Finding Expectation












2












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Karn Watcharasupat
    Dec 24 '18 at 8:21










  • $begingroup$
    @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.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:23










  • $begingroup$
    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 $4$.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:30










  • $begingroup$
    Yes you are right, but still, how can we come to the right answer, be it less than 4?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:32
















2












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Karn Watcharasupat
    Dec 24 '18 at 8:21










  • $begingroup$
    @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.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:23










  • $begingroup$
    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 $4$.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:30










  • $begingroup$
    Yes you are right, but still, how can we come to the right answer, be it less than 4?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:32














2












2








2





$begingroup$


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









$endgroup$




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 Dec 24 '18 at 8:09









user601297user601297

41719




41719












  • $begingroup$
    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.
    $endgroup$
    – Karn Watcharasupat
    Dec 24 '18 at 8:21










  • $begingroup$
    @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.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:23










  • $begingroup$
    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 $4$.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:30










  • $begingroup$
    Yes you are right, but still, how can we come to the right answer, be it less than 4?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:32


















  • $begingroup$
    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.
    $endgroup$
    – Karn Watcharasupat
    Dec 24 '18 at 8:21










  • $begingroup$
    @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.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:23










  • $begingroup$
    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 $4$.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:30










  • $begingroup$
    Yes you are right, but still, how can we come to the right answer, be it less than 4?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:32
















$begingroup$
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.
$endgroup$
– Karn Watcharasupat
Dec 24 '18 at 8:21




$begingroup$
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.
$endgroup$
– Karn Watcharasupat
Dec 24 '18 at 8:21












$begingroup$
@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.
$endgroup$
– user601297
Dec 24 '18 at 8:23




$begingroup$
@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.
$endgroup$
– user601297
Dec 24 '18 at 8:23












$begingroup$
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 $4$.
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:30




$begingroup$
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 $4$.
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:30












$begingroup$
Yes you are right, but still, how can we come to the right answer, be it less than 4?
$endgroup$
– user601297
Dec 24 '18 at 8:32




$begingroup$
Yes you are right, but still, how can we come to the right answer, be it less than 4?
$endgroup$
– user601297
Dec 24 '18 at 8:32










2 Answers
2






active

oldest

votes


















2












$begingroup$

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











$endgroup$













  • $begingroup$
    The question asks for objects chosen by both A and B not by any one of them.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:28










  • $begingroup$
    @user601297 Thank you for attending me. I added something.
    $endgroup$
    – drhab
    Dec 24 '18 at 8:34










  • $begingroup$
    Can you tell me why is $P(U=1)=(4/12)^2$?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:38










  • $begingroup$
    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 $frac4{12}$
    $endgroup$
    – drhab
    Dec 24 '18 at 8:43



















1












$begingroup$

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











$endgroup$













  • $begingroup$
    The answer given is 7.6282 not 4/3, so even this isn’t correct.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:15










  • $begingroup$
    I see, I think i misinterpreted the quesiton then.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:15










  • $begingroup$
    I can't get the answer that you proposed.. hmmm...
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:28






  • 1




    $begingroup$
    +1 I have good trust that the proposed answer is wrong ;-).
    $endgroup$
    – drhab
    Dec 24 '18 at 8:39










  • $begingroup$
    it's either that or both of us misinterpreted something wrongly simultaneosly ;)
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:40












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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











$endgroup$













  • $begingroup$
    The question asks for objects chosen by both A and B not by any one of them.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:28










  • $begingroup$
    @user601297 Thank you for attending me. I added something.
    $endgroup$
    – drhab
    Dec 24 '18 at 8:34










  • $begingroup$
    Can you tell me why is $P(U=1)=(4/12)^2$?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:38










  • $begingroup$
    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 $frac4{12}$
    $endgroup$
    – drhab
    Dec 24 '18 at 8:43
















2












$begingroup$

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











$endgroup$













  • $begingroup$
    The question asks for objects chosen by both A and B not by any one of them.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:28










  • $begingroup$
    @user601297 Thank you for attending me. I added something.
    $endgroup$
    – drhab
    Dec 24 '18 at 8:34










  • $begingroup$
    Can you tell me why is $P(U=1)=(4/12)^2$?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:38










  • $begingroup$
    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 $frac4{12}$
    $endgroup$
    – drhab
    Dec 24 '18 at 8:43














2












2








2





$begingroup$

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











$endgroup$



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 Dec 24 '18 at 8:38

























answered Dec 24 '18 at 8:25









drhabdrhab

104k545136




104k545136












  • $begingroup$
    The question asks for objects chosen by both A and B not by any one of them.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:28










  • $begingroup$
    @user601297 Thank you for attending me. I added something.
    $endgroup$
    – drhab
    Dec 24 '18 at 8:34










  • $begingroup$
    Can you tell me why is $P(U=1)=(4/12)^2$?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:38










  • $begingroup$
    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 $frac4{12}$
    $endgroup$
    – drhab
    Dec 24 '18 at 8:43


















  • $begingroup$
    The question asks for objects chosen by both A and B not by any one of them.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:28










  • $begingroup$
    @user601297 Thank you for attending me. I added something.
    $endgroup$
    – drhab
    Dec 24 '18 at 8:34










