Balls in bins: Prove that the probability that there is some bins of $k+1$ or more balls is at most $frac...











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(Using Chebyshev's inequality and Union bound)



Suppose that we throw $n$ balls into n bins uniformly at random. Let $k≥sqrt n$ be a positive integer Show that with probability at least $1-frac n{k^2}$, no bin has strictly more than $k$ balls.

Prove that the probability that there is some bins of $k+1$ or more balls is at most $frac n{k^2}$.










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    (Using Chebyshev's inequality and Union bound)



    Suppose that we throw $n$ balls into n bins uniformly at random. Let $k≥sqrt n$ be a positive integer Show that with probability at least $1-frac n{k^2}$, no bin has strictly more than $k$ balls.

    Prove that the probability that there is some bins of $k+1$ or more balls is at most $frac n{k^2}$.










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      (Using Chebyshev's inequality and Union bound)



      Suppose that we throw $n$ balls into n bins uniformly at random. Let $k≥sqrt n$ be a positive integer Show that with probability at least $1-frac n{k^2}$, no bin has strictly more than $k$ balls.

      Prove that the probability that there is some bins of $k+1$ or more balls is at most $frac n{k^2}$.










      share|cite|improve this question















      (Using Chebyshev's inequality and Union bound)



      Suppose that we throw $n$ balls into n bins uniformly at random. Let $k≥sqrt n$ be a positive integer Show that with probability at least $1-frac n{k^2}$, no bin has strictly more than $k$ balls.

      Prove that the probability that there is some bins of $k+1$ or more balls is at most $frac n{k^2}$.







      probability balls-in-bins






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      edited Nov 18 at 7:11









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      asked Nov 18 at 2:59









      user8863554

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          Consider of the distribution of balls that land in a particular bin, $X$. This will be a binomial distribution. Its mean is $ncdot$$1over n$$=1$ and its variance is $ncdot$$1over n$$cdot(1-$$1over n$$)$$=1-$$1over n$. Then Chebyshev tells us $P(|X-1|ge kcdot(1-$$1 over n$$)) le$$ 1over k^2$. For $1le klt n$, since $kover n$$lt 1$ this means $P(|X| ge k+1) le $$1over k^2$. Now if $k ge sqrt n$ then the probability of one of the n bins having more than k balls will be at most $nover k^2$ by the union bound (for $k lt sqrt n$ the upper bound we get is greater than 1 and so is meaningless). For the case $k=n$ simply note the $P(X=n)=($$1over n$$)^n$ so the union bound gives $P($n balls land in some box$)=$$n over n^n$$le$$ nover n^2$.






          share|cite|improve this answer





















          • thanks for you answer, but does the denominator in the last fraction should be k^2 ?
            – user8863554
            Nov 20 at 1:41










          • In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
            – Mark
            Nov 21 at 3:59










          • Thank you so much
            – user8863554
            Nov 27 at 15:02











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          oldest

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          active

          oldest

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          up vote
          0
          down vote



          accepted










          Consider of the distribution of balls that land in a particular bin, $X$. This will be a binomial distribution. Its mean is $ncdot$$1over n$$=1$ and its variance is $ncdot$$1over n$$cdot(1-$$1over n$$)$$=1-$$1over n$. Then Chebyshev tells us $P(|X-1|ge kcdot(1-$$1 over n$$)) le$$ 1over k^2$. For $1le klt n$, since $kover n$$lt 1$ this means $P(|X| ge k+1) le $$1over k^2$. Now if $k ge sqrt n$ then the probability of one of the n bins having more than k balls will be at most $nover k^2$ by the union bound (for $k lt sqrt n$ the upper bound we get is greater than 1 and so is meaningless). For the case $k=n$ simply note the $P(X=n)=($$1over n$$)^n$ so the union bound gives $P($n balls land in some box$)=$$n over n^n$$le$$ nover n^2$.






          share|cite|improve this answer





















          • thanks for you answer, but does the denominator in the last fraction should be k^2 ?
            – user8863554
            Nov 20 at 1:41










          • In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
            – Mark
            Nov 21 at 3:59










