Urn problem involving removement of half of the balls












1












$begingroup$


An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?



It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.






probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well...each ball has an equal chance of being the one selected by this process, so...
    $endgroup$
    – lulu
    Jan 10 at 20:02










  • $begingroup$
    To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
    $endgroup$
    – amd
    Jan 10 at 20:04










  • $begingroup$
    Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
    $endgroup$
    – user1337
    Jan 10 at 20:10










  • $begingroup$
    Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
    $endgroup$
    – rtybase
    Jan 10 at 20:29












  • $begingroup$
    Thanks very much. Could you put it as an answer?
    $endgroup$
    – user1337
    Jan 10 at 20:42
















1












$begingroup$


An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?



It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.






probability probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well...each ball has an equal chance of being the one selected by this process, so...
    $endgroup$
    – lulu
    Jan 10 at 20:02










  • $begingroup$
    To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
    $endgroup$
    – amd
    Jan 10 at 20:04










  • $begingroup$
    Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
    $endgroup$
    – user1337
    Jan 10 at 20:10










  • $begingroup$
    Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
    $endgroup$
    – rtybase
    Jan 10 at 20:29












  • $begingroup$
    Thanks very much. Could you put it as an answer?
    $endgroup$
    – user1337
    Jan 10 at 20:42














1












1








1





$begingroup$


An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?



It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.






probability probability-theory







share|cite|improve this question











$endgroup$




An urn contains $n$ heliotrope and $n$ tangerine balls. Half the balls are removed and placed in a box. One of those remaining in the urn is
removed. What is the probability that it is tangerine?



It is not a homework problem. I am just curious about the result. Such problem comes from the book "Elementary probability" from David Stirzaker. Any help or hint is greatly appreciated.






probability probability-theory




probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 20:12







user1337

















asked Jan 10 at 20:01









user1337user1337

48210




48210








  • 1




    $begingroup$
    Well...each ball has an equal chance of being the one selected by this process, so...
    $endgroup$
    – lulu
    Jan 10 at 20:02










  • $begingroup$
    To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
    $endgroup$
    – amd
    Jan 10 at 20:04










  • $begingroup$
    Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
    $endgroup$
    – user1337
    Jan 10 at 20:10










  • $begingroup$
    Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
    $endgroup$
    – rtybase
    Jan 10 at 20:29












  • $begingroup$
    Thanks very much. Could you put it as an answer?
    $endgroup$
    – user1337
    Jan 10 at 20:42














  • 1




    $begingroup$
    Well...each ball has an equal chance of being the one selected by this process, so...
    $endgroup$
    – lulu
    Jan 10 at 20:02










  • $begingroup$
    To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
    $endgroup$
    – amd
    Jan 10 at 20:04










  • $begingroup$
    Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
    $endgroup$
    – user1337
    Jan 10 at 20:10










  • $begingroup$
    Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
    $endgroup$
    – rtybase
    Jan 10 at 20:29












  • $begingroup$
    Thanks very much. Could you put it as an answer?
    $endgroup$
    – user1337
    Jan 10 at 20:42








1




1




$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02




$begingroup$
Well...each ball has an equal chance of being the one selected by this process, so...
$endgroup$
– lulu
Jan 10 at 20:02












$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04




$begingroup$
To put @lulu's comment differently, suppose instead you take the single ball out first. Does this change the result in any way?
$endgroup$
– amd
Jan 10 at 20:04












$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10




$begingroup$
Sorry, I am afraid I have misunderstood the problem description, as it seems really easy to solve it. Could someone provide me a full answer? I'm not an English native speaker.
$endgroup$
– user1337
Jan 10 at 20:10












$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29






$begingroup$
Note the events like $B={text{tangerine ball is extracted with the 2nd extraction}}$ and $A_k={k - text{tangerine balls are extracted with the 1st extraction}}$ And apply total probability $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$
$endgroup$
– rtybase
Jan 10 at 20:29














$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42




$begingroup$
Thanks very much. Could you put it as an answer?
$endgroup$
– user1337
Jan 10 at 20:42










2 Answers
2






active

oldest

votes


















1












$begingroup$

Note the events like





  • $B={text{tangerine ball is extracted with the 2nd extraction}}$

  • $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$


And apply total probability
$$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$



$P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.



And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
Finally
$$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
applying Vandermonde's identity
$$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
    $endgroup$
    – user1337
    Jan 10 at 21:00






  • 1




    $begingroup$
    @user1337 updated ...
    $endgroup$
    – rtybase
    Jan 10 at 21:28





















1












$begingroup$

The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.



Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.






share|cite|improve this answer











$endgroup$














    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%2f3069110%2furn-problem-involving-removement-of-half-of-the-balls%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Note the events like





    • $B={text{tangerine ball is extracted with the 2nd extraction}}$

    • $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$


    And apply total probability
    $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$



    $P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.



    And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
    Finally
    $$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
    applying Vandermonde's identity
    $$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
      $endgroup$
      – user1337
      Jan 10 at 21:00






    • 1




      $begingroup$
      @user1337 updated ...
      $endgroup$
      – rtybase
      Jan 10 at 21:28


















    1












    $begingroup$

    Note the events like





    • $B={text{tangerine ball is extracted with the 2nd extraction}}$

    • $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$


    And apply total probability
    $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$



    $P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.



    And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
    Finally
    $$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
    applying Vandermonde's identity
    $$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
      $endgroup$
      – user1337
      Jan 10 at 21:00






    • 1




      $begingroup$
      @user1337 updated ...
      $endgroup$
      – rtybase
      Jan 10 at 21:28
















    1












    1








    1





    $begingroup$

    Note the events like





    • $B={text{tangerine ball is extracted with the 2nd extraction}}$

    • $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$


    And apply total probability
    $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$



    $P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.



    And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
    Finally
    $$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
    applying Vandermonde's identity
    $$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$






    share|cite|improve this answer











    $endgroup$



    Note the events like





    • $B={text{tangerine ball is extracted with the 2nd extraction}}$

    • $A_k={k - text{tangerine balls exactly are extracted with the 1st extraction}}$


    And apply total probability
    $$P(B)=sumlimits_{k=0}^n P(B mid A_k)cdot P(A_k)$$



    $P(B mid A_k)$ are easy to calculate, $n$ balls were extracted 1st time, $k$ were tangerine, so $n-k$ tangerine balls were left, thus $P(B mid A_k)=frac{n-k}{n}$.



    And $$P(A_k)=frac{binom{n}{n-k}cdot binom{n}{k}}{binom{2n}{n}}$$
    Finally
    $$P(B)=frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nfrac{n-k}{n}binom{n}{n-k}binom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{n-k}{n}cdotfrac{n!}{(n-k)!k!}}cdotbinom{n}{k}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{frac{(n-1)!}{(n-k-1)!k!}}cdotbinom{n}{k}right)=
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^ncolor{blue}{binom{n-1}{k}}cdotcolor{red}{binom{n}{k}}right)=\
    frac{1}{binom{2n}{n}}left(sumlimits_{k=0}^nbinom{n-1}{k}cdotcolor{red}{binom{n}{n-k}}right)=...$$
    applying Vandermonde's identity
    $$...=frac{binom{2n-1}{n}}{binom{2n}{n}}=frac{1}{2}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 10 at 21:36

























    answered Jan 10 at 20:46









    rtybasertybase

    11.5k31534




    11.5k31534












    • $begingroup$
      The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
      $endgroup$
      – user1337
      Jan 10 at 21:00






    • 1




      $begingroup$
      @user1337 updated ...
      $endgroup$
      – rtybase
      Jan 10 at 21:28




















    • $begingroup$
      The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
      $endgroup$
      – user1337
      Jan 10 at 21:00






    • 1




      $begingroup$
      @user1337 updated ...
      $endgroup$
      – rtybase
      Jan 10 at 21:28


















    $begingroup$
    The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
    $endgroup$
    – user1337
    Jan 10 at 21:00




    $begingroup$
    The book gives $1/2$ as an answer. Is there a shortcut in order to obtain it?
    $endgroup$
    – user1337
    Jan 10 at 21:00




    1




    1




    $begingroup$
    @user1337 updated ...
    $endgroup$
    – rtybase
    Jan 10 at 21:28






    $begingroup$
    @user1337 updated ...
    $endgroup$
    – rtybase
    Jan 10 at 21:28













    1












    $begingroup$

    The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.



    Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.



      Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.



        Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.






        share|cite|improve this answer











        $endgroup$



        The answer is $frac 12$. Indeed, each ball is equally likely to be chosen by this process and half the balls have the desired color. Had there been $a$ of the desired color and $2n-a$ of the undesired color, the answer would have been $frac a{2n}$.



        Note: it's clear from symmetry that each ball is likely to be selected this way, but it's not difficult to explicitly compute the probability that a given ball is selected. Indeed, for a given ball to be selected it must first not be deleted (probability $frac 12$) and second it must be chosen from the $n$ survivors (probability $frac 1n$). Thus the probability that a given ball is selected is $frac 1{2n}$ as desired.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 23:53

























        answered Jan 10 at 21:18









        lulululu

        43.6k25081




        43.6k25081






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069110%2furn-problem-involving-removement-of-half-of-the-balls%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more