Probability that $25$ calls are received in the first $5$ minutes.
$begingroup$
Calls are received at a company according to a Poisson process at the
rate of 5 calls per minute. Find the probability that $25$ calls are
received in the first $5$ minutes and six of those calls occur during
the first minute.
Denote the number of calls with $N_t$ at time $t$. We have that $N_tsimtext{Poi}(lambda t),$ where $lambda=5$. We are looking for
$$mathbb{P}(N_5=25 | N_1=6 )=frac{mathbb{P}(N_1=6,N_5-N_1=19)}{mathbb{P}(N_1=6)}=frac{mathbb{P}(N_1=6,tilde{N_4}=19)}{mathbb{P}(N_1=6)}=...$$
by stationary increments. Independent icrements also give that we can proceed with
$$...=frac{mathbb{P}(N_1=6)mathbb{P}(tilde{N_4}=19)}{mathbb{P}(N_1=6)}=mathbb{P}(tilde{N_4}=19)=frac{(5cdot 4)^{19}e^{-5cdot 4}}{19!}approx0.0888.$$
Which is incorrect. However I get the correct answer if I, with the same method using increments, calculate $mathbb{P}(N_5=25 , N_1=6 ).$
Question:
Why is it wrong to calculate $mathbb{P}(N_5=25 | N_1=6 )$? To me this seems intuitive: We want to find the probability that $25$ calls are received given that $6$ calls already have happened in the first minute.
probability probability-theory poisson-distribution poisson-process
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
$endgroup$
add a comment |
$begingroup$
Calls are received at a company according to a Poisson process at the
rate of 5 calls per minute. Find the probability that $25$ calls are
received in the first $5$ minutes and six of those calls occur during
the first minute.
Denote the number of calls with $N_t$ at time $t$. We have that $N_tsimtext{Poi}(lambda t),$ where $lambda=5$. We are looking for
$$mathbb{P}(N_5=25 | N_1=6 )=frac{mathbb{P}(N_1=6,N_5-N_1=19)}{mathbb{P}(N_1=6)}=frac{mathbb{P}(N_1=6,tilde{N_4}=19)}{mathbb{P}(N_1=6)}=...$$
by stationary increments. Independent icrements also give that we can proceed with
$$...=frac{mathbb{P}(N_1=6)mathbb{P}(tilde{N_4}=19)}{mathbb{P}(N_1=6)}=mathbb{P}(tilde{N_4}=19)=frac{(5cdot 4)^{19}e^{-5cdot 4}}{19!}approx0.0888.$$
Which is incorrect. However I get the correct answer if I, with the same method using increments, calculate $mathbb{P}(N_5=25 , N_1=6 ).$
Question:
Why is it wrong to calculate $mathbb{P}(N_5=25 | N_1=6 )$? To me this seems intuitive: We want to find the probability that $25$ calls are received given that $6$ calls already have happened in the first minute.
probability probability-theory poisson-distribution poisson-process
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
$endgroup$
$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30
add a comment |
$begingroup$
Calls are received at a company according to a Poisson process at the
rate of 5 calls per minute. Find the probability that $25$ calls are
received in the first $5$ minutes and six of those calls occur during
the first minute.
Denote the number of calls with $N_t$ at time $t$. We have that $N_tsimtext{Poi}(lambda t),$ where $lambda=5$. We are looking for
$$mathbb{P}(N_5=25 | N_1=6 )=frac{mathbb{P}(N_1=6,N_5-N_1=19)}{mathbb{P}(N_1=6)}=frac{mathbb{P}(N_1=6,tilde{N_4}=19)}{mathbb{P}(N_1=6)}=...$$
by stationary increments. Independent icrements also give that we can proceed with
$$...=frac{mathbb{P}(N_1=6)mathbb{P}(tilde{N_4}=19)}{mathbb{P}(N_1=6)}=mathbb{P}(tilde{N_4}=19)=frac{(5cdot 4)^{19}e^{-5cdot 4}}{19!}approx0.0888.$$
Which is incorrect. However I get the correct answer if I, with the same method using increments, calculate $mathbb{P}(N_5=25 , N_1=6 ).$
Question:
Why is it wrong to calculate $mathbb{P}(N_5=25 | N_1=6 )$? To me this seems intuitive: We want to find the probability that $25$ calls are received given that $6$ calls already have happened in the first minute.
probability probability-theory poisson-distribution poisson-process
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
$endgroup$
Calls are received at a company according to a Poisson process at the
rate of 5 calls per minute. Find the probability that $25$ calls are
received in the first $5$ minutes and six of those calls occur during
the first minute.
