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










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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
    Nov 19 at 10:59















up vote
0
down vote

favorite
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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










share|cite|improve this question
























  • 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
    Nov 19 at 10:59













up vote
0
down vote

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up vote
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down vote

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1





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










share|cite|improve this question















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






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edited Nov 19 at 11:36

























asked Nov 19 at 10:27









Daniel

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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
    Nov 19 at 10:59


















  • 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
    Nov 19 at 10:59
















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
Nov 19 at 10:59




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
Nov 19 at 10:59










1 Answer
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accepted










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




  • $Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$


  • $Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{16}{19}+frac{16}{20}frac{2}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$







share|cite|improve this answer























  • 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
    Nov 19 at 12:06












  • 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
    Nov 19 at 12:41










  • 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
    Nov 19 at 15:57










  • 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
    Nov 19 at 18:16












  • 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
    Nov 19 at 19:12











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1 Answer
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up vote
0
down vote



accepted










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




  • $Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$


  • $Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{16}{19}+frac{16}{20}frac{2}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$







share|cite|improve this answer























  • 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
    Nov 19 at 12:06












  • 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
    Nov 19 at 12:41










  • 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
    Nov 19 at 15:57










  • 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
    Nov 19 at 18:16












  • 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
    Nov 19 at 19:12















up vote
0
down vote



accepted










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




  • $Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$


  • $Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{16}{19}+frac{16}{20}frac{2}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$







share|cite|improve this answer























  • 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
    Nov 19 at 12:06












  • 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
    Nov 19 at 12:41










  • 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
    Nov 19 at 15:57










  • 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
    Nov 19 at 18:16












  • 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
    Nov 19 at 19:12













up vote
0
down vote



accepted







up vote
0
down vote



accepted






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




  • $Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$


  • $Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{16}{19}+frac{16}{20}frac{2}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$







share|cite|improve this answer














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




  • $Pleft(X_{2}=1mid Z_{1}=2right)=1-Pleft(X_{2}=0mid Z_{1}=2right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=0right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=0right)=1-frac{16}{18}frac{15}{17}$


  • $Pleft(X_{2}=1mid Z_{1}=1wedge Y_{1}=1right)=1-Pleft(X_{2}=0mid Z_{1}=1wedge Y_{1}=1right)=1-frac{17}{18}frac{16}{17}$


  • $Pleft(Z_{1}=2right)=frac{2}{20}frac{1}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=0right)=frac{2}{20}frac{16}{19}+frac{16}{20}frac{2}{19}$


  • $Pleft(Z_{1}=1wedge Y_{1}=1right)=frac{2}{20}frac{2}{19}+frac{2}{20}frac{2}{19}$








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edited Nov 20 at 9:39

























answered Nov 19 at 11:37









drhab

95.1k543126




95.1k543126












  • 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
    Nov 19 at 12:06












  • 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
    Nov 19 at 12:41










  • 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
    Nov 19 at 15:57










  • 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
    Nov 19 at 18:16












  • 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
    Nov 19 at 19:12


















  • 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
    Nov 19 at 12:06












  • 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
    Nov 19 at 12:41










  • 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
    Nov 19 at 15:57










  • 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
    Nov 19 at 18:16












  • 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
    Nov 19 at 19:12
















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
Nov 19 at 12:06






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
Nov 19 at 12:06














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
Nov 19 at 12:41




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
Nov 19 at 12:41












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
Nov 19 at 15:57




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
Nov 19 at 15:57












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
Nov 19 at 18:16






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
Nov 19 at 18:16














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
Nov 19 at 19:12




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
Nov 19 at 19:12


















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