Balls and boxes. Average
$begingroup$
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
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
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
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
$endgroup$
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
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
asked Dec 18 '18 at 8:13
AtstovasAtstovas
1139
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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.
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
$endgroup$
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.
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
edited Dec 18 '18 at 12:59
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
answered Dec 18 '18 at 10:03
drhabdrhab
101k545136
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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