Finding Expectation
Suppose that A and B each randomly and independently choose 4 out of 12 objects. Find the expected number of objects chosen by both A and B.
My attempt:
$$X_i = 1$$ when ith is chosen by A
$$X_i = 0$$ otherwise
Similarly I define another indicator $Y_i$ for person B which says the exact same thing as $X_i$ but for B.
Now, $$Z=sum(X_iY_i)$$ $i=0,dots12$
$$mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{1}{12}$$
And $$mathbb{E}[X_iY_i] = 1/12^2$$
And so the expected number chosen by both should be $$frac{12}{12^2} = frac{1}{12}$$
However this is not the correct answer, so can someone give me the solution and tell me where I went wrong.
probability random-variables expected-value
asked 2 days ago
user601297
1176
add a comment |
Suppose that A and B each randomly and independently choose 4 out of 12 objects. Find the expected number of objects chosen by both A and B.
My attempt:
$$X_i = 1$$ when ith is chosen by A
$$X_i = 0$$ otherwise
Similarly I define another indicator $Y_i$ for person B which says the exact same thing as $X_i$ but for B.
Now, $$Z=sum(X_iY_i)$$ $i=0,dots12$
$$mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{1}{12}$$
And $$mathbb{E}[X_iY_i] = 1/12^2$$
And so the expected number chosen by both should be $$frac{12}{12^2} = frac{1}{12}$$
However this is not the correct answer, so can someone give me the solution and tell me where I went wrong.
probability random-variables expected-value
asked 2 days ago
user601297
1176
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago
add a comment |
Suppose that A and B each randomly and independently choose 4 out of 12 objects. Find the expected number of objects chosen by both A and B.
My attempt:
$$X_i = 1$$ when ith is chosen by A
$$X_i = 0$$ otherwise
Similarly I define another indicator $Y_i$ for person B which says the exact same thing as $X_i$ but for B.
Now, $$Z=sum(X_iY_i)$$ $i=0,dots12$
$$mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{1}{12}$$
And $$mathbb{E}[X_iY_i] = 1/12^2$$
And so the expected number chosen by both should be $$frac{12}{12^2} = frac{1}{12}$$
However this is not the correct answer, so can someone give me the solution and tell me where I went wrong.
probability random-variables expected-value
asked 2 days ago
user601297
1176
Suppose that A and B each randomly and independently choose 4 out of 12 objects. Find the expected number of objects chosen by both A and B.
My attempt:
$$X_i = 1$$ when ith is chosen by A
$$X_i = 0$$ otherwise
Similarly I define another indicator $Y_i$ for person B which says the exact same thing as $X_i$ but for B.
Now, $$Z=sum(X_iY_i)$$ $i=0,dots12$
$$mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{1}{12}$$
And $$mathbb{E}[X_iY_i] = 1/12^2$$
And so the expected number chosen by both should be $$frac{12}{12^2} = frac{1}{12}$$
However this is not the correct answer, so can someone give me the solution and tell me where I went wrong.
probability random-variables expected-value
probability random-variables expected-value
asked 2 days ago
user601297
1176
asked 2 days ago
user601297
1176
asked 2 days ago
user601297
1176
asked 2 days ago
user601297
1176
asked 2 days ago
user601297
1176
1176
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago
add a comment |
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$mathbb EZ=12mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1text{ is chosen by }Atext{ or } 1text{ is chosen by }B)=$$$$P(1text{ is chosen by }A)+P(1text{ is chosen by }B)-P(1text{ is chosen by }Atext{ and } B)=$$$$frac13+frac13-frac13frac13=frac59$$so the final answer is: $$12frac59=frac{20}3$$
edit: (I was attended on a misinterpretation of your question)
Let $U_i$ take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$mathbb EU=12mathbb EU_1=12P(U_1=1)=12frac19=frac43$$
Observe that: $$4+4=mathbb EZ+mathbb EU$$as it should.
answered 2 days ago
drhab
97.4k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
add a comment |
Let's compute the number of items not chosen by any of them.
The number of items not chosen by either of them would be
$$12left(frac{8}{12}right)^2= 12left( frac{2}{3}right)^2$$
Hence the number of item chosen by at least one of them is
$$12 left( 1-frac{4}{9}right)= 12left( frac59 right)= frac{20}3$$
Edit:
The expected number of items that are chosen by both of them would be
$$12 left( frac13 right)^2=frac43$$
answered 2 days ago
Siong Thye Goh
98.9k1464116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
add a comment |
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',
autoActivateHeartbeat: false,
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051052%2ffinding-expectation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$mathbb EZ=12mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1text{ is chosen by }Atext{ or } 1text{ is chosen by }B)=$$$$P(1text{ is chosen by }A)+P(1text{ is chosen by }B)-P(1text{ is chosen by }Atext{ and } B)=$$$$frac13+frac13-frac13frac13=frac59$$so the final answer is: $$12frac59=frac{20}3$$
edit: (I was attended on a misinterpretation of your question)
Let $U_i$ take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$mathbb EU=12mathbb EU_1=12P(U_1=1)=12frac19=frac43$$
Observe that: $$4+4=mathbb EZ+mathbb EU$$as it should.
answered 2 days ago
drhab
97.4k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
add a comment |
Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$mathbb EZ=12mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1text{ is chosen by }Atext{ or } 1text{ is chosen by }B)=$$$$P(1text{ is chosen by }A)+P(1text{ is chosen by }B)-P(1text{ is chosen by }Atext{ and } B)=$$$$frac13+frac13-frac13frac13=frac59$$so the final answer is: $$12frac59=frac{20}3$$
edit: (I was attended on a misinterpretation of your question)
Let $U_i$ take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$mathbb EU=12mathbb EU_1=12P(U_1=1)=12frac19=frac43$$
Observe that: $$4+4=mathbb EZ+mathbb EU$$as it should.
