Probability - Expectation value and Covariance











up vote
0
down vote

favorite
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
























  • 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
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
























  • 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
1









up vote
0
down vote

favorite
1






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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 11:36

























asked Nov 19 at 10:27









Daniel

83




83












  • 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
1






active

oldest

votes

















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004761%2fprobability-expectation-value-and-covariance%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








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




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


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


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


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


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


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








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 at 9:39

























answered Nov 19 at 11:37









drhab

95.7k543126




95.7k543126












  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004761%2fprobability-expectation-value-and-covariance%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

ComboBox Display Member on multiple fields

Is it possible to collect Nectar points via Trainline?