Balls and boxes. Average












0












$begingroup$


We have $r$ balls and $n$ boxes. We took all balls to boxes randomly. Find the average of empty boxes.



So I think there are two ways. When $r<n$ and $rgeq n$.



When $r<n$ thank can be $1,2,3,...,n-1$ boxes empty. Than average would be $frac{sum^{n-1}_{i=1}i}{n-1}$



When $rgeq n$ , then can be $0,1,2,...,n-1$ with average $frac{sum^{n-1}_{i=0}i}{n-1}$



And I don't know what to do next. To sum my given averages and divide from two? Or everything here is wrong? How to find the average than?






probability probability-theory average







share|cite|improve this question









$endgroup$












  • $begingroup$
    Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
    $endgroup$
    – David K
    Dec 18 '18 at 14:17


















0












$begingroup$


We have $r$ balls and $n$ boxes. We took all balls to boxes randomly. Find the average of empty boxes.



So I think there are two ways. When $r<n$ and $rgeq n$.



When $r<n$ thank can be $1,2,3,...,n-1$ boxes empty. Than average would be $frac{sum^{n-1}_{i=1}i}{n-1}$



When $rgeq n$ , then can be $0,1,2,...,n-1$ with average $frac{sum^{n-1}_{i=0}i}{n-1}$



And I don't know what to do next. To sum my given averages and divide from two? Or everything here is wrong? How to find the average than?






probability probability-theory average







share|cite|improve this question









$endgroup$












  • $begingroup$
    Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
    $endgroup$
    – David K
    Dec 18 '18 at 14:17
















0












0








0





$begingroup$


We have $r$ balls and $n$ boxes. We took all balls to boxes randomly. Find the average of empty boxes.



So I think there are two ways. When $r<n$ and $rgeq n$.



When $r<n$ thank can be $1,2,3,...,n-1$ boxes empty. Than average would be $frac{sum^{n-1}_{i=1}i}{n-1}$



When $rgeq n$ , then can be $0,1,2,...,n-1$ with average $frac{sum^{n-1}_{i=0}i}{n-1}$



And I don't know what to do next. To sum my given averages and divide from two? Or everything here is wrong? How to find the average than?






probability probability-theory average







share|cite|improve this question









$endgroup$




We have $r$ balls and $n$ boxes. We took all balls to boxes randomly. Find the average of empty boxes.



So I think there are two ways. When $r<n$ and $rgeq n$.



When $r<n$ thank can be $1,2,3,...,n-1$ boxes empty. Than average would be $frac{sum^{n-1}_{i=1}i}{n-1}$



When $rgeq n$ , then can be $0,1,2,...,n-1$ with average $frac{sum^{n-1}_{i=0}i}{n-1}$



And I don't know what to do next. To sum my given averages and divide from two? Or everything here is wrong? How to find the average than?






probability probability-theory average




probability probability-theory average






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 8:13









AtstovasAtstovas

1139




1139












  • $begingroup$
    Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
    $endgroup$
    – David K
    Dec 18 '18 at 14:17




















  • $begingroup$
    Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
    $endgroup$
    – David K
    Dec 18 '18 at 14:17


















$begingroup$
Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
$endgroup$
– David K
Dec 18 '18 at 14:17






$begingroup$
Almost everything here is wrong. It is not true that every time you can divide the possible outcomes of something into $n$ cases, every case must be equally likely as any other. It is true that if $rgeq n$ then all the cases $0,1,ldots,n-1$ are possible, and if $r<n$ then $n-1$ is possible. But take $r=3,n=7,$ then you cannot have exactly $1$ box empty or even only $3$ boxes empty.
$endgroup$
– David K
Dec 18 '18 at 14:17












1 Answer
1






active

oldest

votes


















3












$begingroup$

I preassume that "we took all balls to boxes randomly" can be interpreted as: for every ball the probability that it ends up in box $i$ is $frac1n$.



For $i=1,dots,n$ let $X_i$ take value $1$ if the box $i$ is empty after the placing, and let it take value $0$ otherwise.



Then $X=sum_{i=1}^nX_i$ is the number of empty boxes after placing the balls and with linearity of expectation and symmetry we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(X_1=1)$$



So it remains to find $P(X_1=1)$.



Give that a try.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
    $endgroup$
    – Atstovas
    Dec 18 '18 at 10:44










  • $begingroup$
    $P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
    $endgroup$
    – drhab
    Dec 18 '18 at 10:56










  • $begingroup$
    but what is next? I can't figure out
    $endgroup$
    – Atstovas
    Dec 18 '18 at 11:41










  • $begingroup$
    Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
    $endgroup$
    – drhab
    Dec 18 '18 at 12:44








  • 2




    $begingroup$
    $X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
    $endgroup$
    – drhab
    Dec 18 '18 at 13:02













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

I preassume that "we took all balls to boxes randomly" can be interpreted as: for every ball the probability that it ends up in box $i$ is $frac1n$.



For $i=1,dots,n$ let $X_i$ take value $1$ if the box $i$ is empty after the placing, and let it take value $0$ otherwise.



Then $X=sum_{i=1}^nX_i$ is the number of empty boxes after placing the balls and with linearity of expectation and symmetry we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(X_1=1)$$



So it remains to find $P(X_1=1)$.



