Find the joint distribution with 2 discrete random variables











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How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:



P(X=0, Y=0) = (0.4)(0.8) = 0.32



P(X=1, Y=1) = (1-0.4)(0.9) = 0.54



How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?



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    up vote
    2
    down vote

    favorite












    How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:



    P(X=0, Y=0) = (0.4)(0.8) = 0.32



    P(X=1, Y=1) = (1-0.4)(0.9) = 0.54



    How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?



    enter image description here










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:



      P(X=0, Y=0) = (0.4)(0.8) = 0.32



      P(X=1, Y=1) = (1-0.4)(0.9) = 0.54



      How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?



      enter image description here










      share|cite|improve this question













      How do you find the following joint distribution P(X=r, Y=k)? So far I have gotten:



      P(X=0, Y=0) = (0.4)(0.8) = 0.32



      P(X=1, Y=1) = (1-0.4)(0.9) = 0.54



      How do I go on to find P(X=0, Y=1)and P(X=1, Y=0)?



      enter image description here







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      asked 2 days ago









      Yolanda Hui

      13710




      13710






















          1 Answer
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          Hint:



          $P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$






          share|cite|improve this answer





















          • Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
            – Yolanda Hui
            2 days ago










          • Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
            – drhab
            2 days ago












          • How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
            – Yolanda Hui
            2 days ago












          • $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
            – drhab
            2 days ago










          • P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
            – Yolanda Hui
            2 days ago













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          1 Answer
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          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Hint:



          $P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$






          share|cite|improve this answer





















          • Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
            – Yolanda Hui
            2 days ago










          • Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
            – drhab
            2 days ago












          • How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
            – Yolanda Hui
            2 days ago












          • $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
            – drhab
            2 days ago










          • P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
            – Yolanda Hui
            2 days ago

















          up vote
          1
          down vote



          accepted










          Hint:



          $P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$






          share|cite|improve this answer





















          • Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
            – Yolanda Hui
            2 days ago










          • Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
            – drhab
            2 days ago












          • How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
            – Yolanda Hui
            2 days ago












          • $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
            – drhab
            2 days ago










          • P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
            – Yolanda Hui
            2 days ago















          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Hint:



          $P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$






          share|cite|improve this answer












          Hint:



          $P(X=0,Y=1)+P(X=0,Y=0)=P(X=0)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          drhab

          94.6k543125




          94.6k543125












          • Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
            – Yolanda Hui
            2 days ago










          • Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
            – drhab
            2 days ago












          • How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
            – Yolanda Hui
            2 days ago












          • $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
            – drhab
            2 days ago










          • P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
            – Yolanda Hui
            2 days ago




















          • Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
            – Yolanda Hui
            2 days ago










          • Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
            – drhab
            2 days ago












          • How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
            – Yolanda Hui
            2 days ago












          • $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
            – drhab
            2 days ago










          • P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
            – Yolanda Hui
            2 days ago


















          Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
          – Yolanda Hui
          2 days ago




          Then P(X=1,Y=0) + P(X=1,Y=1) = P(X=1)? Are my calculations for P(X=0, Y=0) and P(X=1, Y=1) correct also? Thanks.
          – Yolanda Hui
          2 days ago












          Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
          – drhab
          2 days ago






          Your calculations are correct. If you have found $P(X=0,Y=1)$ then you can find $P(X=1,Y=0)$ also. This by means of $P(X=0,Y=0)+P(X=0,Y=1)+P(X=1,Y=0)+P(X=1,Y=1)=1$. (Or the method you propose yourself).
          – drhab
          2 days ago














          How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
          – Yolanda Hui
          2 days ago






          How do I find P(Y=0) so I can use P(X=0,Y=0) = P(X=0) x P(Y=0) to see if X and Y are independent?
          – Yolanda Hui
          2 days ago














          $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
          – drhab
          2 days ago




          $P(Y=0)=P(X=0,Y=0)+P(X=1,Y=0)$.
          – drhab
          2 days ago












          P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
          – Yolanda Hui
          2 days ago






          P(Y=0) = 0.32 + 0.06 = 0.38, P(X=0) x P(Y=0)= 0.4 x 0.38 = 0.152, P(X=0,Y=0) = 0.32. They aren't equal, so X and Y are not independent.
          – Yolanda Hui
          2 days ago




















           

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