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 hours ago
user601297
1105
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 hours ago
user601297
1105
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 hours 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 hours 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
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour 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 hours ago
user601297
1105
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 hours ago
user601297
1105
asked 2 hours ago
user601297
1105
asked 2 hours ago
user601297
1105
asked 2 hours ago
user601297
1105
asked 2 hours ago
user601297
1105
1105
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 hours 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 hours 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
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour 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 hours 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 hours 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
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour 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 hours 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 hours 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 hours 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 hours 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
1 hour 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
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour 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 hours ago
drhab
97.1k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour 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 hours ago
Siong Thye Goh
98.4k1463116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour ago
add a comment |
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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 hours ago
drhab
97.1k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour 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 hours ago
drhab
97.1k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour 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 hours ago
drhab
97.1k544128
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 hours ago
drhab
97.1k544128
edited 1 hour ago
answered 2 hours ago
drhab
97.1k544128
answered 2 hours ago
drhab
97.1k544128
answered 2 hours ago
drhab
97.1k544128
97.1k544128
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour ago
add a comment |
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour ago
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
The question asks for objects chosen by both A and B not by any one of them.
– user601297
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
@user601297 Thank you for attending me. I added something.
– drhab
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour ago
Can you tell me why is $P(U=1)=(4/12)^2$?
– user601297
1 hour 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
1 hour 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
1 hour 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 hours ago
Siong Thye Goh
98.4k1463116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour 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 hours ago
Siong Thye Goh
98.4k1463116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour 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 hours ago
Siong Thye Goh
98.4k1463116
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 hours ago
Siong Thye Goh
98.4k1463116
edited 1 hour ago
answered 2 hours ago
Siong Thye Goh
98.4k1463116
answered 2 hours ago
Siong Thye Goh
98.4k1463116
answered 2 hours ago
Siong Thye Goh
98.4k1463116
98.4k1463116
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour ago
add a comment |
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour ago
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
The answer given is 7.6282 not 4/3, so even this isn’t correct.
– user601297
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I see, I think i misinterpreted the quesiton then.
– Siong Thye Goh
2 hours ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
I can't get the answer that you proposed.. hmmm...
– Siong Thye Goh
1 hour ago
1
1
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
+1 I have good trust that the proposed answer is wrong ;-).
– drhab
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour ago
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
– Siong Thye Goh
1 hour ago
add a comment |
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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 hours 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 hours 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
1 hour ago
Yes you are right, but still, how can we come to the right answer, be it less than 4?
– user601297
1 hour ago