Mixing
Probability - Expectation value and Covariance
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A group of $n = 10$ men exchange their gloves randomly (altogether there are 20 gloves).
Let $X_i$ be the random variable which attains the value 1 if the $i-th$ man got at least one of his gloves back and it attains the value 0 otherwise. Set $p = E(X_i)$ and $σ^2 = Var(X_i)$.
Observe that $p = frac{37}{190}$ and that $Var (X_i) = frac{5661}{190^2}$.
Let $b$ be the probability that for $ine j$ both the $i-th$ and $j-th$ men got at least one of their gloves back.
Show that $b = frac {1157}{{20 choose 2}{18 choose 2}}$.
1.) Let $X$ be the total number of men who received at least one of their gloves. Find $E(X)$.
2.) Calculate $Var(X).$ You may write it as an expression in $p$, $σ^2$ and $b.$
At first I just thought that $E(X)=np=10frac{37}{190}$ and then $Var(X)=npq=10cdot 37cdotfrac{153}{190^2}$ but now I am not sure about that, as the events do intersect: if someone takes gloves of someone else, then someone else has nothing to choose from. However, I do not know how to count these. Could you help me, please? Thanks
probability probability-distributions variance expected-value
asked 18 hours ago
Daniel
63
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A group of $n = 10$ men exchange their gloves randomly (altogether there are 20 gloves).
Let $X_i$ be the random variable which attains the value 1 if the $i-th$ man got at least one of his gloves back and it attains the value 0 otherwise. Set $p = E(X_i)$ and $σ^2 = Var(X_i)$.
Observe that $p = frac{37}{190}$ and that $Var (X_i) = frac{5661}{190^2}$.
Let $b$ be the probability that for $ine j$ both the $i-th$ and $j-th$ men got at least one of their gloves back.
Show that $b = frac {1157}{{20 choose 2}{18 choose 2}}$.
1.) Let $X$ be the total number of men who received at least one of their gloves. Find $E(X)$.
2.) Calculate $Var(X).$ You may write it as an expression in $p$, $σ^2$ and $b.$
At first I just thought that $E(X)=np=10frac{37}{190}$ and then $Var(X)=npq=10cdot 37cdotfrac{153}{190^2}$ but now I am not sure about that, as the events do intersect: if someone takes gloves of someone else, then someone else has nothing to choose from. However, I do not know how to count these. Could you help me, please? Thanks
probability probability-distributions variance expected-value
asked 18 hours ago
Daniel
63
What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A group of $n = 10$ men exchange their gloves randomly (altogether there are 20 gloves).
Let $X_i$ be the random variable which attains the value 1 if the $i-th$ man got at least one of his gloves back and it attains the value 0 otherwise. Set $p = E(X_i)$ and $σ^2 = Var(X_i)$.
Observe that $p = frac{37}{190}$ and that $Var (X_i) = frac{5661}{190^2}$.
Let $b$ be the probability that for $ine j$ both the $i-th$ and $j-th$ men got at least one of their gloves back.
Show that $b = frac {1157}{{20 choose 2}{18 choose 2}}$.
1.) Let $X$ be the total number of men who received at least one of their gloves. Find $E(X)$.
2.) Calculate $Var(X).$ You may write it as an expression in $p$, $σ^2$ and $b.$
At first I just thought that $E(X)=np=10frac{37}{190}$ and then $Var(X)=npq=10cdot 37cdotfrac{153}{190^2}$ but now I am not sure about that, as the events do intersect: if someone takes gloves of someone else, then someone else has nothing to choose from. However, I do not know how to count these. Could you help me, please? Thanks
probability probability-distributions variance expected-value
asked 18 hours ago
Daniel
63
A group of $n = 10$ men exchange their gloves randomly (altogether there are 20 gloves).
Let $X_i$ be the random variable which attains the value 1 if the $i-th$ man got at least one of his gloves back and it attains the value 0 otherwise. Set $p = E(X_i)$ and $σ^2 = Var(X_i)$.
Observe that $p = frac{37}{190}$ and that $Var (X_i) = frac{5661}{190^2}$.
