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-

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 ?

Answered by aleks041103 last updated on 10/Aug/22

i f s h e d i d N t r i a l s t h e n s h e f a i l e d N - 1 t i m e s a n d s u c c e d e d 1 t i m e .

T h e r e f o r e, t h e p r o b a b i l i t y o f g e t t i n g a n o f f e r

o n t h e N - t h t r i a l i s

p_{N} = {(1 - 0.35)}^{N - 1} 0.35 = 0.35 {(0.65)}^{N - 1}

c h e c k i f t h i s i s i n d e e d a p r o b a b i l i t y d i s t r i b u t i o n

\underset{i = 1}{\sum^{\infty}} p_{i} = 0.35 \underset{i = 0}{\sum^{\infty}} {(0.65)}^{i} = 0.35 \frac{1}{1 - 0.65} = 1

\Rightarrow i t i s!

\Rightarrow ⟨ N ⟩ = \underset{i = 1}{\sum^{\infty}} {i p}_{i} = 0.35 \underset{i = 1}{\sum^{\infty}} {i x}^{i - 1} =

= 0.35 \underset{i = 1}{\sum^{\infty}} {(x^{i})}^{'} = 0.35 {(\underset{i = 1}{\sum^{\infty}} x^{i})}^{'} =

= 0.35 {(\frac{x}{1 - x})}^{'} = \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

thanks a lot

Commented by MathsFan last updated on 10/Aug/22

but which method is this

Commented by aleks041103 last updated on 10/Aug/22

w d y m ?

t h i s j u s t f i n d i n g t h e e x p e c t e d v a l u e

o f a q u a n t i t y w i t h a g i v e n p r o b a b i l i t y

d i s t r i b u t i o n

Commented by Tawa11 last updated on 10/Aug/22

Great sir

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