Give that a try.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
    $endgroup$
    – Atstovas
    Dec 18 '18 at 10:44










  • $begingroup$
    $P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
    $endgroup$
    – drhab
    Dec 18 '18 at 10:56










  • $begingroup$
    but what is next? I can't figure out
    $endgroup$
    – Atstovas
    Dec 18 '18 at 11:41










  • $begingroup$
    Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
    $endgroup$
    – drhab
    Dec 18 '18 at 12:44








  • 2




    $begingroup$
    $X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
    $endgroup$
    – drhab
    Dec 18 '18 at 13:02


















3












$begingroup$

I preassume that "we took all balls to boxes randomly" can be interpreted as: for every ball the probability that it ends up in box $i$ is $frac1n$.



For $i=1,dots,n$ let $X_i$ take value $1$ if the box $i$ is empty after the placing, and let it take value $0$ otherwise.



Then $X=sum_{i=1}^nX_i$ is the number of empty boxes after placing the balls and with linearity of expectation and symmetry we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(X_1=1)$$



So it remains to find $P(X_1=1)$.



Give that a try.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
    $endgroup$
    – Atstovas
    Dec 18 '18 at 10:44










  • $begingroup$
    $P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
    $endgroup$
    – drhab
    Dec 18 '18 at 10:56










  • $begingroup$
    but what is next? I can't figure out
    $endgroup$
    – Atstovas
    Dec 18 '18 at 11:41










  • $begingroup$
    Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
    $endgroup$
    – drhab
    Dec 18 '18 at 12:44








  • 2




    $begingroup$
    $X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
    $endgroup$
    – drhab
    Dec 18 '18 at 13:02
















3












3








3





$begingroup$

I preassume that "we took all balls to boxes randomly" can be interpreted as: for every ball the probability that it ends up in box $i$ is $frac1n$.



For $i=1,dots,n$ let $X_i$ take value $1$ if the box $i$ is empty after the placing, and let it take value $0$ otherwise.



Then $X=sum_{i=1}^nX_i$ is the number of empty boxes after placing the balls and with linearity of expectation and symmetry we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(X_1=1)$$



So it remains to find $P(X_1=1)$.



Give that a try.






share|cite|improve this answer











$endgroup$



I preassume that "we took all balls to boxes randomly" can be interpreted as: for every ball the probability that it ends up in box $i$ is $frac1n$.



For $i=1,dots,n$ let $X_i$ take value $1$ if the box $i$ is empty after the placing, and let it take value $0$ otherwise.



Then $X=sum_{i=1}^nX_i$ is the number of empty boxes after placing the balls and with linearity of expectation and symmetry we find:$$mathbb EX=sum_{i=1}^nmathbb EX_i=nmathbb EX_1=nP(X_1=1)$$



So it remains to find $P(X_1=1)$.



Give that a try.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 12:59

























answered Dec 18 '18 at 10:03









drhabdrhab

101k545136




101k545136












  • $begingroup$
    I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
    $endgroup$
    – Atstovas
    Dec 18 '18 at 10:44










  • $begingroup$
    $P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
    $endgroup$
    – drhab
    Dec 18 '18 at 10:56










  • $begingroup$
    but what is next? I can't figure out
    $endgroup$
    – Atstovas
    Dec 18 '18 at 11:41










  • $begingroup$
    Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
    $endgroup$
    – drhab
    Dec 18 '18 at 12:44








  • 2




    $begingroup$
    $X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
    $endgroup$
    – drhab
    Dec 18 '18 at 13:02




















  • $begingroup$
    I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
    $endgroup$
    – Atstovas
    Dec 18 '18 at 10:44










  • $begingroup$
    $P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
    $endgroup$
    – drhab
    Dec 18 '18 at 10:56










  • $begingroup$
    but what is next? I can't figure out
    $endgroup$
    – Atstovas
    Dec 18 '18 at 11:41










  • $begingroup$
    Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
    $endgroup$
    – drhab
    Dec 18 '18 at 12:44








  • 2




    $begingroup$
    $X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
    $endgroup$
    – drhab
    Dec 18 '18 at 13:02


















$begingroup$
I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
$endgroup$
– Atstovas
Dec 18 '18 at 10:44




$begingroup$
I you sure that we need to find $mathbb{P}(X_1=1)$ not a $mathbb{P}(X_i=1)$?
$endgroup$
– Atstovas
Dec 18 '18 at 10:44












$begingroup$
$P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
$endgroup$
– drhab
Dec 18 '18 at 10:56




$begingroup$
$P(X_i=1)=P(X_1=1)$ for every $i$, so there is no difference.
$endgroup$
– drhab
Dec 18 '18 at 10:56












$begingroup$
but what is next? I can't figure out
$endgroup$
– Atstovas
Dec 18 '18 at 11:41




$begingroup$
but what is next? I can't figure out
$endgroup$
– Atstovas
Dec 18 '18 at 11:41












$begingroup$
Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
$endgroup$
– drhab
Dec 18 '18 at 12:44






$begingroup$
Next? If you have found $P(X_1=1)$ then you are ready. The average you are looking for is $nP(X_1=1)$ as stated in my answer. Can you find $P(X_1=1)$?
$endgroup$
– drhab
Dec 18 '18 at 12:44






2




2




$begingroup$
$X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
$endgroup$
– drhab
Dec 18 '18 at 13:02






$begingroup$
$X_1=1$ iff the first box stays empty. The probability on that is $left(frac{n-1}{n}right)^r$ (i.e. the probability that each of the $r$ balls is placed in one of the $n-1$ other boxes).
$endgroup$
– drhab
Dec 18 '18 at 13:02




















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