Mixing
Variance of number of married couples sitting together around a table
$begingroup$
The question: 10 married couples are being sat around a (round) table.
Denote $Y$ - the random variable that equals the number of women sitting next to their husbands. Calculate $V(Y)$.
I tried figuring what is the probabilty that $i$ couples will sit next to each other, but I think this way requires the inclusion and exclusion principle, which I'm pretty sure unnecessary here.
I also tried marking $Y = Y_1 + Y_2 + ... + Y_{10}$ (Where $Y_i$ is 1 if the i-th couple sit next to one another, and 0 otherwise). But this brought up some trouble as well.
I guess I'm stuck more with the combinatorics part of the question, I was hoping you could help me out.
Thanks.
probability combinatorics
edited Jan 31 at 14:24
awkward
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
$endgroup$
|
show 9 more comments
$begingroup$
The question: 10 married couples are being sat around a (round) table.
Denote $Y$ - the random variable that equals the number of women sitting next to their husbands. Calculate $V(Y)$.
I tried figuring what is the probabilty that $i$ couples will sit next to each other, but I think this way requires the inclusion and exclusion principle, which I'm pretty sure unnecessary here.
I also tried marking $Y = Y_1 + Y_2 + ... + Y_{10}$ (Where $Y_i$ is 1 if the i-th couple sit next to one another, and 0 otherwise). But this brought up some trouble as well.
I guess I'm stuck more with the combinatorics part of the question, I was hoping you could help me out.
Thanks.
probability combinatorics
edited Jan 31 at 14:24
awkward
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
$endgroup$
$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
1
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
1
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
1
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09
|
show 9 more comments
$begingroup$
The question: 10 married couples are being sat around a (round) table.
Denote $Y$ - the random variable that equals the number of women sitting next to their husbands. Calculate $V(Y)$.
I tried figuring what is the probabilty that $i$ couples will sit next to each other, but I think this way requires the inclusion and exclusion principle, which I'm pretty sure unnecessary here.
I also tried marking $Y = Y_1 + Y_2 + ... + Y_{10}$ (Where $Y_i$ is 1 if the i-th couple sit next to one another, and 0 otherwise). But this brought up some trouble as well.
I guess I'm stuck more with the combinatorics part of the question, I was hoping you could help me out.
Thanks.
probability combinatorics
edited Jan 31 at 14:24
awkward
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
$endgroup$
The question: 10 married couples are being sat around a (round) table.
Denote $Y$ - the random variable that equals the number of women sitting next to their husbands. Calculate $V(Y)$.
I tried figuring what is the probabilty that $i$ couples will sit next to each other, but I think this way requires the inclusion and exclusion principle, which I'm pretty sure unnecessary here.
I also tried marking $Y = Y_1 + Y_2 + ... + Y_{10}$ (Where $Y_i$ is 1 if the i-th couple sit next to one another, and 0 otherwise). But this brought up some trouble as well.
I guess I'm stuck more with the combinatorics part of the question, I was hoping you could help me out.
Thanks.
probability combinatorics
probability combinatorics
edited Jan 31 at 14:24
awkward
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
edited Jan 31 at 14:24
awkward
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
edited Jan 31 at 14:24
awkward
6,89011026
edited Jan 31 at 14:24
awkward
6,89011026
edited Jan 31 at 14:24
awkward
6,89011026
6,89011026
asked Jan 31 at 14:18
Barak BBarak B
265
asked Jan 31 at 14:18
Barak BBarak B
265
asked Jan 31 at 14:18
Barak BBarak B
265
265
$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
1
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
1
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
1
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09
|
show 9 more comments
$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
1
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
1
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
1
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09
$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
1
1
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
1
1
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
1
1
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09
|
show 9 more comments
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$begingroup$
I think indicator variables work better. To get the variance, you'll have to compute both $E[Y]$ and $E[Y^2]$. For the latter you'll need to work with terms of the form $E[Y_iY_j]$, but that's not that difficult.
$endgroup$
– lulu
Jan 31 at 14:30
$begingroup$
It gets hard since when calculation $P(Y_i = 1)$ I can't assume anything on the couples 1,2...,i-1, so there are some problematic spots there.
$endgroup$
– Barak B
Jan 31 at 14:36
1
$begingroup$
Please read up on Linearity of Expectation. The most important feature is that you can ignore dependency.
$endgroup$
– lulu
Jan 31 at 14:49
1
$begingroup$
I don't understand. What's $n$? And why would you imagine that $P(Y_i=1)$ depends on $i$? Isn't it clear that $P(Y_i=1)$ is independent of $i$?
$endgroup$
– lulu
Jan 31 at 15:00
1
$begingroup$
Note: When you deal with terms like $E[Y_1Y_2]$ then you will have to consider dependence since now you need to compute the probability that both of those couples are seated next to each other.
$endgroup$
– lulu
Jan 31 at 15:09