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







share|cite|improve this question






















  • 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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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






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














    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





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006456%2fsum-of-poisson-distribution-why-is-my-solution-incorrect%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006456%2fsum-of-poisson-distribution-why-is-my-solution-incorrect%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                How to change which sound is reproduced for terminal bell?

                                Image
                                3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                                Read more

                                Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                Image
                                0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                                Read more

                                Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                                Image
                                up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                                Read more