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Is there a big chance that Ithaca helps Dryden?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Conditional probability problem and Alias Methodhelp with poissonPoisson process/probabilityPoisson(Exponential) Distribution questionPhone calls are received at Janice’s house according to a Poisson Process with parameter $lambda= 2$ per hour.Call Waiting Time process; questions on probabilities and expectations.The number of calls to receptionist per one hour has Poisson distribution with mean 1.Poisson Distribution question solvingtime between poisson eventswriting a poisson distribution
$begingroup$
Suppose calls to Dryden Fire Department arrive according to a Poisson Process with rate $0.5$ per hour. Suppose the time $T$ needed to respond to a call, return to the station and be ready for the next call is uniformly distributed between $0.5 $ hour and $1$ hour. If a new call comes before Dryden is ready to respond, then Ithaca fire station is requested to help. Suppose the Dryden fire station is ready to respond now. Then, find the probability distribution of the number of calls they will handle before they have to request Ithaca to help.
Okay, so i gathered that letting $N(t)$ be the number of calls by time $t$, $N(t)in Poisson(0.5t)$. Hence the interarrival time is $Exponential(0.5)$.
If $S$ is the interarrival time, then we "should be" interested in $P(S<T)$, which is the probability that the time between two consecutive calls is less than the time taken to respond to the previous call.
Now denoting the number of calls taken by Dryden before having to ask Ithaca for help, by $N$, we want to find $P(N=n)$ where $ngeq1$.
It seems that it has a Geometric distribution with $p=P(S<T)$ i.e. first success occurs when $S<T$.
However, I am getting $P(S<T)=(1-2e^{-1/4})^2$ which is not at all the answer. Answer is $P(N=n)=(1-p)^{n-1}p$ for $ngeq1$ where $p=e^{-1}-e^{-2}$. Also, is my reasoning right?
probability probability-distributions random-variables poisson-distribution poisson-process
edited Nov 20 '15 at 22:20
BCLC
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
$endgroup$
add a comment |
$begingroup$
Suppose calls to Dryden Fire Department arrive according to a Poisson Process with rate $0.5$ per hour. Suppose the time $T$ needed to respond to a call, return to the station and be ready for the next call is uniformly distributed between $0.5 $ hour and $1$ hour. If a new call comes before Dryden is ready to respond, then Ithaca fire station is requested to help. Suppose the Dryden fire station is ready to respond now. Then, find the probability distribution of the number of calls they will handle before they have to request Ithaca to help.
Okay, so i gathered that letting $N(t)$ be the number of calls by time $t$, $N(t)in Poisson(0.5t)$. Hence the interarrival time is $Exponential(0.5)$.
If $S$ is the interarrival time, then we "should be" interested in $P(S<T)$, which is the probability that the time between two consecutive calls is less than the time taken to respond to the previous call.
Now denoting the number of calls taken by Dryden before having to ask Ithaca for help, by $N$, we want to find $P(N=n)$ where $ngeq1$.
It seems that it has a Geometric distribution with $p=P(S<T)$ i.e. first success occurs when $S<T$.
However, I am getting $P(S<T)=(1-2e^{-1/4})^2$ which is not at all the answer. Answer is $P(N=n)=(1-p)^{n-1}p$ for $ngeq1$ where $p=e^{-1}-e^{-2}$. Also, is my reasoning right?
probability probability-distributions random-variables poisson-distribution poisson-process
edited Nov 20 '15 at 22:20
BCLC
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
$endgroup$
1
$begingroup$
I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50
add a comment |
$begingroup$
Suppose calls to Dryden Fire Department arrive according to a Poisson Process with rate $0.5$ per hour. Suppose the time $T$ needed to respond to a call, return to the station and be ready for the next call is uniformly distributed between $0.5 $ hour and $1$ hour. If a new call comes before Dryden is ready to respond, then Ithaca fire station is requested to help. Suppose the Dryden fire station is ready to respond now. Then, find the probability distribution of the number of calls they will handle before they have to request Ithaca to help.
