Probability: the numeric value of a drawn ball












1












$begingroup$


An urn contains N balls. Let $X$ be a random variable to specify the numeric value of a drawn ball twice for the first time (with replacement).
I would like to find the following probability: $P(X=k)$ for $ k in mathbb{N}$.
Can you help me, please? I don't know how to start.



Is the probability the same when $X$ is the numeric value of the drawing when a ball is drawn twice for the first time (with replacement)? (this is what I am actually looking for).






probability random-variables







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    An urn contains N balls. Let $X$ be a random variable to specify the numeric value of a drawn ball twice for the first time (with replacement).
    I would like to find the following probability: $P(X=k)$ for $ k in mathbb{N}$.
    Can you help me, please? I don't know how to start.



    Is the probability the same when $X$ is the numeric value of the drawing when a ball is drawn twice for the first time (with replacement)? (this is what I am actually looking for).






    probability random-variables







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      An urn contains N balls. Let $X$ be a random variable to specify the numeric value of a drawn ball twice for the first time (with replacement).
      I would like to find the following probability: $P(X=k)$ for $ k in mathbb{N}$.
      Can you help me, please? I don't know how to start.



      Is the probability the same when $X$ is the numeric value of the drawing when a ball is drawn twice for the first time (with replacement)? (this is what I am actually looking for).






      probability random-variables







      share|cite|improve this question











      $endgroup$




      An urn contains N balls. Let $X$ be a random variable to specify the numeric value of a drawn ball twice for the first time (with replacement).
      I would like to find the following probability: $P(X=k)$ for $ k in mathbb{N}$.
      Can you help me, please? I don't know how to start.



      Is the probability the same when $X$ is the numeric value of the drawing when a ball is drawn twice for the first time (with replacement)? (this is what I am actually looking for).






      probability random-variables




      probability random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 22:29







      tommy_m

















      asked Dec 9 '18 at 22:09









      tommy_mtommy_m

      1115




      1115






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: One ball drawn equal to $i$:



          $$P(text{first ball} = i) = frac{1}{N}.$$



          This is true for any $i in {1, ldots, N}$.



          Moreover, since there is replacement, notice that:



          $$P(text{second ball} = i) = frac{1}{N}.$$



          Again, this is true for any $i in {1, ldots, N}$.



          Replacement is very important since:




          1. Each drawn is the same

          2. Each drawn is independent


          Then, for the independence:



          $$P(text{first ball} = i ~text{AND}~ text{second ball} = i) = frac{1}{N} cdot frac{1}{N} = frac{1}{N^2}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:21










          • $begingroup$
            In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:35










          • $begingroup$
            @GenericMathematician the OP did not specify how many time a ball is drawn...
            $endgroup$
            – the_candyman
            Dec 9 '18 at 23:44



















          0












          $begingroup$

          If the urn only contains $N=1$ ball, then $P(X=1)=1$.



          If the urn contains $N=2$ balls, then what is $P(X=1)$? I claim that it is $1/2$, based on the following argument: What are the possible outcomes of drawing until the first repetition? It could be $11$, $121$, $122$, $211$, $212$, or $22$; half of these result in the number $1$ being the first repetition. Thus, $P(X=1)=P(X=2)=1/2$.



          I don't think we need to go further (thankfully) in order to wrap our head around a construct that will illuminate the problem. We want to count the number of strings in an $N$ letter alphabet that don't have any repeats, starting from size $1$ all the way to size $N$. Each string that has a specific value $k$ in it will contribute to the count of the total number of ways to have $k$ be the first repetition, and then after dividing by the total number we will have our probability, since everything is uniformly random.



          So how many strings are there of length $1$ in an $N$ letter alphabet that avoid repetition? There are $N$ of them. How about length $2$? There are $N(N-1)$ of them. Indeed, the number of such strings of length $L$ is $N!/(N-L)!$ So the total number of strings is $sum_L N!/(N-L)!$.



          How many of these strings contain the number $k$? ... Wait a second, the number $k$ doesn't matter! Whatever this count turns out to be for $1$ must be the same as the count for $2$ and the count for $3$, and so on. But then what can we see? All numbers have an equal probability of being the first repeat!



          Thus, for an urn with $N$ balls in it, $P(X=k)=1/N$.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

            oldest

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            active

            oldest

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            1












            $begingroup$

            Hint: One ball drawn equal to $i$:



            $$P(text{first ball} = i) = frac{1}{N}.$$



            This is true for any $i in {1, ldots, N}$.



            Moreover, since there is replacement, notice that:



            $$P(text{second ball} = i) = frac{1}{N}.$$



            Again, this is true for any $i in {1, ldots, N}$.



