Joint conditional distribution












1














Hey I have a question where i think i am doing everything correct but I don’t get the expected solution, I think the solution may be wrong - or I am....



enter image description here



Calculate P(0I1) and P(1I1)



for $$P(X=1|Y=1) =P(X=1,Y=1)/P(Y=1)=0.3/0.5=3/5$$



And $$P(X=0|Y=1) = P(X=0,Y=1)/P(Y=1)= 2/5$$



But this is apparently not correct.... maybe you could tell me if i am wrong or not...



Many thanks










share|cite|improve this question
























  • What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
    – snarski
    2 days ago






  • 1




    It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
    – David K
    2 days ago






  • 1




    You are plugging wrong numbers in your last calculations ...
    – Math-fun
    2 days ago






  • 1




    Looks good to me. Obvious check, sum of answers = 1.
    – herb steinberg
    2 days ago






  • 1




    $0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
    – David K
    2 days ago
















1














Hey I have a question where i think i am doing everything correct but I don’t get the expected solution, I think the solution may be wrong - or I am....



enter image description here



Calculate P(0I1) and P(1I1)



for $$P(X=1|Y=1) =P(X=1,Y=1)/P(Y=1)=0.3/0.5=3/5$$



And $$P(X=0|Y=1) = P(X=0,Y=1)/P(Y=1)= 2/5$$



But this is apparently not correct.... maybe you could tell me if i am wrong or not...



Many thanks










share|cite|improve this question
























  • What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
    – snarski
    2 days ago






  • 1




    It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
    – David K
    2 days ago






  • 1




    You are plugging wrong numbers in your last calculations ...
    – Math-fun
    2 days ago






  • 1




    Looks good to me. Obvious check, sum of answers = 1.
    – herb steinberg
    2 days ago






  • 1




    $0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
    – David K
    2 days ago














1












1








1







Hey I have a question where i think i am doing everything correct but I don’t get the expected solution, I think the solution may be wrong - or I am....



enter image description here



Calculate P(0I1) and P(1I1)



for $$P(X=1|Y=1) =P(X=1,Y=1)/P(Y=1)=0.3/0.5=3/5$$



And $$P(X=0|Y=1) = P(X=0,Y=1)/P(Y=1)= 2/5$$



But this is apparently not correct.... maybe you could tell me if i am wrong or not...



Many thanks










share|cite|improve this question















Hey I have a question where i think i am doing everything correct but I don’t get the expected solution, I think the solution may be wrong - or I am....



enter image description here



Calculate P(0I1) and P(1I1)



for $$P(X=1|Y=1) =P(X=1,Y=1)/P(Y=1)=0.3/0.5=3/5$$



And $$P(X=0|Y=1) = P(X=0,Y=1)/P(Y=1)= 2/5$$



But this is apparently not correct.... maybe you could tell me if i am wrong or not...



Many thanks







conditional-probability






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked 2 days ago









Lillys

758




758












  • What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
    – snarski
    2 days ago






  • 1




    It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
    – David K
    2 days ago






  • 1




    You are plugging wrong numbers in your last calculations ...
    – Math-fun
    2 days ago






  • 1




    Looks good to me. Obvious check, sum of answers = 1.
    – herb steinberg
    2 days ago






  • 1




    $0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
    – David K
    2 days ago


















  • What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
    – snarski
    2 days ago






  • 1




    It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
    – David K
    2 days ago






  • 1




    You are plugging wrong numbers in your last calculations ...
    – Math-fun
    2 days ago






  • 1




    Looks good to me. Obvious check, sum of answers = 1.
    – herb steinberg
    2 days ago






  • 1




    $0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
    – David K
    2 days ago
















What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
– snarski
2 days ago




What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
– snarski
2 days ago




1




1




It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
– David K
2 days ago




It might help if you also wrote your own working in more detail. For example in $0.3/0.5$ there is no indication why you wrote those particular numbers $0.3$ and $0.5.$ It looks like they should be probabilities of something, but you should say what.
– David K
2 days ago




1




1




You are plugging wrong numbers in your last calculations ...
– Math-fun
2 days ago




You are plugging wrong numbers in your last calculations ...
– Math-fun
2 days ago




1




1




Looks good to me. Obvious check, sum of answers = 1.
– herb steinberg
2 days ago




Looks good to me. Obvious check, sum of answers = 1.
– herb steinberg
2 days ago




1




1




$0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
– David K
2 days ago




$0.8$ and $0.2$ would make sense if it were $P(X=0mid Y=0)=0.8$ and $P(X=1mid Y=0)=0.2.$ But if it says $Y=1$ then I think your results are the correct ones. It is possible the answer key is wrong--doesn't happen very often, but it does happen sometimes.
– David K
2 days ago










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