probability of choosing the i-th red ball from a basket containing $m$ and $n$ indexed red and black balls...












0












$begingroup$


A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$



We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).



Does it matter that the black balls are indexed or not?



The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$



    We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).



    Does it matter that the black balls are indexed or not?



    The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$



      We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).



      Does it matter that the black balls are indexed or not?



      The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?










      share|cite|improve this question









      $endgroup$




      A basket contains $m$ and $n$ red and black balls resp. indexed ${R_1,...,R_m,B_1,...,B_n}$



      We perform the action of withdrawing $k$ balls from the basket, putting each ball aside (without putting it back).



      Does it matter that the black balls are indexed or not?



      The probability of the $i-th$ ball being chosen is $frac{{n+m-1choose k}}{n+mchoose k-1}$ if the black balls are indistinguishable, because there are ${n+m-1choose k-1}$ possibilities of success (picking the $R_i$ ball and $k-1$ other balls) and ${n+mchoose k}$ total possibilities. But what if the black balls are indexed from $1$ to $n$? Does that make a difference?







      probability






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      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 5 '18 at 14:30









      John CataldoJohn Cataldo

      1,1771316




      1,1771316






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.



          This because $k$ of the $m+n$ balls are selected.



          Btw, this is true for every fixed ball.



          No matter whether the black balls are numbered or not.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I changed my answer.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:06










          • $begingroup$
            Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:09










          • $begingroup$
            Yes. It does not matter whether the black balls wear a label or not.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:10










          • $begingroup$
            Could you explain a little bit why that's true? Because that was really my question
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:11












          • $begingroup$
            It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:29



















          0












          $begingroup$

          The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
          For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$



          If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
          For example,
          ${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
          ${R_1, R_3, B_3, B_5}$
          count as three separate outcomes if we index the black balls,
          but if we don't distinguish the black balls these are all part of the single outcome
          ${R_1, R_3, 2times B}.$



          The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.



            This because $k$ of the $m+n$ balls are selected.



            Btw, this is true for every fixed ball.



            No matter whether the black balls are numbered or not.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I changed my answer.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:06










            • $begingroup$
              Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:09










            • $begingroup$
              Yes. It does not matter whether the black balls wear a label or not.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:10










            • $begingroup$
              Could you explain a little bit why that's true? Because that was really my question
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:11












            • $begingroup$
              It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:29
















            1












            $begingroup$

            The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.



            This because $k$ of the $m+n$ balls are selected.



            Btw, this is true for every fixed ball.



            No matter whether the black balls are numbered or not.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I changed my answer.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:06










            • $begingroup$
              Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:09










            • $begingroup$
              Yes. It does not matter whether the black balls wear a label or not.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:10










            • $begingroup$
              Could you explain a little bit why that's true? Because that was really my question
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:11












            • $begingroup$
              It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:29














            1












            1








            1





            $begingroup$

            The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.



            This because $k$ of the $m+n$ balls are selected.



            Btw, this is true for every fixed ball.



            No matter whether the black balls are numbered or not.






            share|cite|improve this answer











            $endgroup$



            The probability that the $i$-th red ball is one of the chosen balls is $frac{k}{m+n}$.



            This because $k$ of the $m+n$ balls are selected.



            Btw, this is true for every fixed ball.



            No matter whether the black balls are numbered or not.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 5 '18 at 15:05

























            answered Dec 5 '18 at 14:52









            drhabdrhab

            99.7k544130




            99.7k544130












            • $begingroup$
              I changed my answer.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:06










            • $begingroup$
              Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:09










            • $begingroup$
              Yes. It does not matter whether the black balls wear a label or not.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:10










            • $begingroup$
              Could you explain a little bit why that's true? Because that was really my question
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:11












            • $begingroup$
              It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:29


















            • $begingroup$
              I changed my answer.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:06










            • $begingroup$
              Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:09










            • $begingroup$
              Yes. It does not matter whether the black balls wear a label or not.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:10










            • $begingroup$
              Could you explain a little bit why that's true? Because that was really my question
              $endgroup$
              – John Cataldo
              Dec 5 '18 at 15:11












            • $begingroup$
              It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
              $endgroup$
              – drhab
              Dec 5 '18 at 15:29
















            $begingroup$
            I changed my answer.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:06




            $begingroup$
            I changed my answer.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:06












            $begingroup$
            Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:09




            $begingroup$
            Yeah thanks, does that probability also correspond to the event where the black balls are indexed?
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:09












            $begingroup$
            Yes. It does not matter whether the black balls wear a label or not.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:10




            $begingroup$
            Yes. It does not matter whether the black balls wear a label or not.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:10












            $begingroup$
            Could you explain a little bit why that's true? Because that was really my question
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:11






            $begingroup$
            Could you explain a little bit why that's true? Because that was really my question
            $endgroup$
            – John Cataldo
            Dec 5 '18 at 15:11














            $begingroup$
            It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:29




            $begingroup$
            It is just plain thinking. If there are $m+n$ balls and one of them is fixed. Then the probability that this fixed ball will belong to a selection of $k$ balls will be $frac{k}{m+n}$. It is like buying a lot in a lottery with $m+n$ lots and $k$ prices. The probability that you win is price is $frac{k}{m+n}$, right? Labeling of the objects (lots, balls) plays not really a part in this.
            $endgroup$
            – drhab
            Dec 5 '18 at 15:29











            0












            $begingroup$

            The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
            For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$



            If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
            For example,
            ${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
            ${R_1, R_3, B_3, B_5}$
            count as three separate outcomes if we index the black balls,
            but if we don't distinguish the black balls these are all part of the single outcome
            ${R_1, R_3, 2times B}.$



            The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
              For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$



              If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
              For example,
              ${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
              ${R_1, R_3, B_3, B_5}$
              count as three separate outcomes if we index the black balls,
              but if we don't distinguish the black balls these are all part of the single outcome
              ${R_1, R_3, 2times B}.$



              The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
                For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$



                If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
                For example,
                ${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
                ${R_1, R_3, B_3, B_5}$
                count as three separate outcomes if we index the black balls,
                but if we don't distinguish the black balls these are all part of the single outcome
                ${R_1, R_3, 2times B}.$



                The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.






                share|cite|improve this answer











                $endgroup$



                The counting you've already done is the counting you get when the black balls are indexed but we don't care about the sequence in which the $k$ balls were drawn.
                For example, with $m = n = 5,$ $k = 4,$ and $i = 1,$ one of the $binom{n+m-1}{k-1}$ subsets of balls including $R_i$ is ${R_1, R_3, B_2, B_3}.$



                If you truly consider the black balls indistinguishable, but the red balls are indexed, and you only want to count distinguishable configurations, then the total number of configurations and the number that contain $R_i$ both will be smaller than you computed.
                For example,
                ${R_1, R_3, B_1, B_2},$ ${R_1, R_3, B_1, B_5},$ and
                ${R_1, R_3, B_3, B_5}$
                count as three separate outcomes if we index the black balls,
                but if we don't distinguish the black balls these are all part of the single outcome
                ${R_1, R_3, 2times B}.$



                The configurations with indistinguishable black balls also won't have uniform probability except under a very contrived mechanism for choosing the balls, so it would be difficult to justify a probability based on the ratio of those two numbers.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 5 '18 at 14:49

























                answered Dec 5 '18 at 14:43









                David KDavid K

                53.6k342116




                53.6k342116






























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