Discrete Probability: Expected Value and Random Variable independence [closed]












0












$begingroup$


enter image description here



For this, I took n=2 which makes the set: {1,2,3,4}



Set will contain: {C1,C2,B1,B2}



X = 1 if the position of first cider bottle is 1



P(X=1) = 6/24 = 1/4



E(X) = 2 * 1/4 = 1/2



The general form will be: n*1/2n = 1/n.



This is my attempt, I'm not sure if I'm correct on this.



For this question:




You roll a fair die repeatedly and independently until the result is an
even number. Define the random variables
X = the number of times you roll the die
and
Y = the result of the last roll.
For example, if the results of the rolls are 5; 1; 3; 3; 5; 2, then X = 6 and
Y = 2.
Prove that the random variables X and Y are independent.




I defined X = 1 if number of times roll die is 1 time



and Y =1 if result of last roll is even



So, Pr(X) = 3/6 = 1/2 = Pr(Y)



Pr(X and Y) = 1/2



This gives me 1/2 = 1/4 which is not independent but the question is asking to prove independence










share|cite|improve this question









$endgroup$



closed as unclear what you're asking by Did, NCh, Leucippus, Tianlalu, KReiser Dec 19 '18 at 7:10


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    Please don't ask multiple unrelated questions in one post.
    $endgroup$
    – Bungo
    Dec 17 '18 at 6:22
















0












$begingroup$


enter image description here



For this, I took n=2 which makes the set: {1,2,3,4}



Set will contain: {C1,C2,B1,B2}



X = 1 if the position of first cider bottle is 1



P(X=1) = 6/24 = 1/4



E(X) = 2 * 1/4 = 1/2



The general form will be: n*1/2n = 1/n.



This is my attempt, I'm not sure if I'm correct on this.



For this question:




You roll a fair die repeatedly and independently until the result is an
even number. Define the random variables
X = the number of times you roll the die
and
Y = the result of the last roll.
For example, if the results of the rolls are 5; 1; 3; 3; 5; 2, then X = 6 and
Y = 2.
Prove that the random variables X and Y are independent.




I defined X = 1 if number of times roll die is 1 time



and Y =1 if result of last roll is even



So, Pr(X) = 3/6 = 1/2 = Pr(Y)



Pr(X and Y) = 1/2



This gives me 1/2 = 1/4 which is not independent but the question is asking to prove independence










share|cite|improve this question









$endgroup$



closed as unclear what you're asking by Did, NCh, Leucippus, Tianlalu, KReiser Dec 19 '18 at 7:10


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • 1




    $begingroup$
    Please don't ask multiple unrelated questions in one post.
    $endgroup$
    – Bungo
    Dec 17 '18 at 6:22














0












0








0





$begingroup$


enter image description here



For this, I took n=2 which makes the set: {1,2,3,4}



Set will contain: {C1,C2,B1,B2}



X = 1 if the position of first cider bottle is 1



P(X=1) = 6/24 = 1/4



E(X) = 2 * 1/4 = 1/2



The general form will be: n*1/2n = 1/n.



This is my attempt, I'm not sure if I'm correct on this.



For this question:




You roll a fair die repeatedly and independently until the result is an
even number. Define the random variables
X = the number of times you roll the die
and
Y = the result of the last roll.
For example, if the results of the rolls are 5; 1; 3; 3; 5; 2, then X = 6 and
Y = 2.
Prove that the random variables X and Y are independent.




I defined X = 1 if number of times roll die is 1 time



and Y =1 if result of last roll is even



So, Pr(X) = 3/6 = 1/2 = Pr(Y)



Pr(X and Y) = 1/2



This gives me 1/2 = 1/4 which is not independent but the question is asking to prove independence










share|cite|improve this question









$endgroup$




enter image description here



For this, I took n=2 which makes the set: {1,2,3,4}



Set will contain: {C1,C2,B1,B2}



X = 1 if the position of first cider bottle is 1



P(X=1) = 6/24 = 1/4



E(X) = 2 * 1/4 = 1/2



The general form will be: n*1/2n = 1/n.



This is my attempt, I'm not sure if I'm correct on this.



For this question:




You roll a fair die repeatedly and independently until the result is an
even number. Define the random variables
X = the number of times you roll the die
and
Y = the result of the last roll.
For example, if the results of the rolls are 5; 1; 3; 3; 5; 2, then X = 6 and
Y = 2.
Prove that the random variables X and Y are independent.