            Replacement is very important since:




            1. Each drawn is the same

            2. Each drawn is independent


            Then, for the independence:



            $$P(text{first ball} = i ~text{AND}~ text{second ball} = i) = frac{1}{N} cdot frac{1}{N} = frac{1}{N^2}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:21










            • $begingroup$
              In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:35










            • $begingroup$
              @GenericMathematician the OP did not specify how many time a ball is drawn...
              $endgroup$
              – the_candyman
              Dec 9 '18 at 23:44
















            1












            $begingroup$

            Hint: One ball drawn equal to $i$:



            $$P(text{first ball} = i) = frac{1}{N}.$$



            This is true for any $i in {1, ldots, N}$.



            Moreover, since there is replacement, notice that:



            $$P(text{second ball} = i) = frac{1}{N}.$$



            Again, this is true for any $i in {1, ldots, N}$.



            Replacement is very important since:




            1. Each drawn is the same

            2. Each drawn is independent


            Then, for the independence:



            $$P(text{first ball} = i ~text{AND}~ text{second ball} = i) = frac{1}{N} cdot frac{1}{N} = frac{1}{N^2}.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:21










            • $begingroup$
              In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:35










            • $begingroup$
              @GenericMathematician the OP did not specify how many time a ball is drawn...
              $endgroup$
              – the_candyman
              Dec 9 '18 at 23:44














            1












            1








            1





            $begingroup$

            Hint: One ball drawn equal to $i$:



            $$P(text{first ball} = i) = frac{1}{N}.$$



            This is true for any $i in {1, ldots, N}$.



            Moreover, since there is replacement, notice that:



            $$P(text{second ball} = i) = frac{1}{N}.$$



            Again, this is true for any $i in {1, ldots, N}$.



            Replacement is very important since:




            1. Each drawn is the same

            2. Each drawn is independent


            Then, for the independence:



            $$P(text{first ball} = i ~text{AND}~ text{second ball} = i) = frac{1}{N} cdot frac{1}{N} = frac{1}{N^2}.$$






            share|cite|improve this answer









            $endgroup$



            Hint: One ball drawn equal to $i$:



            $$P(text{first ball} = i) = frac{1}{N}.$$



            This is true for any $i in {1, ldots, N}$.



            Moreover, since there is replacement, notice that:



            $$P(text{second ball} = i) = frac{1}{N}.$$



            Again, this is true for any $i in {1, ldots, N}$.



            Replacement is very important since:




            1. Each drawn is the same

            2. Each drawn is independent


            Then, for the independence:



            $$P(text{first ball} = i ~text{AND}~ text{second ball} = i) = frac{1}{N} cdot frac{1}{N} = frac{1}{N^2}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 22:15









            the_candymanthe_candyman

            8,85632145




            8,85632145












            • $begingroup$
              I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:21










            • $begingroup$
              In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:35










            • $begingroup$
              @GenericMathematician the OP did not specify how many time a ball is drawn...
              $endgroup$
              – the_candyman
              Dec 9 '18 at 23:44


















            • $begingroup$
              I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:21










            • $begingroup$
              In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
              $endgroup$
              – GenericMathematician
              Dec 9 '18 at 22:35










            • $begingroup$
              @GenericMathematician the OP did not specify how many time a ball is drawn...
              $endgroup$
              – the_candyman
              Dec 9 '18 at 23:44
















            $begingroup$
            I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:21




            $begingroup$
            I think you have answered the wrong question... don't we want the probability that the first ball to be repeated is $k$, and not the probability that you draw the same ball twice in a row?
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:21












            $begingroup$
            In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:35




            $begingroup$
            In fact, here is a proof that you have solved the wrong problem; if it were the case that $P(X=k)=1/N^2$ then the total probability would only be $1/N$, which is impossible by definition.
            $endgroup$
            – GenericMathematician
            Dec 9 '18 at 22:35












            $begingroup$
            @GenericMathematician the OP did not specify how many time a ball is drawn...
            $endgroup$
            – the_candyman
            Dec 9 '18 at 23:44




            $begingroup$
            @GenericMathematician the OP did not specify how many time a ball is drawn...
            $endgroup$
            – the_candyman
            Dec 9 '18 at 23:44











            0












            $begingroup$

            If the urn only contains $N=1$ ball, then $P(X=1)=1$.



            If the urn contains $N=2$ balls, then what is $P(X=1)$? I claim that it is $1/2$, based on the following argument: What are the possible outcomes of drawing until the first repetition? It could be $11$, $121$, $122$, $211$, $212$, or $22$; half of these result in the number $1$ being the first repetition. Thus, $P(X=1)=P(X=2)=1/2$.



            I don't think we need to go further (thankfully) in order to wrap our head around a construct that will illuminate the problem. We want to count the number of strings in an $N$ letter alphabet that don't have any repeats, starting from size $1$ all the way to size $N$. Each string that has a specific value $k$ in it will contribute to the count of the total number of ways to have $k$ be the first repetition, and then after dividing by the total number we will have our probability, since everything is uniformly random.



            So how many strings are there of length $1$ in an $N$ letter alphabet that avoid repetition? There are $N$ of them. How about length $2$? There are $N(N-1)$ of them. Indeed, the number of such strings of length $L$ is $N!/(N-L)!$ So the total number of strings is $sum_L N!/(N-L)!$.