Okay, so i gathered that letting $N(t)$ be the number of calls by time $t$, $N(t)in Poisson(0.5t)$. Hence the interarrival time is $Exponential(0.5)$.
If $S$ is the interarrival time, then we "should be" interested in $P(S<T)$, which is the probability that the time between two consecutive calls is less than the time taken to respond to the previous call.
Now denoting the number of calls taken by Dryden before having to ask Ithaca for help, by $N$, we want to find $P(N=n)$ where $ngeq1$.
It seems that it has a Geometric distribution with $p=P(S<T)$ i.e. first success occurs when $S<T$.
However, I am getting $P(S<T)=(1-2e^{-1/4})^2$ which is not at all the answer. Answer is $P(N=n)=(1-p)^{n-1}p$ for $ngeq1$ where $p=e^{-1}-e^{-2}$. Also, is my reasoning right?
probability probability-distributions random-variables poisson-distribution poisson-process
edited Nov 20 '15 at 22:20
BCLC
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
$endgroup$
Suppose calls to Dryden Fire Department arrive according to a Poisson Process with rate $0.5$ per hour. Suppose the time $T$ needed to respond to a call, return to the station and be ready for the next call is uniformly distributed between $0.5 $ hour and $1$ hour. If a new call comes before Dryden is ready to respond, then Ithaca fire station is requested to help. Suppose the Dryden fire station is ready to respond now. Then, find the probability distribution of the number of calls they will handle before they have to request Ithaca to help.
Okay, so i gathered that letting $N(t)$ be the number of calls by time $t$, $N(t)in Poisson(0.5t)$. Hence the interarrival time is $Exponential(0.5)$.
If $S$ is the interarrival time, then we "should be" interested in $P(S<T)$, which is the probability that the time between two consecutive calls is less than the time taken to respond to the previous call.
Now denoting the number of calls taken by Dryden before having to ask Ithaca for help, by $N$, we want to find $P(N=n)$ where $ngeq1$.
It seems that it has a Geometric distribution with $p=P(S<T)$ i.e. first success occurs when $S<T$.
However, I am getting $P(S<T)=(1-2e^{-1/4})^2$ which is not at all the answer. Answer is $P(N=n)=(1-p)^{n-1}p$ for $ngeq1$ where $p=e^{-1}-e^{-2}$. Also, is my reasoning right?
probability probability-distributions random-variables poisson-distribution poisson-process
probability probability-distributions random-variables poisson-distribution poisson-process
edited Nov 20 '15 at 22:20
BCLC
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
edited Nov 20 '15 at 22:20
BCLC
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
edited Nov 20 '15 at 22:20
BCLC
1
edited Nov 20 '15 at 22:20
BCLC
1
edited Nov 20 '15 at 22:20
BCLC
1
1
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
asked Nov 12 '15 at 15:15
Landon CarterLandon Carter
7,51811645
7,51811645
1
$begingroup$
I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50
add a comment |
1
$begingroup$
I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50
1
1
$begingroup$
I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50
$begingroup$
I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $Ssim exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
$endgroup$
add a comment |
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$begingroup$
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $Ssim exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
$endgroup$
add a comment |
$begingroup$
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $Ssim exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
$endgroup$
add a comment |
$begingroup$
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $Ssim exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
$endgroup$
Indeed, as @MickA pointed out, you made two mistakes here.
In this problem, you have the call at Dryden less than the prepared time for the first $(n-1)$ and greater than the prepared time for the $n$th call. Therefore, the success is defined as $S>T$, which is unusual.
Also, $Ssim exp(2)$ because $1/2$ is the rate of the poisson distribution of arrivals. Therefore, the mean of this poisson distribution is $2$. It follows the interarrival time should have an exponential distribution with rate $2$.
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
answered Oct 15 '18 at 22:07
James WangJames Wang
50728
50728
add a comment |
add a comment |
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I agree with your answer (and reasoning). The supposedly correct answer seems to (a) suppose $Ssim Exp(2)$ instead of $Exp(1/2)$ and (b) swap $p$ and $1-p$. I can't see why they do either of those things.
$endgroup$
– Mick A
Nov 13 '15 at 4:50