I defined X = 1 if number of times roll die is 1 time



and Y =1 if result of last roll is even



So, Pr(X) = 3/6 = 1/2 = Pr(Y)



Pr(X and Y) = 1/2



This gives me 1/2 = 1/4 which is not independent but the question is asking to prove independence







probability probability-theory discrete-mathematics 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 17 '18 at 6:18









TobyToby

1577




1577




closed as unclear what you're asking by Did, NCh, Leucippus, Tianlalu, KReiser Dec 19 '18 at 7:10


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Did, NCh, Leucippus, Tianlalu, KReiser Dec 19 '18 at 7:10


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $begingroup$
    Please don't ask multiple unrelated questions in one post.
    $endgroup$
    – Bungo
    Dec 17 '18 at 6:22














  • 1




    $begingroup$
    Please don't ask multiple unrelated questions in one post.
    $endgroup$
    – Bungo
    Dec 17 '18 at 6:22








1




1




$begingroup$
Please don't ask multiple unrelated questions in one post.
$endgroup$
– Bungo
Dec 17 '18 at 6:22




$begingroup$
Please don't ask multiple unrelated questions in one post.
$endgroup$
– Bungo
Dec 17 '18 at 6:22










2 Answers
2






active

oldest

votes


















2












$begingroup$

${X=1}$ is the event that the first bottle is a cider bottle.



Probability on that: $P(text{first cider})=frac{n}{n+2}$



${X=2}$ is the event that the first bottle contains beer and the second bottle contains cider.



Probability on that: $P(text{first beer})P(text{second cider}midtext{ first beer})=frac2{n+2}frac{n}{n+1}$.



${X=3}$ is the event that the first bottle contains beer and the second bottle contains beer.



Probability on that: $P(text{first beer})P(text{second beer}midtext{ first beer})=frac2{n+2}frac{1}{n+1}$.



Now we are ready to find:$$mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=frac{n}{n+2}+2frac2{n+2}frac{n}{n+1}+3frac2{n+2}frac1{n+1}=frac{n+3}{n+1}$$





There are $2$ bottles that have index $1$ so that $P(Y=1)=frac2{n+2}$.



${X=1,Y=1}$ is the event that the first bottle is the cider bottle with index $1$.



Probability on that: $P(X=1,Y=1)=frac1{n+2}$.



So a necessary condition for independence is: $$frac{n}{n+2}frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=frac1{n+2}$$



leading to $n=2$.



So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,jin{1,2}$.



I leave that to you.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.



    For $X = 1$



    $C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.



    For X =2



    $B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s



    For X = 3



    $B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.



    Number of ways X = 1 can happen is = ${nchoose1} (n+1)!$



    Number of ways X = 2 can happen is = ${2choose1}{nchoose1} n!$



    Number of ways X = 3 can happen is = ${2choose1} n!$



    Total number of ways =$(n+2)!$



    Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$



    $$frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$



    Thus the expected value $$ E(X) = frac{1.{nchoose1}(n+1)!+2.{2choose1}{nchoose1} n!+3.{2choose1} n!}{(n+2)!} = frac{n+3}{n+1}$$



    Second Part



    For $Y = 1$



    $(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways



    $(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.



    Thus $P(Y=1) = frac{(2(n+1)!)}{(n+2)!}$.



    For Y =2



    $-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



    $-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



    Thus$P(Y=2) = frac{(2n) n!}{(n+2)!}$



    and so on for (Y=n+1) for which the probability = $P(Y=n+1) = frac{(2) n!}{(n+2)!}$



    Thus $$E(Y) = frac{2(n+1).n! times 1 + 2n.n!times 2 + 2(n-1)n!times 3 +cdots + 2(2).n!times n+ 2(1).n!times (n+1)}{(n+2)!} $$ $$= frac{(n+3)}{3}$$






    share|cite|improve this answer











    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      ${X=1}$ is the event that the first bottle is a cider bottle.



      Probability on that: $P(text{first cider})=frac{n}{n+2}$



      ${X=2}$ is the event that the first bottle contains beer and the second bottle contains cider.



      Probability on that: $P(text{first beer})P(text{second cider}midtext{ first beer})=frac2{n+2}frac{n}{n+1}$.



      ${X=3}$ is the event that the first bottle contains beer and the second bottle contains beer.



      Probability on that: $P(text{first beer})P(text{second beer}midtext{ first beer})=frac2{n+2}frac{1}{n+1}$.



      Now we are ready to find:$$mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=frac{n}{n+2}+2frac2{n+2}frac{n}{n+1}+3frac2{n+2}frac1{n+1}=frac{n+3}{n+1}$$





      There are $2$ bottles that have index $1$ so that $P(Y=1)=frac2{n+2}$.



      ${X=1,Y=1}$ is the event that the first bottle is the cider bottle with index $1$.



      Probability on that: $P(X=1,Y=1)=frac1{n+2}$.



      So a necessary condition for independence is: $$frac{n}{n+2}frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=frac1{n+2}$$



      leading to $n=2$.



      So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,jin{1,2}$.



      I leave that to you.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        ${X=1}$ is the event that the first bottle is a cider bottle.



