Sum of Poisson Distribution, why is my solution incorrect












0














From SOA #212:




The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.




I calculated the probabilities as follows



$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$



$$=P(text{employee is sick more than two days in a three-month period})$$



This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$



$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$



$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$



$$=.8005$$



This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.






probability statistics probability-distributions







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  • Read my answer to see what went wrong.
    – callculus
    Nov 20 at 16:23
















0














From SOA #212:




The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.




I calculated the probabilities as follows



$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$



$$=P(text{employee is sick more than two days in a three-month period})$$



This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$



$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$



$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$



$$=.8005$$



This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.






probability statistics probability-distributions







share|cite|improve this question






















  • Read my answer to see what went wrong.
    – callculus
    Nov 20 at 16:23














0












0








0







From SOA #212:




The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.




I calculated the probabilities as follows



$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$



$$=P(text{employee is sick more than two days in a three-month period})$$



This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$



$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$



$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$



$$=.8005$$



This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.






probability statistics probability-distributions







share|cite|improve this question













From SOA #212:




The number of days an employee is sick each month is modeled by a Poisson distribution
with mean $1$. The numbers of sick days in different months are mutually independent.
Calculate the probability that an employee is sick more than two days in a three-month
period.




I calculated the probabilities as follows



$${3choose1}P(X=0)P(X=0)P(Xge2) + $$
$${3choose2}P(X=0)P(Xge1)P(Xge1) + $$
$${3choose3}P(Xge1)P(Xge1)P(Xge1) $$



$$=P(text{employee is sick more than two days in a three-month period})$$



This should come out to
$${3choose1}left({1^0e^{-1}over0!}right)left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}-{1^1e^{-1}over1!}right) + $$



$${3choose2}left({1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) + $$



$${3choose3}left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right)left(1- {1^0e^{-1}over0!}right) $$



$$=.8005$$



This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.






probability statistics probability-distributions




probability statistics probability-distributions






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asked Nov 20 at 15:26









agblt

16214




16214












  • Read my answer to see what went wrong.
    – callculus
    Nov 20 at 16:23


















  • Read my answer to see what went wrong.
    – callculus
    Nov 20 at 16:23
















Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23




Read my answer to see what went wrong.
– callculus
Nov 20 at 16:23










3 Answers
3






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oldest

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1














You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).






share|cite|improve this answer





























    1














    Your calculation does not match mine. It is structurally different. We want



    $1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$



    For $X=2$ we have several combinations



    $110 quad 200$



    $011quad 020$



    $101quad 002$



    $X=1:$ $100,$ $ 010,$$ 001$



    $X=0$: $000$



    Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$






    share|cite|improve this answer































      1














      We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1














        You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).






        share|cite|improve this answer


























          1














          You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).






          share|cite|improve this answer
























            1












            1








            1






            You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).






            share|cite|improve this answer












            You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 at 15:35









            Michael Lugo

            17.9k33576




            17.9k33576























                1














                Your calculation does not match mine. It is structurally different. We want



                $1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$



                For $X=2$ we have several combinations



                $110 quad 200$



                $011quad 020$



                $101quad 002$



                $X=1:$ $100,$ $ 010,$$ 001$



                $X=0$: $000$



                Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$






                share|cite|improve this answer




























                  1














                  Your calculation does not match mine. It is structurally different. We want



                  $1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$



                  For $X=2$ we have several combinations



                  $110 quad 200$



                  $011quad 020$



                  $101quad 002$



                  $X=1:$ $100,$ $ 010,$$ 001$



                  $X=0$: $000$



                  Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$






                  share|cite|improve this answer


























                    1












                    1








                    1






                    Your calculation does not match mine. It is structurally different. We want



                    $1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$



                    For $X=2$ we have several combinations



                    $110 quad 200$



                    $011quad 020$



                    $101quad 002$



                    $X=1:$ $100,$ $ 010,$$ 001$



                    $X=0$: $000$



                    Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$






                    share|cite|improve this answer














                    Your calculation does not match mine. It is structurally different. We want



                    $1-P(Xleq 2)=1-P(X=2)-P(X=1)-P(X=0)$



                    For $X=2$ we have several combinations



                    $110 quad 200$



                    $011quad 020$



                    $101quad 002$



                    $X=1:$ $100,$ $ 010,$$ 001$



                    $X=0$: $000$



                    Therefore the calculation is $1-7cdot (1/e)^3-3cdot (1/(2cdot e))cdot (1/e)^2=0.57681approx 0.577$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 20 at 16:13

























                    answered Nov 20 at 15:59









                    callculus

                    17.8k31427




                    17.8k31427























                        1














                        We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.






                        share|cite|improve this answer


























                          1














                          We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.






                            share|cite|improve this answer












                            We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)approx 0.577$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 20 at 16:47









                            J.G.

                            22.4k22035




                            22.4k22035






























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