There is a bag of 8 candies, and 3 are chocolates. You eat candy until the chocolates are gone. What is the...












1












$begingroup$


You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?






probability statistics







share|cite|improve this question









New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    3 hours ago










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    2 hours ago












  • $begingroup$
    Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
    $endgroup$
    – fleablood
    2 hours ago
















1












$begingroup$


You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?






probability statistics







share|cite|improve this question









New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    3 hours ago










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    2 hours ago












  • $begingroup$
    Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
    $endgroup$
    – fleablood
    2 hours ago














1












1








1





$begingroup$


You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?






probability statistics







share|cite|improve this question









New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




You buy a bag of $8$ candies, of which $3$ are chocolates, but all candies look alike. You eat candies from the bag until you have eaten all three chocolates. What is the probability you will have eaten exactly $7$ of the candies in the bag?






probability statistics




probability statistics






share|cite|improve this question









New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









dantopa

6,53942244




6,53942244






New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









The RangsterThe Rangster

111




111




New contributor




The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






The Rangster is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    3 hours ago










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    2 hours ago












  • $begingroup$
    Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
    $endgroup$
    – fleablood
    2 hours ago














  • 1




    $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    3 hours ago










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    2 hours ago












  • $begingroup$
    Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
    $endgroup$
    – fleablood
    2 hours ago








1




1




$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
3 hours ago




$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. This post explains how to write a good question. For equations, please use MathJax.
$endgroup$
– dantopa
3 hours ago












$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
3 hours ago




$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
3 hours ago












$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
2 hours ago






$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
2 hours ago














$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
2 hours ago




$begingroup$
Oh, yes... there are 8 candies. Yes, $0%$. (This is a joke, of course. The question assumes we are eating them randomly. I am making a joke that the question should have stated that.)
$endgroup$
– fleablood
2 hours ago










4 Answers
4






active

oldest

votes


















4












$begingroup$

A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:



What is the probability that the first pick is not a chocolate, and the second pick is?



So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a clever way of approaching this!
    $endgroup$
    – Remy
    1 hour ago



















1












$begingroup$

As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.



    If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is



      $$frac{{3choose2} {5choose4}}{8choose6}$$



      This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is



      $$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$






      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
        });


        }
        });






        The Rangster is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3126856%2fthere-is-a-bag-of-8-candies-and-3-are-chocolates-you-eat-candy-until-the-choco%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:



        What is the probability that the first pick is not a chocolate, and the second pick is?



        So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          1 hour ago
















        4












        $begingroup$

        A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:



        What is the probability that the first pick is not a chocolate, and the second pick is?



        So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          1 hour ago














        4












        4








        4





        $begingroup$

        A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:



        What is the probability that the first pick is not a chocolate, and the second pick is?



        So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.






        share|cite|improve this answer









        $endgroup$



        A particular sequence of picks is just as likely as the same sequence in reverse. But now the question becomes:



        What is the probability that the first pick is not a chocolate, and the second pick is?



        So the answer is obviously $dfrac{5}{8}timesdfrac{3}{7}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        TonyKTonyK

        42.8k355134




        42.8k355134












        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          1 hour ago


















        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          1 hour ago
















        $begingroup$
        This is a clever way of approaching this!
        $endgroup$
        – Remy
        1 hour ago




        $begingroup$
        This is a clever way of approaching this!
        $endgroup$
        – Remy
        1 hour ago











        1












        $begingroup$

        As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.






            share|cite|improve this answer









            $endgroup$



            As D.R. pointed out, there must be $2$ chocolates among the first $6$ candies. Choose these $2$ positions in $binom{6}{2}$ ways. The $7^{th}$ position must be a chocolate, so in total, the positions of the chocolates can be chosen in $binom{6}{2}$ ways. Hence, your required probability is $binom{6}{2}/binom{8}{3} = 0.2679$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            abcdabcd

            1188




            1188























                0












                $begingroup$

                HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.



                If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.



                  If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.



                    If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.






                    share|cite|improve this answer









                    $endgroup$



                    HINT: if the person stops after the 7th candy, then that one must have been the third chocolate. The other two can be anywhere in the first 6.



                    If the question is asking “at least 7 candies”, consider the case where the 8th candy is the third chocolate, and sum the two answers together.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    D.R.D.R.

                    1,514722




                    1,514722























                        0












                        $begingroup$

                        As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is



                        $$frac{{3choose2} {5choose4}}{8choose6}$$



                        This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is



                        $$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is



                          $$frac{{3choose2} {5choose4}}{8choose6}$$



                          This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is



                          $$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is



                            $$frac{{3choose2} {5choose4}}{8choose6}$$



                            This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is



                            $$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$






                            share|cite|improve this answer









                            $endgroup$



                            As others have pointed out, we must select $2$ chocolates and $4$ non-chocolates in the first $6$ selections. The probability that this occurs is



                            $$frac{{3choose2} {5choose4}}{8choose6}$$



                            This comes from the hypergeometric distribution. Then there is one chocolate and one non-chocolate remaining so we then select the third chocolate with probability $frac{1}{2}$. Hence the desired probability is



                            $$frac{{3choose2} {5choose4}}{8choose6}cdotfrac{1}{2}approx0.268$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            RemyRemy

                            6,504822




                            6,504822






















                                The Rangster is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                The Rangster is a new contributor. Be nice, and check out our Code of Conduct.













                                The Rangster is a new contributor. Be nice, and check out our Code of Conduct.












                                The Rangster is a new contributor. Be nice, and check out our Code of Conduct.
















                                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%2f3126856%2fthere-is-a-bag-of-8-candies-and-3-are-chocolates-you-eat-candy-until-the-choco%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

                                Усть-Каменогорск

                                Image
                                У этого термина существуют и другие значения, см. Усть-Каменогорск. Город Усть-Каменогорск каз. Өскемен Флаг Герб 49°57′ с. ш. 82°37′ в. д. H G Я O L Страна Казахстан   Казахстан Статус областной центр Область Восточно-Казахстанская Аким Жаксылык Омар [1] История и география Основан 1720 Город с 1868 Площадь 303,15 [2] км² Высота центра 283 ± 1 м Тип климата резко-континентальный Часовой пояс UTC+6 Население Население 329 090 [3]  человек ( 2018 ) Национальности русские 53,56 % казахи 42,99 % немцы 0,82 % украинцы 0,57 % татары 0,79 % белорусы 0,14 % азербайджанцы 0,23 % корейцы 0,20 % прочие 1,08 % -1 шт [4] Этнохороним усть-каменогорцы, усть-каменогорец Цифровые идентификаторы Телефонный код +7 (7232) Почтовый индекс F0***** [5] Автомобильный код 16 (ранее F) oskemen.kz Усть-Каменогорск Усть-Каменогорск ...
                                Read more

                                Халкинская богословская школа

                                Image
                                Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
                                Read more

                                Высокополье (Харьковская область)

                                Image
                                У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
                                Read more