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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    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
    3 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
    3 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






  • 1




    $begingroup$
    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
    3 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
    3 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






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asked 3 hours ago









The RangsterThe Rangster

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  • 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






  • 1




    $begingroup$
    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
    3 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
    3 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






  • 1




    $begingroup$
    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
    3 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
    3 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




1




1




$begingroup$
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
3 hours ago






$begingroup$
@fleablood: So $0%$ then. (And they all look alike, so you have to sniff them all.)
$endgroup$
– TonyK
3 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
3 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
3 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









      $endgroup$













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        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 3 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






















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