$\Rightarrow\langle{N}\rangle=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ip}_{{i}} =\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ix}^{{i}−\mathrm{1}} = \\ $$$$=\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left({x}^{{i}} \right)'=\mathrm{0}.\mathrm{35}\left(\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{i}} \right)'= \\ $$$$=\mathrm{0}.\mathrm{35}\left(\frac{{x}}{\mathrm{1}−{x}}\right)'=\frac{\mathrm{0}.\mathrm{35}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{0}.\mathrm{35}}{\mathrm{0}.\mathrm{35}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{0}.\mathrm{35}} \\ $$$$\Rightarrow\langle{N}\rangle=\frac{\mathrm{100}}{\mathrm{35}}=\frac{\mathrm{20}}{\mathrm{7}}\approx\mathrm{3} \\ $$
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
$${wdym}? \\ $$$${this}\:{just}\:{finding}\:{the}\:{expected}\:{value} \\ $$$${of}\:{a}\:{quantity}\:{with}\:{a}\:{given}\:{probability} \\ $$$${distribution} \\ $$
Commented by Tawa11 last updated on 10/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$