answered 2 days ago
drhab
97.4k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
add a comment |
Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$mathbb EZ=12mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1text{ is chosen by }Atext{ or } 1text{ is chosen by }B)=$$$$P(1text{ is chosen by }A)+P(1text{ is chosen by }B)-P(1text{ is chosen by }Atext{ and } B)=$$$$frac13+frac13-frac13frac13=frac59$$so the final answer is: $$12frac59=frac{20}3$$
edit: (I was attended on a misinterpretation of your question)
Let $U_i$ take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$mathbb EU=12mathbb EU_1=12P(U_1=1)=12frac19=frac43$$
Observe that: $$4+4=mathbb EZ+mathbb EU$$as it should.
answered 2 days ago
drhab
97.4k544128
Let $Z_i$ take value $1$ if object $i$ is chosen by $A$ or by $B$.
Then $$Z=Z_1+cdots+Z_{12}$$is the number of objects chosen by $A$ or by $B$.
With linearity of expectation and symmetry we find:$$mathbb EZ=12mathbb EZ_1=12P(Z_1=1)$$
Here $$P(Z_1=1)=$$$$P(1text{ is chosen by }Atext{ or } 1text{ is chosen by }B)=$$$$P(1text{ is chosen by }A)+P(1text{ is chosen by }B)-P(1text{ is chosen by }Atext{ and } B)=$$$$frac13+frac13-frac13frac13=frac59$$so the final answer is: $$12frac59=frac{20}3$$
edit: (I was attended on a misinterpretation of your question)
Let $U_i$ take value $1$ if object $i$ is chosen by $A$ and $B$.
Then $$U=U_1+cdots+U_{12}$$is the number of objects chosen by $A$ and by $B$.
With linearity of expectation and symmetry we find:$$mathbb EU=12mathbb EU_1=12P(U_1=1)=12frac19=frac43$$
Observe that: $$4+4=mathbb EZ+mathbb EU$$as it should.
answered 2 days ago
drhab
97.4k544128
edited 2 days ago
answered 2 days ago
drhab
97.4k544128
answered 2 days ago
drhab
97.4k544128
answered 2 days ago
drhab
97.4k544128
97.4k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
add a comment |
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
The question asks for objects chosen by both A and B not by any one of them.
– user601297
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
@user601297 Thank you for attending me. I added something.
– drhab
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
Not $P(U=1)=frac19$ but $P(U_1=1)=frac19$. Two independent events take place: $4$ object are chosen out of $12$ by $A$ and $4$ object are chosen out of $12$ by $B$. In both cases the probability that object $1$ will be among the chosen objects is $frac4{12}$
– drhab
2 days ago
add a comment |
Let's compute the number of items not chosen by any of them.
The number of items not chosen by either of them would be
$$12left(frac{8}{12}right)^2= 12left( frac{2}{3}right)^2$$
Hence the number of item chosen by at least one of them is
$$12 left( 1-frac{4}{9}right)= 12left( frac59 right)= frac{20}3$$
Edit:
The expected number of items that are chosen by both of them would be
$$12 left( frac13 right)^2=frac43$$
answered 2 days ago
Siong Thye Goh
98.9k1464116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
add a comment |
Let's compute the number of items not chosen by any of them.
The number of items not chosen by either of them would be
$$12left(frac{8}{12}right)^2= 12left( frac{2}{3}right)^2$$
Hence the number of item chosen by at least one of them is
$$12 left( 1-frac{4}{9}right)= 12left( frac59 right)= frac{20}3$$
Edit:
The expected number of items that are chosen by both of them would be
$$12 left( frac13 right)^2=frac43$$
answered 2 days ago
Siong Thye Goh
98.9k1464116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
add a comment |
Let's compute the number of items not chosen by any of them.
The number of items not chosen by either of them would be
$$12left(frac{8}{12}right)^2= 12left( frac{2}{3}right)^2$$
Hence the number of item chosen by at least one of them is
$$12 left( 1-frac{4}{9}right)= 12left( frac59 right)= frac{20}3$$
Edit:
The expected number of items that are chosen by both of them would be
$$12 left( frac13 right)^2=frac43$$
answered 2 days ago
Siong Thye Goh
98.9k1464116
Let's compute the number of items not chosen by any of them.
The number of items not chosen by either of them would be
$$12left(frac{8}{12}right)^2= 12left( frac{2}{3}right)^2$$
Hence the number of item chosen by at least one of them is
$$12 left( 1-frac{4}{9}right)= 12left( frac59 right)= frac{20}3$$
Edit:
The expected number of items that are chosen by both of them would be
$$12 left( frac13 right)^2=frac43$$
answered 2 days ago
Siong Thye Goh
98.9k1464116
edited 2 days ago
answered 2 days ago
Siong Thye Goh
98.9k1464116
answered 2 days ago
Siong Thye Goh
98.9k1464116
answered 2 days ago
Siong Thye Goh
98.9k1464116
98.9k1464116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
add a comment |
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
2 days ago
1
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
2 days ago
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051052%2ffinding-expectation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
Shouldn't $mathbb{E}(X_i)=mathbb{E}(Y_i)= frac{4}{12} = frac{1}{3}$? since 4 out of 12 is being chosen each time.
– Karn Watcharasupat
2 days ago
@KarnWatcharasupat $mathbb{E}(X_i)$ is defined as the probability of ith object being chosen so that is why i think it’ll be 1/12.
– user601297
2 days ago
note that if the question is asking for items that are chosen by both people simultaneously, the answer can't exceed $4$ but your proposed answer is bigger than $4$.
– Siong Thye Goh
2 days ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
2 days ago