        Probability on that: $P(text{first cider})=frac{n}{n+2}$



        ${X=2}$ is the event that the first bottle contains beer and the second bottle contains cider.



        Probability on that: $P(text{first beer})P(text{second cider}midtext{ first beer})=frac2{n+2}frac{n}{n+1}$.



        ${X=3}$ is the event that the first bottle contains beer and the second bottle contains beer.



        Probability on that: $P(text{first beer})P(text{second beer}midtext{ first beer})=frac2{n+2}frac{1}{n+1}$.



        Now we are ready to find:$$mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=frac{n}{n+2}+2frac2{n+2}frac{n}{n+1}+3frac2{n+2}frac1{n+1}=frac{n+3}{n+1}$$





        There are $2$ bottles that have index $1$ so that $P(Y=1)=frac2{n+2}$.



        ${X=1,Y=1}$ is the event that the first bottle is the cider bottle with index $1$.



        Probability on that: $P(X=1,Y=1)=frac1{n+2}$.



        So a necessary condition for independence is: $$frac{n}{n+2}frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=frac1{n+2}$$



        leading to $n=2$.



        So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,jin{1,2}$.



        I leave that to you.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          ${X=1}$ is the event that the first bottle is a cider bottle.



          Probability on that: $P(text{first cider})=frac{n}{n+2}$



          ${X=2}$ is the event that the first bottle contains beer and the second bottle contains cider.



          Probability on that: $P(text{first beer})P(text{second cider}midtext{ first beer})=frac2{n+2}frac{n}{n+1}$.



          ${X=3}$ is the event that the first bottle contains beer and the second bottle contains beer.



          Probability on that: $P(text{first beer})P(text{second beer}midtext{ first beer})=frac2{n+2}frac{1}{n+1}$.



          Now we are ready to find:$$mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=frac{n}{n+2}+2frac2{n+2}frac{n}{n+1}+3frac2{n+2}frac1{n+1}=frac{n+3}{n+1}$$





          There are $2$ bottles that have index $1$ so that $P(Y=1)=frac2{n+2}$.



          ${X=1,Y=1}$ is the event that the first bottle is the cider bottle with index $1$.



          Probability on that: $P(X=1,Y=1)=frac1{n+2}$.



          So a necessary condition for independence is: $$frac{n}{n+2}frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=frac1{n+2}$$



          leading to $n=2$.



          So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,jin{1,2}$.



          I leave that to you.






          share|cite|improve this answer









          $endgroup$



          ${X=1}$ is the event that the first bottle is a cider bottle.



          Probability on that: $P(text{first cider})=frac{n}{n+2}$



          ${X=2}$ is the event that the first bottle contains beer and the second bottle contains cider.



          Probability on that: $P(text{first beer})P(text{second cider}midtext{ first beer})=frac2{n+2}frac{n}{n+1}$.



          ${X=3}$ is the event that the first bottle contains beer and the second bottle contains beer.



          Probability on that: $P(text{first beer})P(text{second beer}midtext{ first beer})=frac2{n+2}frac{1}{n+1}$.



          Now we are ready to find:$$mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=frac{n}{n+2}+2frac2{n+2}frac{n}{n+1}+3frac2{n+2}frac1{n+1}=frac{n+3}{n+1}$$





          There are $2$ bottles that have index $1$ so that $P(Y=1)=frac2{n+2}$.



          ${X=1,Y=1}$ is the event that the first bottle is the cider bottle with index $1$.



          Probability on that: $P(X=1,Y=1)=frac1{n+2}$.



          So a necessary condition for independence is: $$frac{n}{n+2}frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=frac1{n+2}$$



          leading to $n=2$.



          So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,jin{1,2}$.



          I leave that to you.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 10:32









          drhabdrhab

          102k545136




          102k545136























              1












              $begingroup$

              For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.



              For $X = 1$



              $C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.



              For X =2



              $B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s



              For X = 3



              $B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.



              Number of ways X = 1 can happen is = ${nchoose1} (n+1)!$



              Number of ways X = 2 can happen is = ${2choose1}{nchoose1} n!$



              Number of ways X = 3 can happen is = ${2choose1} n!$



              Total number of ways =$(n+2)!$



              Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$



              $$frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$



              Thus the expected value $$ E(X) = frac{1.{nchoose1}(n+1)!+2.{2choose1}{nchoose1} n!+3.{2choose1} n!}{(n+2)!} = frac{n+3}{n+1}$$



              Second Part



              For $Y = 1$



              $(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways



              $(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.



              Thus $P(Y=1) = frac{(2(n+1)!)}{(n+2)!}$.



