Tossing a triple of coins












1












$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14


















1












$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14
















1












1








1





$begingroup$


There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability







share|cite|improve this question











$endgroup$




There is a red coin for which P(Heads) = 0.4, a green coin for which P(Heads) = 0.5 and a yellow coin for which P(Heads) = 0.6.



Let N be the number of heads in 300 tosses. Before each toss, I choose 1 of the 3 coins at random (each coin is equally likely to be selected).



What is the mean and variance of N?



I model N as follows: $$N = T_1 + T_2 + cdots + T_{300}$$



$$mathbb{E}[T_i] = frac{0.4 + 0.5 + 0.6}{3} = 0.5$$



$$begin{eqnarray}
Var(T_i) &=& frac{1}{3}(0.4-mathbb{E}[T_i])^2 + frac{1}{3}(0.5-0.5)^2 + frac{1}{3}(0.6-0.5)^2
\ &=& frac{(0.4-0.5)^2 + (0.5-0.5)^2 + (0.6-0.5)^2}{3}
\ &=& frac{0.02}{3}
\ &=& frac{1}{150}
end{eqnarray}$$



By linearity of expectations and making use of the fact that all the Ts are iid,
$$begin{eqnarray}
mathbb{E}[N] &=& mathbb{E}[T_1] + mathbb{E}[T_2] + cdots + mathbb{E}[T_{300}]
\ &=& 300 cdot mathbb{E}[T_1]
\ &=& 150
end{eqnarray}$$



Similarly, since the Ts are independent,
$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& Var(T_1) + Var(T_2) + cdots + Var(T_{300})
\ &=& 300 cdot Var(T_1)
\ &=& 300 cdot frac{1}{150}
\ &=& 2
end{eqnarray}$$



I know that my calculation of the mean is correct. The variance is incorrect, but I haven't been able to figure out why.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 12:33







bard

















asked Nov 27 '18 at 10:40









bardbard

1886




1886








  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14
















  • 1




    $begingroup$
    Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
    $endgroup$
    – nicola
    Nov 27 '18 at 11:16








  • 1




    $begingroup$
    The variance Var(T1) above is the variance of the expectation, not the total variance.
    $endgroup$
    – KRKirov
    Nov 29 '18 at 19:14










1




1




$begingroup$
Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
$endgroup$
– nicola
Nov 27 '18 at 11:16






$begingroup$
Not getting how you are determining the variance. You should have something like $P(1)(1-0.5)^2 + P(0)(0-0.5)^2$, where $P(1)$ is the probability of having Heads. BTW, this is not any different from having just one single, fair, coin.
$endgroup$
– nicola
Nov 27 '18 at 11:16






1




1




$begingroup$
The variance Var(T1) above is the variance of the expectation, not the total variance.
$endgroup$
– KRKirov
Nov 29 '18 at 19:14






$begingroup$
The variance Var(T1) above is the variance of the expectation, not the total variance.
$endgroup$
– KRKirov
Nov 29 '18 at 19:14












1 Answer
1






active

oldest

votes


















0












$begingroup$

The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



By the law of total variance,
$$begin{eqnarray}
Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
\ &=& mathbb{E}[P(1-P)] + Var(P)
\ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
\ &=& 0.5 - 0.5^2
\ &=& 0.25
end{eqnarray}$$



$$begin{eqnarray}
Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
\ &=& 300 cdot Var(T_i)
\ &=& 300 cdot 0.25
\ &=& 75
end{eqnarray}$$



Still don't understand why my first approach is wrong though.






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%2f3015626%2ftossing-a-triple-of-coins%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



    By the law of total variance,
    $$begin{eqnarray}
    Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
    \ &=& mathbb{E}[P(1-P)] + Var(P)
    \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
    \ &=& 0.5 - 0.5^2
    \ &=& 0.25
    end{eqnarray}$$



    $$begin{eqnarray}
    Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
    \ &=& 300 cdot Var(T_i)
    \ &=& 300 cdot 0.25
    \ &=& 75
    end{eqnarray}$$



    Still don't understand why my first approach is wrong though.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



      By the law of total variance,
      $$begin{eqnarray}
      Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
      \ &=& mathbb{E}[P(1-P)] + Var(P)
      \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
      \ &=& 0.5 - 0.5^2
      \ &=& 0.25
      end{eqnarray}$$



      $$begin{eqnarray}
      Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
      \ &=& 300 cdot Var(T_i)
      \ &=& 300 cdot 0.25
      \ &=& 75
      end{eqnarray}$$



      Still don't understand why my first approach is wrong though.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



        By the law of total variance,
        $$begin{eqnarray}
        Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
        \ &=& mathbb{E}[P(1-P)] + Var(P)
        \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
        \ &=& 0.5 - 0.5^2
        \ &=& 0.25
        end{eqnarray}$$



        $$begin{eqnarray}
        Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
        \ &=& 300 cdot Var(T_i)
        \ &=& 300 cdot 0.25
        \ &=& 75
        end{eqnarray}$$



        Still don't understand why my first approach is wrong though.






        share|cite|improve this answer









        $endgroup$



        The problem appears to be in my calculation of the variance of T. I was able to get the right answer using a different approach.



        By the law of total variance,
        $$begin{eqnarray}
        Var(T_i) &=& mathbb{E}[Var(T_i|coin)] + Var(mathbb{E}[T_i|coin])
        \ &=& mathbb{E}[P(1-P)] + Var(P)
        \ &=& mathbb{E}[P] - mathbb{E}[P^2] + mathbb{E}[P^2] - (mathbb{E}[P])^2
        \ &=& 0.5 - 0.5^2
        \ &=& 0.25
        end{eqnarray}$$



        $$begin{eqnarray}
        Var(N) &=& Var(T_1 + T_2 + cdots + T_{300})
        \ &=& 300 cdot Var(T_i)
        \ &=& 300 cdot 0.25
        \ &=& 75
        end{eqnarray}$$



        Still don't understand why my first approach is wrong though.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 12:57









        bardbard

        1886




        1886






























            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%2f3015626%2ftossing-a-triple-of-coins%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