          • Thank you so much
            – user8863554
            Nov 27 at 15:02















          up vote
          0
          down vote



          accepted










          Consider of the distribution of balls that land in a particular bin, $X$. This will be a binomial distribution. Its mean is $ncdot$$1over n$$=1$ and its variance is $ncdot$$1over n$$cdot(1-$$1over n$$)$$=1-$$1over n$. Then Chebyshev tells us $P(|X-1|ge kcdot(1-$$1 over n$$)) le$$ 1over k^2$. For $1le klt n$, since $kover n$$lt 1$ this means $P(|X| ge k+1) le $$1over k^2$. Now if $k ge sqrt n$ then the probability of one of the n bins having more than k balls will be at most $nover k^2$ by the union bound (for $k lt sqrt n$ the upper bound we get is greater than 1 and so is meaningless). For the case $k=n$ simply note the $P(X=n)=($$1over n$$)^n$ so the union bound gives $P($n balls land in some box$)=$$n over n^n$$le$$ nover n^2$.






          share|cite|improve this answer





















          • thanks for you answer, but does the denominator in the last fraction should be k^2 ?
            – user8863554
            Nov 20 at 1:41










          • In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
            – Mark
            Nov 21 at 3:59










          • Thank you so much
            – user8863554
            Nov 27 at 15:02













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          Consider of the distribution of balls that land in a particular bin, $X$. This will be a binomial distribution. Its mean is $ncdot$$1over n$$=1$ and its variance is $ncdot$$1over n$$cdot(1-$$1over n$$)$$=1-$$1over n$. Then Chebyshev tells us $P(|X-1|ge kcdot(1-$$1 over n$$)) le$$ 1over k^2$. For $1le klt n$, since $kover n$$lt 1$ this means $P(|X| ge k+1) le $$1over k^2$. Now if $k ge sqrt n$ then the probability of one of the n bins having more than k balls will be at most $nover k^2$ by the union bound (for $k lt sqrt n$ the upper bound we get is greater than 1 and so is meaningless). For the case $k=n$ simply note the $P(X=n)=($$1over n$$)^n$ so the union bound gives $P($n balls land in some box$)=$$n over n^n$$le$$ nover n^2$.






          share|cite|improve this answer












          Consider of the distribution of balls that land in a particular bin, $X$. This will be a binomial distribution. Its mean is $ncdot$$1over n$$=1$ and its variance is $ncdot$$1over n$$cdot(1-$$1over n$$)$$=1-$$1over n$. Then Chebyshev tells us $P(|X-1|ge kcdot(1-$$1 over n$$)) le$$ 1over k^2$. For $1le klt n$, since $kover n$$lt 1$ this means $P(|X| ge k+1) le $$1over k^2$. Now if $k ge sqrt n$ then the probability of one of the n bins having more than k balls will be at most $nover k^2$ by the union bound (for $k lt sqrt n$ the upper bound we get is greater than 1 and so is meaningless). For the case $k=n$ simply note the $P(X=n)=($$1over n$$)^n$ so the union bound gives $P($n balls land in some box$)=$$n over n^n$$le$$ nover n^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 7:15









          Mark

          3916




          3916












          • thanks for you answer, but does the denominator in the last fraction should be k^2 ?
            – user8863554
            Nov 20 at 1:41










          • In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
            – Mark
            Nov 21 at 3:59










          • Thank you so much
            – user8863554
            Nov 27 at 15:02


















          • thanks for you answer, but does the denominator in the last fraction should be k^2 ?
            – user8863554
            Nov 20 at 1:41










          • In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
            – Mark
            Nov 21 at 3:59










          • Thank you so much
            – user8863554
            Nov 27 at 15:02
















          thanks for you answer, but does the denominator in the last fraction should be k^2 ?
          – user8863554
          Nov 20 at 1:41




          thanks for you answer, but does the denominator in the last fraction should be k^2 ?
          – user8863554
          Nov 20 at 1:41












          In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
          – Mark
          Nov 21 at 3:59




          In the last inequality we are only considering the case k=n so the denominator does equal k^2. Sorry for the confusion.
          – Mark
          Nov 21 at 3:59












          Thank you so much
          – user8863554
          Nov 27 at 15:02




          Thank you so much
          – user8863554
          Nov 27 at 15:02


















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