Finding Expectation












2














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







share|cite|improve this question






















  • 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
















2














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







share|cite|improve this question






















  • 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














2












2








2







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










2 Answers
2






active

oldest

votes


















2














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.






share|cite|improve this answer























  • 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



















1














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



$$12 left( frac13 right)^2=frac43$$






share|cite|improve this answer























  • 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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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.






share|cite|improve this answer























  • 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
















2














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.






share|cite|improve this answer























  • 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














2












2








2






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























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


















  • 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











1














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



$$12 left( frac13 right)^2=frac43$$






share|cite|improve this answer























  • 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
















1














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



$$12 left( frac13 right)^2=frac43$$






share|cite|improve this answer























  • 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














1












1








1






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



$$12 left( frac13 right)^2=frac43$$






share|cite|improve this answer














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



$$12 left( frac13 right)^2=frac43$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























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


















  • 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


















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