Denote the number of calls with $N_t$ at time $t$. We have that $N_tsimtext{Poi}(lambda t),$ where $lambda=5$. We are looking for
$$mathbb{P}(N_5=25 | N_1=6 )=frac{mathbb{P}(N_1=6,N_5-N_1=19)}{mathbb{P}(N_1=6)}=frac{mathbb{P}(N_1=6,tilde{N_4}=19)}{mathbb{P}(N_1=6)}=...$$
by stationary increments. Independent icrements also give that we can proceed with
$$...=frac{mathbb{P}(N_1=6)mathbb{P}(tilde{N_4}=19)}{mathbb{P}(N_1=6)}=mathbb{P}(tilde{N_4}=19)=frac{(5cdot 4)^{19}e^{-5cdot 4}}{19!}approx0.0888.$$
Which is incorrect. However I get the correct answer if I, with the same method using increments, calculate $mathbb{P}(N_5=25 , N_1=6 ).$
Question:
Why is it wrong to calculate $mathbb{P}(N_5=25 | N_1=6 )$? To me this seems intuitive: We want to find the probability that $25$ calls are received given that $6$ calls already have happened in the first minute.
probability probability-theory poisson-distribution poisson-process
probability probability-theory poisson-distribution poisson-process
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
asked Jan 15 at 16:44
ParsevalParseval
2,8581718
2,8581718
$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30
add a comment |
$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30
$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30
$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $Acap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
$endgroup$
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
add a comment |
$begingroup$
The key word is highlighted below:
Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.
Since they used the word and, you want $P(N_5=25 cap N_1=6)$, and not $P(N_5=25 mid N_1=6)$.
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $Acap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
$endgroup$
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
add a comment |
$begingroup$
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $Acap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
$endgroup$
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
add a comment |
$begingroup$
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $Acap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
$endgroup$
We weren't given a conditional probability question. It isn't previously known or given that $6$ calls happened in the first minute. Had it said something along those lines then your conditional probability approach would have been correct.
Instead we have two events $A$ and $B$, and we want the probability of $Acap B$, and because events over disjoint time intervals are independent in the Poisson process, we can find $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
edited Jan 15 at 18:27
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
answered Jan 15 at 17:05
WaveXWaveX
2,5822722
2,5822722
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
add a comment |
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
$begingroup$
I thought the exponential distribution was memorylessa not the Poisson? That formula holds if $A$ and $B$ are disjoint and/or independent. I don't see how memorylessness implies it. Could you elaborate on this?
$endgroup$
– Parseval
Jan 15 at 18:13
1
1
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Both distributions have this property (in fact the exponential distribution and Poisson distribution are closely related). If we let event $A$ be that we get $6$ calls in the first minute and let $B$ be the event that we get $19$ calls in minutes 2 through 5 (as Math1000 points out) then event $B$ will have no memory of event $A$, thus making them independent events
$endgroup$
– WaveX
Jan 15 at 18:17
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
$begingroup$
Perhaps calling the Poisson distribution "memoryless" may have been somewhat of a stretch; hopefully my comment and edited answer makes more sense
$endgroup$
– WaveX
Jan 15 at 18:28
add a comment |
$begingroup$
The key word is highlighted below:
Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.
Since they used the word and, you want $P(N_5=25 cap N_1=6)$, and not $P(N_5=25 mid N_1=6)$.
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
$endgroup$
add a comment |
$begingroup$
The key word is highlighted below:
Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.
Since they used the word and, you want $P(N_5=25 cap N_1=6)$, and not $P(N_5=25 mid N_1=6)$.
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
$endgroup$
add a comment |
$begingroup$
The key word is highlighted below:
Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.
Since they used the word and, you want $P(N_5=25 cap N_1=6)$, and not $P(N_5=25 mid N_1=6)$.
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
$endgroup$
The key word is highlighted below:
Find the probability that 25 calls are received in the first 5 minutes and six of those calls occur during the first minute.
Since they used the word and, you want $P(N_5=25 cap N_1=6)$, and not $P(N_5=25 mid N_1=6)$.
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
answered Jan 15 at 18:09
Mike EarnestMike Earnest
22.3k12051
22.3k12051
add a comment |
add a comment |
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$begingroup$
$25$ calls occuring during the first five minutes, with $6$ of those occurring during the first minute, means how many calls occurred during minutes two through five?
$endgroup$
– Math1000
Jan 15 at 17:30