Mixing
Joint conditional distribution
1
$begingroup$
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....
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
asked Jan 3 at 19:03
LillysLillys
778
$endgroup$
|
show 5 more comments
1
$begingroup$
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....
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
asked Jan 3 at 19:03
LillysLillys
778
$endgroup$
$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
1
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
1
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
1
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
1
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00
|
show 5 more comments
1
1
1
$begingroup$
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....
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
asked Jan 3 at 19:03
LillysLillys
778
$endgroup$
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....
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
conditional-probability
asked Jan 3 at 19:03
LillysLillys
778
asked Jan 3 at 19:03
LillysLillys
778
edited Jan 3 at 19:17
Lillys
asked Jan 3 at 19:03
LillysLillys
778
asked Jan 3 at 19:03
LillysLillys
778
asked Jan 3 at 19:03
LillysLillys
778
778
$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
1
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
1
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
1
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
1
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00
|
show 5 more comments
$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
1
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
1
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
1
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
1
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00
$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
1
1
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
1
1
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
1
1
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
1
1
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00
|
show 5 more comments
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$begingroup$
What is $P(0 | 1)$? You have to specify which random variable gave $0$ and which gave $1$, e.g. $P(X = 0 | Y = 1)$.
$endgroup$
– snarski
Jan 3 at 19:07
1
$begingroup$
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.
$endgroup$
– David K
Jan 3 at 19:11
1
$begingroup$
You are plugging wrong numbers in your last calculations ...
$endgroup$
– Math-fun
Jan 3 at 19:15
1
$begingroup$
Looks good to me. Obvious check, sum of answers = 1.
$endgroup$
– herb steinberg
Jan 3 at 19:25
1
$begingroup$
$0.8$ 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.
$endgroup$
– David K
Jan 3 at 20:00