Let $b$ be the probability that for $ine j$ both the $i-th$ and $j-th$ men got at least one of their gloves back.
Show that $b = frac {1157}{{20 choose 2}{18 choose 2}}$.
1.) Let $X$ be the total number of men who received at least one of their gloves. Find $E(X)$.
2.) Calculate $Var(X).$ You may write it as an expression in $p$, $σ^2$ and $b.$
At first I just thought that $E(X)=np=10frac{37}{190}$ and then $Var(X)=npq=10cdot 37cdotfrac{153}{190^2}$ but now I am not sure about that, as the events do intersect: if someone takes gloves of someone else, then someone else has nothing to choose from. However, I do not know how to count these. Could you help me, please? Thanks
probability probability-distributions variance expected-value
probability probability-distributions variance expected-value
asked 18 hours ago
Daniel
63
asked 18 hours ago
Daniel
63
edited 17 hours ago
asked 18 hours ago
Daniel
63
asked 18 hours ago
Daniel
63
asked 18 hours ago
Daniel
63
63
What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago
add a comment |
What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago
What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago
What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago
add a comment |
1 Answer
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Your calculations of $p$, $sigma^2$ and $mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$mathsf{Var}(X)=mathsf{Cov}(sum_{i=1}^{10}X_i,sum_{j=1}^{10}X_j)=sum_{i=1}^{10}sum_{j=1}^{10}mathsf{Cov}(X_i,X_j)=10mathsf{Var}(X_1)+90mathsf{Cov}(X_1,X_2)=$$$$10sigma^2+90(mathbb EX_1X_2-mathbb EX_1mathbb EX_2)=10sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that
belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong
to person 2.
Then:
$$Pleft(X_{1}=X_{2}=1right)=$$$$Pleft(X_{2}=1mid Z_{1}=2right)Pleft(Z_{1}=2right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)Pleft(Z_{1}=1wedge Y_{1}=0right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)Pleft(Z_{1}=1wedge Y_{1}=1right)$$
This with
$Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$
$Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{18}{19}+frac{18}{20}frac{2}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$
answered 17 hours ago
drhab
94.2k543125
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your calculations of $p$, $sigma^2$ and $mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$mathsf{Var}(X)=mathsf{Cov}(sum_{i=1}^{10}X_i,sum_{j=1}^{10}X_j)=sum_{i=1}^{10}sum_{j=1}^{10}mathsf{Cov}(X_i,X_j)=10mathsf{Var}(X_1)+90mathsf{Cov}(X_1,X_2)=$$$$10sigma^2+90(mathbb EX_1X_2-mathbb EX_1mathbb EX_2)=10sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that
belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong
to person 2.
Then:
$$Pleft(X_{1}=X_{2}=1right)=$$$$Pleft(X_{2}=1mid Z_{1}=2right)Pleft(Z_{1}=2right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)Pleft(Z_{1}=1wedge Y_{1}=0right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)Pleft(Z_{1}=1wedge Y_{1}=1right)$$
This with
$Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$
$Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{18}{19}+frac{18}{20}frac{2}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$
answered 17 hours ago
drhab
94.2k543125
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
add a comment |
up vote
0
down vote
Your calculations of $p$, $sigma^2$ and $mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$mathsf{Var}(X)=mathsf{Cov}(sum_{i=1}^{10}X_i,sum_{j=1}^{10}X_j)=sum_{i=1}^{10}sum_{j=1}^{10}mathsf{Cov}(X_i,X_j)=10mathsf{Var}(X_1)+90mathsf{Cov}(X_1,X_2)=$$$$10sigma^2+90(mathbb EX_1X_2-mathbb EX_1mathbb EX_2)=10sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that
belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong
to person 2.