  • $begingroup$
    Can you tell me why is $P(U=1)=(4/12)^2$?
    $endgroup$
    – user601297
    Dec 24 '18 at 8:38










  • $begingroup$
    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 $frac4{12}$
    $endgroup$
    – drhab
    Dec 24 '18 at 8:43
















$begingroup$
The question asks for objects chosen by both A and B not by any one of them.
$endgroup$
– user601297
Dec 24 '18 at 8:28




$begingroup$
The question asks for objects chosen by both A and B not by any one of them.
$endgroup$
– user601297
Dec 24 '18 at 8:28












$begingroup$
@user601297 Thank you for attending me. I added something.
$endgroup$
– drhab
Dec 24 '18 at 8:34




$begingroup$
@user601297 Thank you for attending me. I added something.
$endgroup$
– drhab
Dec 24 '18 at 8:34












$begingroup$
Can you tell me why is $P(U=1)=(4/12)^2$?
$endgroup$
– user601297
Dec 24 '18 at 8:38




$begingroup$
Can you tell me why is $P(U=1)=(4/12)^2$?
$endgroup$
– user601297
Dec 24 '18 at 8:38












$begingroup$
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 $frac4{12}$
$endgroup$
– drhab
Dec 24 '18 at 8:43




$begingroup$
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 $frac4{12}$
$endgroup$
– drhab
Dec 24 '18 at 8:43











1












$begingroup$

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











$endgroup$













  • $begingroup$
    The answer given is 7.6282 not 4/3, so even this isn’t correct.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:15










  • $begingroup$
    I see, I think i misinterpreted the quesiton then.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:15










  • $begingroup$
    I can't get the answer that you proposed.. hmmm...
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:28






  • 1




    $begingroup$
    +1 I have good trust that the proposed answer is wrong ;-).
    $endgroup$
    – drhab
    Dec 24 '18 at 8:39










  • $begingroup$
    it's either that or both of us misinterpreted something wrongly simultaneosly ;)
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:40
















1












$begingroup$

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











$endgroup$













  • $begingroup$
    The answer given is 7.6282 not 4/3, so even this isn’t correct.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:15










  • $begingroup$
    I see, I think i misinterpreted the quesiton then.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:15










  • $begingroup$
    I can't get the answer that you proposed.. hmmm...
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:28






  • 1




    $begingroup$
    +1 I have good trust that the proposed answer is wrong ;-).
    $endgroup$
    – drhab
    Dec 24 '18 at 8:39










  • $begingroup$
    it's either that or both of us misinterpreted something wrongly simultaneosly ;)
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:40














1












1








1





$begingroup$

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











$endgroup$



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 Dec 24 '18 at 8:34

























answered Dec 24 '18 at 8:13









Siong Thye GohSiong Thye Goh

103k1468120




103k1468120












  • $begingroup$
    The answer given is 7.6282 not 4/3, so even this isn’t correct.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:15










  • $begingroup$
    I see, I think i misinterpreted the quesiton then.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:15










  • $begingroup$
    I can't get the answer that you proposed.. hmmm...
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:28






  • 1




    $begingroup$
    +1 I have good trust that the proposed answer is wrong ;-).
    $endgroup$
    – drhab
    Dec 24 '18 at 8:39










  • $begingroup$
    it's either that or both of us misinterpreted something wrongly simultaneosly ;)
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:40


















  • $begingroup$
    The answer given is 7.6282 not 4/3, so even this isn’t correct.
    $endgroup$
    – user601297
    Dec 24 '18 at 8:15










  • $begingroup$
    I see, I think i misinterpreted the quesiton then.
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:15










  • $begingroup$
    I can't get the answer that you proposed.. hmmm...
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:28






  • 1




    $begingroup$
    +1 I have good trust that the proposed answer is wrong ;-).
    $endgroup$
    – drhab
    Dec 24 '18 at 8:39










  • $begingroup$
    it's either that or both of us misinterpreted something wrongly simultaneosly ;)
    $endgroup$
    – Siong Thye Goh
    Dec 24 '18 at 8:40
















$begingroup$
The answer given is 7.6282 not 4/3, so even this isn’t correct.
$endgroup$
– user601297
Dec 24 '18 at 8:15




$begingroup$
The answer given is 7.6282 not 4/3, so even this isn’t correct.
$endgroup$
– user601297
Dec 24 '18 at 8:15












$begingroup$
I see, I think i misinterpreted the quesiton then.
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:15




$begingroup$
I see, I think i misinterpreted the quesiton then.
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:15












$begingroup$
I can't get the answer that you proposed.. hmmm...
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:28




$begingroup$
I can't get the answer that you proposed.. hmmm...
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:28




1




1




$begingroup$
+1 I have good trust that the proposed answer is wrong ;-).
$endgroup$
– drhab
Dec 24 '18 at 8:39




$begingroup$
+1 I have good trust that the proposed answer is wrong ;-).
$endgroup$
– drhab
Dec 24 '18 at 8:39












$begingroup$
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:40




$begingroup$
it's either that or both of us misinterpreted something wrongly simultaneosly ;)
$endgroup$
– Siong Thye Goh
Dec 24 '18 at 8:40


















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