Mixing
Poisson Distribution Consecutive Event Disjointed by Less Than x Time
$begingroup$
In a network of packages there are two transmission lines.One has λ=10 packages/s and Number Two has λ=20p/s.In an interval of 5 seconds 100 packages arrived.
So the first question is whats the probability of 40p coming from line 1 and 60 from line 2.
I figured it would be
Pt=(P(5)12=40)+(P(5)22=60)
And the second problem is the calculate the probability of two consecutive packages being separated by less than 10 ms.
I Have no idea how to solve this once since the lines overlap right?Should I calculate a poisson distribution for two consecutive packages being one from line 1 and another from line 2, and another one with 2 from line 2?
Thank you for your time
probability probability-distributions poisson-distribution
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
$endgroup$
add a comment |
$begingroup$
In a network of packages there are two transmission lines.One has λ=10 packages/s and Number Two has λ=20p/s.In an interval of 5 seconds 100 packages arrived.
So the first question is whats the probability of 40p coming from line 1 and 60 from line 2.
I figured it would be
Pt=(P(5)12=40)+(P(5)22=60)
And the second problem is the calculate the probability of two consecutive packages being separated by less than 10 ms.
I Have no idea how to solve this once since the lines overlap right?Should I calculate a poisson distribution for two consecutive packages being one from line 1 and another from line 2, and another one with 2 from line 2?
Thank you for your time
probability probability-distributions poisson-distribution
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
$endgroup$
add a comment |
$begingroup$
In a network of packages there are two transmission lines.One has λ=10 packages/s and Number Two has λ=20p/s.In an interval of 5 seconds 100 packages arrived.
So the first question is whats the probability of 40p coming from line 1 and 60 from line 2.
I figured it would be
Pt=(P(5)12=40)+(P(5)22=60)
And the second problem is the calculate the probability of two consecutive packages being separated by less than 10 ms.
I Have no idea how to solve this once since the lines overlap right?Should I calculate a poisson distribution for two consecutive packages being one from line 1 and another from line 2, and another one with 2 from line 2?
Thank you for your time
probability probability-distributions poisson-distribution
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
$endgroup$
In a network of packages there are two transmission lines.One has λ=10 packages/s and Number Two has λ=20p/s.In an interval of 5 seconds 100 packages arrived.
So the first question is whats the probability of 40p coming from line 1 and 60 from line 2.
I figured it would be
Pt=(P(5)12=40)+(P(5)22=60)
And the second problem is the calculate the probability of two consecutive packages being separated by less than 10 ms.
I Have no idea how to solve this once since the lines overlap right?Should I calculate a poisson distribution for two consecutive packages being one from line 1 and another from line 2, and another one with 2 from line 2?
Thank you for your time
probability probability-distributions poisson-distribution
probability probability-distributions poisson-distribution
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
asked Feb 1 at 23:26
IHaveNoIdeaWhatImDoingIHaveNoIdeaWhatImDoing
133
133
add a comment |
add a comment |
1 Answer
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$begingroup$
Each packet has a $10/(10+20)=1/3$ chance of being from line $1$. So the probability that of the first $100$ packets, $40$ of them are from line $1$ is given by a binomial distribution:
$$
binom{100}{40}(1/3)^{40}(1-1/3)^{100-40} approx 0.03075091.
$$
For the second question, recall that the superposition of two Poisson processes is again a Poisson process with its rate the sum of the rates of the constituent processes. So we may consider a single Poisson process with rate $lambda = 30$ (per second). The time between consecutive arrivals is exponentially distributed with the same rate $lambda$, so the probability that two arrivals are separated by less than $1/100$ seconds is given by
$$
1-exp(30cdot 1/100) approx 0.1845155.
$$
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
$endgroup$
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
Each packet has a $10/(10+20)=1/3$ chance of being from line $1$. So the probability that of the first $100$ packets, $40$ of them are from line $1$ is given by a binomial distribution:
$$
binom{100}{40}(1/3)^{40}(1-1/3)^{100-40} approx 0.03075091.
$$
For the second question, recall that the superposition of two Poisson processes is again a Poisson process with its rate the sum of the rates of the constituent processes. So we may consider a single Poisson process with rate $lambda = 30$ (per second). The time between consecutive arrivals is exponentially distributed with the same rate $lambda$, so the probability that two arrivals are separated by less than $1/100$ seconds is given by
$$
1-exp(30cdot 1/100) approx 0.1845155.
$$
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
$endgroup$
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
add a comment |
$begingroup$
Each packet has a $10/(10+20)=1/3$ chance of being from line $1$. So the probability that of the first $100$ packets, $40$ of them are from line $1$ is given by a binomial distribution:
$$
binom{100}{40}(1/3)^{40}(1-1/3)^{100-40} approx 0.03075091.
$$
For the second question, recall that the superposition of two Poisson processes is again a Poisson process with its rate the sum of the rates of the constituent processes. So we may consider a single Poisson process with rate $lambda = 30$ (per second). The time between consecutive arrivals is exponentially distributed with the same rate $lambda$, so the probability that two arrivals are separated by less than $1/100$ seconds is given by
$$
1-exp(30cdot 1/100) approx 0.1845155.
$$
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
$endgroup$
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
add a comment |
$begingroup$
Each packet has a $10/(10+20)=1/3$ chance of being from line $1$. So the probability that of the first $100$ packets, $40$ of them are from line $1$ is given by a binomial distribution:
$$
binom{100}{40}(1/3)^{40}(1-1/3)^{100-40} approx 0.03075091.
$$
For the second question, recall that the superposition of two Poisson processes is again a Poisson process with its rate the sum of the rates of the constituent processes. So we may consider a single Poisson process with rate $lambda = 30$ (per second). The time between consecutive arrivals is exponentially distributed with the same rate $lambda$, so the probability that two arrivals are separated by less than $1/100$ seconds is given by
$$
1-exp(30cdot 1/100) approx 0.1845155.
$$
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
$endgroup$
Each packet has a $10/(10+20)=1/3$ chance of being from line $1$. So the probability that of the first $100$ packets, $40$ of them are from line $1$ is given by a binomial distribution:
$$
binom{100}{40}(1/3)^{40}(1-1/3)^{100-40} approx 0.03075091.
$$
For the second question, recall that the superposition of two Poisson processes is again a Poisson process with its rate the sum of the rates of the constituent processes. So we may consider a single Poisson process with rate $lambda = 30$ (per second). The time between consecutive arrivals is exponentially distributed with the same rate $lambda$, so the probability that two arrivals are separated by less than $1/100$ seconds is given by
$$
1-exp(30cdot 1/100) approx 0.1845155.
$$
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
answered Feb 2 at 1:11
Math1000Math1000
19.4k31746
19.4k31746
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
add a comment |
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
Would it be possible to solve question 1 using a poisson process?I say this because this exercice was in a Poisson Chapter of my book
$endgroup$
– IHaveNoIdeaWhatImDoing
Feb 2 at 23:58
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
$begingroup$
The time between arrivals in a Poisson process is exponentially distributed. This is a fundamental property of Poisson processes...
$endgroup$
– Math1000
Feb 3 at 21:39
add a comment |
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