            How many of these strings contain the number $k$? ... Wait a second, the number $k$ doesn't matter! Whatever this count turns out to be for $1$ must be the same as the count for $2$ and the count for $3$, and so on. But then what can we see? All numbers have an equal probability of being the first repeat!



            Thus, for an urn with $N$ balls in it, $P(X=k)=1/N$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              If the urn only contains $N=1$ ball, then $P(X=1)=1$.



              If the urn contains $N=2$ balls, then what is $P(X=1)$? I claim that it is $1/2$, based on the following argument: What are the possible outcomes of drawing until the first repetition? It could be $11$, $121$, $122$, $211$, $212$, or $22$; half of these result in the number $1$ being the first repetition. Thus, $P(X=1)=P(X=2)=1/2$.



              I don't think we need to go further (thankfully) in order to wrap our head around a construct that will illuminate the problem. We want to count the number of strings in an $N$ letter alphabet that don't have any repeats, starting from size $1$ all the way to size $N$. Each string that has a specific value $k$ in it will contribute to the count of the total number of ways to have $k$ be the first repetition, and then after dividing by the total number we will have our probability, since everything is uniformly random.



              So how many strings are there of length $1$ in an $N$ letter alphabet that avoid repetition? There are $N$ of them. How about length $2$? There are $N(N-1)$ of them. Indeed, the number of such strings of length $L$ is $N!/(N-L)!$ So the total number of strings is $sum_L N!/(N-L)!$.



              How many of these strings contain the number $k$? ... Wait a second, the number $k$ doesn't matter! Whatever this count turns out to be for $1$ must be the same as the count for $2$ and the count for $3$, and so on. But then what can we see? All numbers have an equal probability of being the first repeat!



              Thus, for an urn with $N$ balls in it, $P(X=k)=1/N$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                If the urn only contains $N=1$ ball, then $P(X=1)=1$.



                If the urn contains $N=2$ balls, then what is $P(X=1)$? I claim that it is $1/2$, based on the following argument: What are the possible outcomes of drawing until the first repetition? It could be $11$, $121$, $122$, $211$, $212$, or $22$; half of these result in the number $1$ being the first repetition. Thus, $P(X=1)=P(X=2)=1/2$.



                I don't think we need to go further (thankfully) in order to wrap our head around a construct that will illuminate the problem. We want to count the number of strings in an $N$ letter alphabet that don't have any repeats, starting from size $1$ all the way to size $N$. Each string that has a specific value $k$ in it will contribute to the count of the total number of ways to have $k$ be the first repetition, and then after dividing by the total number we will have our probability, since everything is uniformly random.



                So how many strings are there of length $1$ in an $N$ letter alphabet that avoid repetition? There are $N$ of them. How about length $2$? There are $N(N-1)$ of them. Indeed, the number of such strings of length $L$ is $N!/(N-L)!$ So the total number of strings is $sum_L N!/(N-L)!$.



                How many of these strings contain the number $k$? ... Wait a second, the number $k$ doesn't matter! Whatever this count turns out to be for $1$ must be the same as the count for $2$ and the count for $3$, and so on. But then what can we see? All numbers have an equal probability of being the first repeat!



                Thus, for an urn with $N$ balls in it, $P(X=k)=1/N$.






                share|cite|improve this answer











                $endgroup$



                If the urn only contains $N=1$ ball, then $P(X=1)=1$.



                If the urn contains $N=2$ balls, then what is $P(X=1)$? I claim that it is $1/2$, based on the following argument: What are the possible outcomes of drawing until the first repetition? It could be $11$, $121$, $122$, $211$, $212$, or $22$; half of these result in the number $1$ being the first repetition. Thus, $P(X=1)=P(X=2)=1/2$.



                I don't think we need to go further (thankfully) in order to wrap our head around a construct that will illuminate the problem. We want to count the number of strings in an $N$ letter alphabet that don't have any repeats, starting from size $1$ all the way to size $N$. Each string that has a specific value $k$ in it will contribute to the count of the total number of ways to have $k$ be the first repetition, and then after dividing by the total number we will have our probability, since everything is uniformly random.



                So how many strings are there of length $1$ in an $N$ letter alphabet that avoid repetition? There are $N$ of them. How about length $2$? There are $N(N-1)$ of them. Indeed, the number of such strings of length $L$ is $N!/(N-L)!$ So the total number of strings is $sum_L N!/(N-L)!$.



                How many of these strings contain the number $k$? ... Wait a second, the number $k$ doesn't matter! Whatever this count turns out to be for $1$ must be the same as the count for $2$ and the count for $3$, and so on. But then what can we see? All numbers have an equal probability of being the first repeat!



                Thus, for an urn with $N$ balls in it, $P(X=k)=1/N$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 22:43

























                answered Dec 9 '18 at 22:31









                GenericMathematicianGenericMathematician

                863




                863






























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