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probability interseting question not able to solve it [closed]

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0

$begingroup$

A random variable X can take all non-negative integer values and the probability that $X$ take the value $r$ is proportional to $a^r$ where a is between 0 to 1. Find $Pr{X=0}$

probability

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edited Mar 19 at 16:06

Mostafa Ayaz

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asked Mar 19 at 15:52

Shivam Shivam

32

$endgroup$

closed as off-topic by Saad, Lee David Chung Lin, Leucippus, Eevee Trainer, Cesareo Mar 22 at 8:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Leucippus, Eevee Trainer, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

0

$begingroup$

A random variable X can take all non-negative integer values and the probability that $X$ take the value $r$ is proportional to $a^r$ where a is between 0 to 1. Find $Pr{X=0}$

probability

share|cite|improve this question

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

asked Mar 19 at 15:52

Shivam Shivam

32

$endgroup$

closed as off-topic by Saad, Lee David Chung Lin, Leucippus, Eevee Trainer, Cesareo Mar 22 at 8:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Leucippus, Eevee Trainer, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

A random variable X can take all non-negative integer values and the probability that $X$ take the value $r$ is proportional to $a^r$ where a is between 0 to 1. Find $Pr{X=0}$

probability

share|cite|improve this question

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

asked Mar 19 at 15:52

Shivam Shivam

32

$endgroup$

share|cite|improve this question

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

asked Mar 19 at 15:52

Shivam Shivam

32

share|cite|improve this question

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

asked Mar 19 at 15:52

Shivam Shivam

32

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

edited Mar 19 at 16:06

Mostafa Ayaz

18.2k31040

asked Mar 19 at 15:52

Shivam Shivam

32

asked Mar 19 at 15:52

Shivam Shivam

32

1 Answer
1

active

oldest

votes

1

$begingroup$

The general form of pmf goes like this$$Pr{X=r}=kcdot a^r$$where $k$ is a constant. As any pmf $f(n)$ must satisfy $sum f(n)=1$ we must have $$sum_{r=0}^{infty}Pr{X=r}=sum_{r=0}^{infty}kcdot a^r=ksum_{r=0}^{infty} a^r=1$$therefore $$k={1over sum_{r=0}^{infty} a^r}$$and we need to find $sum_{r=0}^{infty}a^r$ which equals to ${1over 1-a}$ according to geometric series sum formula. Therefore $k=1-a$ and the final answer becomes $$Pr{X=r}={a^rcdot (1-a)}$$so we have$$Pr{X=0}=1-a$$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  answer again its answer is 1-a
  $endgroup$
  – Shivam
  Mar 19 at 16:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks great answer
  $endgroup$
  – Shivam
  Mar 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your welcome. Good luck!
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 16:43
  

  
  
  
  

add a comment | 

1

$begingroup$

The general form of pmf goes like this$$Pr{X=r}=kcdot a^r$$where $k$ is a constant. As any pmf $f(n)$ must satisfy $sum f(n)=1$ we must have $$sum_{r=0}^{infty}Pr{X=r}=sum_{r=0}^{infty}kcdot a^r=ksum_{r=0}^{infty} a^r=1$$therefore $$k={1over sum_{r=0}^{infty} a^r}$$and we need to find $sum_{r=0}^{infty}a^r$ which equals to ${1over 1-a}$ according to geometric series sum formula. Therefore $k=1-a$ and the final answer becomes $$Pr{X=r}={a^rcdot (1-a)}$$so we have$$Pr{X=0}=1-a$$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  answer again its answer is 1-a
  $endgroup$
  – Shivam
  Mar 19 at 16:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks great answer
  $endgroup$
  – Shivam
  Mar 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your welcome. Good luck!
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 16:43
  

  
  
  
  

add a comment | 

1

$begingroup$

The general form of pmf goes like this$$Pr{X=r}=kcdot a^r$$where $k$ is a constant. As any pmf $f(n)$ must satisfy $sum f(n)=1$ we must have $$sum_{r=0}^{infty}Pr{X=r}=sum_{r=0}^{infty}kcdot a^r=ksum_{r=0}^{infty} a^r=1$$therefore $$k={1over sum_{r=0}^{infty} a^r}$$and we need to find $sum_{r=0}^{infty}a^r$ which equals to ${1over 1-a}$ according to geometric series sum formula. Therefore $k=1-a$ and the final answer becomes $$Pr{X=r}={a^rcdot (1-a)}$$so we have$$Pr{X=0}=1-a$$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  answer again its answer is 1-a
  $endgroup$
  – Shivam
  Mar 19 at 16:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks great answer
  $endgroup$
  – Shivam
  Mar 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your welcome. Good luck!
  $endgroup$
  – Mostafa Ayaz
  Mar 19 at 16:43
  

  
  
  
  

add a comment | 

$begingroup$

The general form of pmf goes like this$$Pr{X=r}=kcdot a^r$$where $k$ is a constant. As any pmf $f(n)$ must satisfy $sum f(n)=1$ we must have $$sum_{r=0}^{infty}Pr{X=r}=sum_{r=0}^{infty}kcdot a^r=ksum_{r=0}^{infty} a^r=1$$therefore $$k={1over sum_{r=0}^{infty} a^r}$$and we need to find $sum_{r=0}^{infty}a^r$ which equals to ${1over 1-a}$ according to geometric series sum formula. Therefore $k=1-a$ and the final answer becomes $$Pr{X=r}={a^rcdot (1-a)}$$so we have$$Pr{X=0}=1-a$$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 19 at 16:04

Mostafa AyazMostafa Ayaz

18.2k31040