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












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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







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closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.












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    $begingroup$
    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.
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    – dantopa
    Feb 26 at 0:40






  • 1




    $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    Feb 26 at 0:53










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    Feb 26 at 1:08












  • $begingroup$
    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
    Feb 26 at 1:13
















2












$begingroup$


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











$endgroup$



closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 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
    Feb 26 at 0:40






  • 1




    $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    Feb 26 at 0:53










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    Feb 26 at 1:08












  • $begingroup$
    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
    Feb 26 at 1:13














2












2








2





$begingroup$


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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 26 at 0:40









dantopa

6,64442245




6,64442245










asked Feb 26 at 0:36









The RangsterThe Rangster

161




161




closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup Feb 26 at 4:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Kemono Chen, Adrian Keister, Theo Bendit, heropup

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 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
    Feb 26 at 0:40






  • 1




    $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    Feb 26 at 0:53










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    Feb 26 at 1:08












  • $begingroup$
    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
    Feb 26 at 1:13














  • 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
    Feb 26 at 0:40






  • 1




    $begingroup$
    100%. I'd purposely save a chocolate for last.
    $endgroup$
    – fleablood
    Feb 26 at 0:53










  • $begingroup$
    @fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
    $endgroup$
    – TonyK
    Feb 26 at 1:08












  • $begingroup$
    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
    Feb 26 at 1:13








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
Feb 26 at 0:40




$begingroup$
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
Feb 26 at 0:40




1




1




$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53




$begingroup$
100%. I'd purposely save a chocolate for last.
$endgroup$
– fleablood
Feb 26 at 0:53












$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08






$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
Feb 26 at 1:08














$begingroup$
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
Feb 26 at 1:13




$begingroup$
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
Feb 26 at 1:13










4 Answers
4






active

oldest

votes


















5












$begingroup$

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
    Feb 26 at 2:48



















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 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












      $begingroup$

      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









      $endgroup$




















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        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
          Feb 26 at 2:48
















        5












        $begingroup$

        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
          Feb 26 at 2:48














        5












        5








        5





        $begingroup$

        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 Feb 26 at 1:07









        TonyKTonyK

        43.7k358137




        43.7k358137












        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          Feb 26 at 2:48


















        • $begingroup$
          This is a clever way of approaching this!
          $endgroup$
          – Remy
          Feb 26 at 2:48
















        $begingroup$
        This is a clever way of approaching this!
        $endgroup$
        – Remy
        Feb 26 at 2:48




        $begingroup$
        This is a clever way of approaching this!
        $endgroup$
        – Remy
        Feb 26 at 2:48











        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 Feb 26 at 0:54









            abcdabcd

            1188




            1188























                1












                $begingroup$

                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









                $endgroup$


















                  1












                  $begingroup$

                  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









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    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 Feb 26 at 2:52









                    RemyRemy

                    6,642922




                    6,642922























                        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 Feb 26 at 0:45









                            D.R.D.R.

                            1,776823




                            1,776823















                                -

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