Then:
$$Pleft(X_{1}=X_{2}=1right)=$$$$Pleft(X_{2}=1mid Z_{1}=2right)Pleft(Z_{1}=2right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)Pleft(Z_{1}=1wedge Y_{1}=0right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)Pleft(Z_{1}=1wedge Y_{1}=1right)$$
This with
$Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$
$Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{18}{19}+frac{18}{20}frac{2}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$
answered 17 hours ago
drhab
94.2k543125
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Your calculations of $p$, $sigma^2$ and $mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$mathsf{Var}(X)=mathsf{Cov}(sum_{i=1}^{10}X_i,sum_{j=1}^{10}X_j)=sum_{i=1}^{10}sum_{j=1}^{10}mathsf{Cov}(X_i,X_j)=10mathsf{Var}(X_1)+90mathsf{Cov}(X_1,X_2)=$$$$10sigma^2+90(mathbb EX_1X_2-mathbb EX_1mathbb EX_2)=10sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that
belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong
to person 2.
Then:
$$Pleft(X_{1}=X_{2}=1right)=$$$$Pleft(X_{2}=1mid Z_{1}=2right)Pleft(Z_{1}=2right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)Pleft(Z_{1}=1wedge Y_{1}=0right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)Pleft(Z_{1}=1wedge Y_{1}=1right)$$
This with
$Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$
$Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{18}{19}+frac{18}{20}frac{2}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$
answered 17 hours ago
drhab
94.2k543125
Your calculations of $p$, $sigma^2$ and $mathbb EX$ are okay.
Covariance is bilinear, so that by symmetry:
$$mathsf{Var}(X)=mathsf{Cov}(sum_{i=1}^{10}X_i,sum_{j=1}^{10}X_j)=sum_{i=1}^{10}sum_{j=1}^{10}mathsf{Cov}(X_i,X_j)=10mathsf{Var}(X_1)+90mathsf{Cov}(X_1,X_2)=$$$$10sigma^2+90(mathbb EX_1X_2-mathbb EX_1mathbb EX_2)=10sigma^{10}+90(P(X_1=X_2=1)-P(X_1=1)P(X_2=1))=$$$$10sigma^2+90(b-p^2)$$
Here $b=P(X_1=X_2=1)$ so it remains to find $P(X_1=X_2=1)$
edit:
Let $Z_{1}$ denote the number of gloves taken by person 1 that
belong to person 1.
Let $Y_{1}$ denote the number of gloves taken by person 1 that belong
to person 2.
Then:
$$Pleft(X_{1}=X_{2}=1right)=$$$$Pleft(X_{2}=1mid Z_{1}=2right)Pleft(Z_{1}=2right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)Pleft(Z_{1}=1wedge Y_{1}=0right)+$$$$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)Pleft(Z_{1}=1wedge Y_{1}=1right)$$
This with
$Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$
$Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$
$Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{18}{19}+frac{18}{20}frac{2}{19}$
$Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$
answered 17 hours ago
drhab
94.2k543125
edited 10 hours ago
answered 17 hours ago
drhab
94.2k543125
answered 17 hours ago
drhab
94.2k543125
answered 17 hours ago
drhab
94.2k543125
94.2k543125
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
add a comment |
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
So than P(X1=1|X2=1) is equal to probability of intersection of these events divided by p. But isn't the probability of intersection defined as b=1157/(190*153)?
– Daniel
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I have made an edit (because finding $P(X_1=2mid X_1=1)$ was a bit more complicated than I thought). Does this help?
– drhab
16 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
I still do not understand it completely. I counted this: P(Z=1)=36/190 and P(Z=2)=1/190. P(X2=1∣Z=2)=(2*16+1)/153 But P(X2=1∣Z=1) makes me problem. I think that it is similar to problem above, just there are four cases when the first man picked one his glove and one glove of the second man, so that it is equal to 33-4=29. But then I got in total less than 1157. Could you tell me what is wrong, please?
– Daniel
12 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Again I underestimated the complexity. I edited to provide a way to calculate $P(X_1=X_2=1)$. Do check me on mistakes, though.
– drhab
10 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
Thank you very much! I calculated it and I got 1289/((20C2)(18C2)). But it should be 1157/((20C2)(18C2)), according to the task. Do you think it might be just wrong?
– Daniel
9 hours ago
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What's the definition of $b$? The expectation is correct, as it's just linearity. For the variance, consider the formula using the covariance, as it's not clear if the $X_i$ are independent
– Stockfish
17 hours ago