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

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

$$\mathrm{A}\:\mathrm{graduating}\:\mathrm{student}\:\mathrm{keeps}\:\mathrm{applying} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{job}\:\mathrm{until}\:\mathrm{she}\:\mathrm{gets}\:\mathrm{an}\:\mathrm{offer}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{an}\:\mathrm{offer}\:\mathrm{at}\:\mathrm{any} \\ $$$$\mathrm{trial}\:\mathrm{is}\:\mathrm{0}.\mathrm{35}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{expected}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{applications}? \\ $$

Answered by aleks041103 last updated on 10/Aug/22

if she did N trials then she failed N−1 times and succeded 1 time.  Therefore, the probability of getting an offer  on the N−th trial is  p_N =(1−0.35)^(N−1) 0.35=0.35(0.65)^(N−1)   check if this is indeed a probability distribution  Σ_(i=1) ^∞ p_i =0.35Σ_(i=0) ^∞ (0.65)^i =0.35(1/(1−0.65))=1  ⇒it is!  ⇒⟨N⟩=Σ_(i=1) ^∞ ip_i =0.35Σ_(i=1) ^∞ ix^(i−1) =  =0.35Σ_(i=1) ^∞ (x^i )′=0.35(Σ_(i=1) ^∞ x^i )′=  =0.35((x/(1−x)))′=((0.35)/((1−x)^2 ))=((0.35)/(0.35^2 ))=(1/(0.35))  ⇒⟨N⟩=((100)/(35))=((20)/7)≈3

$${if}\:{she}\:{did}\:{N}\:{trials}\:{then}\:{she}\:{failed}\:{N}−\mathrm{1}\:{times}\:{and}\:{succeded}\:\mathrm{1}\:{time}. \\ $$$${Therefore},\:{the}\:{probability}\:{of}\:{getting}\:{an}\:{offer} \\ $$$${on}\:{the}\:{N}−{th}\:{trial}\:{is} \\ $$$${p}_{{N}} =\left(\mathrm{1}−\mathrm{0}.\mathrm{35}\right)^{{N}−\mathrm{1}} \mathrm{0}.\mathrm{35}=\mathrm{0}.\mathrm{35}\left(\mathrm{0}.\mathrm{65}\right)^{{N}−\mathrm{1}} \\ $$$${check}\:{if}\:{this}\:{is}\:{indeed}\:{a}\:{probability}\:{distribution} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{p}_{{i}} =\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{0}.\mathrm{65}\right)^{{i}} =\mathrm{0}.\mathrm{35}\frac{\mathrm{1}}{\mathrm{1}−\mathrm{0}.\mathrm{65}}=\mathrm{1} \\ $$$$\Rightarrow{it}\:{is}! \\ $$$$\Rightarrow\langle{N}\rangle=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ip}_{{i}} =\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{ix}^{{i}−\mathrm{1}} = \\ $$$$=\mathrm{0}.\mathrm{35}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left({x}^{{i}} \right)'=\mathrm{0}.\mathrm{35}\left(\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{i}} \right)'= \\ $$$$=\mathrm{0}.\mathrm{35}\left(\frac{{x}}{\mathrm{1}−{x}}\right)'=\frac{\mathrm{0}.\mathrm{35}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{0}.\mathrm{35}}{\mathrm{0}.\mathrm{35}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{0}.\mathrm{35}} \\ $$$$\Rightarrow\langle{N}\rangle=\frac{\mathrm{100}}{\mathrm{35}}=\frac{\mathrm{20}}{\mathrm{7}}\approx\mathrm{3} \\ $$

Commented by MathsFan last updated on 10/Aug/22

thanks a lot

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

Commented by MathsFan last updated on 10/Aug/22

but which method is this

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

Commented by aleks041103 last updated on 10/Aug/22

wdym?  this just finding the expected value  of a quantity with a given probability  distribution

$${wdym}? \\ $$$${this}\:{just}\:{finding}\:{the}\:{expected}\:{value} \\ $$$${of}\:{a}\:{quantity}\:{with}\:{a}\:{given}\:{probability} \\ $$$${distribution} \\ $$

Commented by Tawa11 last updated on 10/Aug/22

Great sir

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

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