Probability that $25$ calls are received in the first $5$ minutes.












1












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $25$ 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
















1












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $25$ 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














1












1








1


1



$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










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


















  • $begingroup$
    $25$ 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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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 $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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 $P(N_5=25 mid N_1=6)$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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 $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    2












    $begingroup$

    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 $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    2












    2








    2





    $begingroup$

    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 $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$






    share|cite|improve this answer











    $endgroup$



    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 $mathbb{P}(Acap B) = mathbb{P}(A) mathbb{P}(B)$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 15 at 18:27

























    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


















    • $begingroup$
      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











    1












    $begingroup$

    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 $P(N_5=25 mid N_1=6)$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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 $P(N_5=25 mid N_1=6)$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 $P(N_5=25 mid N_1=6)$.






        share|cite|improve this answer









        $endgroup$



        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 $P(N_5=25 mid N_1=6)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 18:09









        Mike EarnestMike Earnest

        22.3k12051




        22.3k12051






























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