Discrete Probability: Random Variable Independent or Dependent?












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Question a):



You flip a fair coin $7$ times; these coin flips are independent of each other.



Define the random variables:



$X$ = the number of heads in these $7$ coin flips, and



$Y$ = the number of tails in these $7$ coin flips.



Are the random variables independent or not? Show why or why not?




Attempt:



I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.



{HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}



If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$



The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.



The Probability of just 1 head and 1 tail occurring together is $0$



So, they are not equal and hence dependent. Is this correct way?




Question b):



Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.



Define the random variables:



$X$ = $0$ if z is even and $1$ is z odd



$Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}



Are the random variables independent or not? Show why or why not?




Attempt:



$Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.



$Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets



$Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$



$frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$



They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?






probability probability-theory discrete-mathematics random-variables







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    0












    $begingroup$



    Question a):



    You flip a fair coin $7$ times; these coin flips are independent of each other.



    Define the random variables:



    $X$ = the number of heads in these $7$ coin flips, and



    $Y$ = the number of tails in these $7$ coin flips.



    Are the random variables independent or not? Show why or why not?




    Attempt:



    I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.



    {HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}



    If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$



    The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.



    The Probability of just 1 head and 1 tail occurring together is $0$



    So, they are not equal and hence dependent. Is this correct way?




    Question b):



    Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.



    Define the random variables:



    $X$ = $0$ if z is even and $1$ is z odd



    $Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}



    Are the random variables independent or not? Show why or why not?




    Attempt:



    $Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.



    $Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets



    $Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$



    $frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$



    They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?






    probability probability-theory discrete-mathematics random-variables







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Question a):



      You flip a fair coin $7$ times; these coin flips are independent of each other.



      Define the random variables:



      $X$ = the number of heads in these $7$ coin flips, and



      $Y$ = the number of tails in these $7$ coin flips.



      Are the random variables independent or not? Show why or why not?




      Attempt:



      I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.



      {HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}



      If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$



      The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.



      The Probability of just 1 head and 1 tail occurring together is $0$



      So, they are not equal and hence dependent. Is this correct way?




      Question b):



      Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.



      Define the random variables:



      $X$ = $0$ if z is even and $1$ is z odd



      $Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}



      Are the random variables independent or not? Show why or why not?




      Attempt:



      $Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.



      $Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets



      $Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$



      $frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$



      They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?






      probability probability-theory discrete-mathematics random-variables







      share|cite|improve this question









      $endgroup$





      Question a):



      You flip a fair coin $7$ times; these coin flips are independent of each other.



      Define the random variables:



      $X$ = the number of heads in these $7$ coin flips, and



      $Y$ = the number of tails in these $7$ coin flips.



      Are the random variables independent or not? Show why or why not?




      Attempt:



      I know there are $2^{7}$ = $128$ possible sequences for the con toss $7$ times.



      {HHHHHHT, HHHHHTT, HHHHTTT, HHHTTTT, HHTTTTT, HTTTTTT, HHHHHHH}



      If I take $P(X=1 $ and $ Y=1) = P(X=1)P(Y=1)$



      The Probability of just 1 head and 1 tail would be $frac{7}{128}$ for both.



      The Probability of just 1 head and 1 tail occurring together is $0$



      So, they are not equal and hence dependent. Is this correct way?




      Question b):



      Consider the set $(1,2,3,…10)$. You choose a uniformly random element z in S.



      Define the random variables:



      $X$ = $0$ if z is even and $1$ is z odd



      $Y$ = $0$ if z {1,2}, $1$ if z {3,4,5,6}, $2$ if z {7,8,9,20}



      Are the random variables independent or not? Show why or why not?




      Attempt:



      $Pr(X = 0 $ or $ 1)$ for both odd and even is $2^{5} / 2^{10}$ = $frac{1}{32}$.



      $Pr(Y = 0) =$ $1$ subset {1,2} / $2^{10}$ possible subsets



      $Pr$($X=0$ and $Y=0$) = $P$($X=0)P(Y=0)$



      $frac{1}{2^{10}}$ $.$ $frac{1}{32}$. = $frac{1}{32}$ $.$ $frac{1}{2^{10}}$



      They are equal and hence independent. Is my approach to finding the probabilities correct to come to this conclusion of independence?






      probability probability-theory discrete-mathematics random-variables




      probability probability-theory discrete-mathematics random-variables






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      asked Dec 1 '18 at 23:44









      TobyToby

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          The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"



          a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)



          b) Do the same analysis for your other cases.



          Yes... you can also check this through:



          Is $P(X|Y) = P(X)$.






          share|cite|improve this answer









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            $begingroup$

            The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"



            a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)



            b) Do the same analysis for your other cases.



            Yes... you can also check this through:



            Is $P(X|Y) = P(X)$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"



              a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)



              b) Do the same analysis for your other cases.



              Yes... you can also check this through:



              Is $P(X|Y) = P(X)$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"



                a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)



                b) Do the same analysis for your other cases.



                Yes... you can also check this through:



                Is $P(X|Y) = P(X)$.






                share|cite|improve this answer









                $endgroup$



                The way to understand independence is to ask yourself: "If I know that variable $X$ has a particular value, does that provide any information whatsoever about the value of variable $Y$?"



                a) So imagine $X = 7$. Does that provide any information whatsoever about the value of variable $Y$? Of course it does! You can be certain that $Y neq 7$, for instance. (We can only have $Y=0$.)



                b) Do the same analysis for your other cases.



                Yes... you can also check this through:



                Is $P(X|Y) = P(X)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 0:05









                David G. StorkDavid G. Stork

                10.7k31332




                10.7k31332






























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