              For Y =2



              $-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



              $-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



              Thus$P(Y=2) = frac{(2n) n!}{(n+2)!}$



              and so on for (Y=n+1) for which the probability = $P(Y=n+1) = frac{(2) n!}{(n+2)!}$



              Thus $$E(Y) = frac{2(n+1).n! times 1 + 2n.n!times 2 + 2(n-1)n!times 3 +cdots + 2(2).n!times n+ 2(1).n!times (n+1)}{(n+2)!} $$ $$= frac{(n+3)}{3}$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.



                For $X = 1$



                $C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.



                For X =2



                $B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s



                For X = 3



                $B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.



                Number of ways X = 1 can happen is = ${nchoose1} (n+1)!$



                Number of ways X = 2 can happen is = ${2choose1}{nchoose1} n!$



                Number of ways X = 3 can happen is = ${2choose1} n!$



                Total number of ways =$(n+2)!$



                Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$



                $$frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$



                Thus the expected value $$ E(X) = frac{1.{nchoose1}(n+1)!+2.{2choose1}{nchoose1} n!+3.{2choose1} n!}{(n+2)!} = frac{n+3}{n+1}$$



                Second Part



                For $Y = 1$



                $(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways



                $(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.



                Thus $P(Y=1) = frac{(2(n+1)!)}{(n+2)!}$.



                For Y =2



                $-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                $-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                Thus$P(Y=2) = frac{(2n) n!}{(n+2)!}$



                and so on for (Y=n+1) for which the probability = $P(Y=n+1) = frac{(2) n!}{(n+2)!}$



                Thus $$E(Y) = frac{2(n+1).n! times 1 + 2n.n!times 2 + 2(n-1)n!times 3 +cdots + 2(2).n!times n+ 2(1).n!times (n+1)}{(n+2)!} $$ $$= frac{(n+3)}{3}$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.



                  For $X = 1$



                  $C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.



                  For X =2



                  $B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s



                  For X = 3



                  $B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.



                  Number of ways X = 1 can happen is = ${nchoose1} (n+1)!$



                  Number of ways X = 2 can happen is = ${2choose1}{nchoose1} n!$



                  Number of ways X = 3 can happen is = ${2choose1} n!$



                  Total number of ways =$(n+2)!$



                  Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$



                  $$frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$



                  Thus the expected value $$ E(X) = frac{1.{nchoose1}(n+1)!+2.{2choose1}{nchoose1} n!+3.{2choose1} n!}{(n+2)!} = frac{n+3}{n+1}$$



                  Second Part



                  For $Y = 1$



                  $(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways



                  $(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.



                  Thus $P(Y=1) = frac{(2(n+1)!)}{(n+2)!}$.



                  For Y =2



                  $-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                  $-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                  Thus$P(Y=2) = frac{(2n) n!}{(n+2)!}$



                  and so on for (Y=n+1) for which the probability = $P(Y=n+1) = frac{(2) n!}{(n+2)!}$



                  Thus $$E(Y) = frac{2(n+1).n! times 1 + 2n.n!times 2 + 2(n-1)n!times 3 +cdots + 2(2).n!times n+ 2(1).n!times (n+1)}{(n+2)!} $$ $$= frac{(n+3)}{3}$$






                  share|cite|improve this answer











                  $endgroup$



                  For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.



                  For $X = 1$



                  $C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.



                  For X =2



                  $B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s



                  For X = 3



                  $B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.



                  Number of ways X = 1 can happen is = ${nchoose1} (n+1)!$



                  Number of ways X = 2 can happen is = ${2choose1}{nchoose1} n!$



                  Number of ways X = 3 can happen is = ${2choose1} n!$



                  Total number of ways =$(n+2)!$



                  Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$



                  $$frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$



                  Thus the expected value $$ E(X) = frac{1.{nchoose1}(n+1)!+2.{2choose1}{nchoose1} n!+3.{2choose1} n!}{(n+2)!} = frac{n+3}{n+1}$$



                  Second Part



                  For $Y = 1$



                  $(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways



                  $(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.



                  Thus $P(Y=1) = frac{(2(n+1)!)}{(n+2)!}$.



                  For Y =2



                  $-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                  $-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${nchoose1}n!$ ways



                  Thus$P(Y=2) = frac{(2n) n!}{(n+2)!}$



                  and so on for (Y=n+1) for which the probability = $P(Y=n+1) = frac{(2) n!}{(n+2)!}$



                  Thus $$E(Y) = frac{2(n+1).n! times 1 + 2n.n!times 2 + 2(n-1)n!times 3 +cdots + 2(2).n!times n+ 2(1).n!times (n+1)}{(n+2)!} $$ $$= frac{(n+3)}{3}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 18 '18 at 6:54

























                  answered Dec 17 '18 at 9:35









                  Satish RamanathanSatish Ramanathan

                  10k31323




                  10k31323















                      Popular posts from this blog

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Aardman Animations

                      Are they similar matrix