Statistics with Baye's theorem and partitions

[rock.freak667]

Well the questions and solutions are in one for some and I will type out the rest.

Q6
http://img249.imageshack.us/img249/2757/47026197.jpg

Q4
http://img237.imageshack.us/img237/1802/93774799.jpg
Q3
http://img3.imageshack.us/img3/5919/34981698.jpg

Q1
http://img19.imageshack.us/img19/7307/25729239.jpg

I need some help doing Q3 and I need some clarification with Q1.

With Q1, if Y=2, then there are only two possibilities, (1,2) and (1,2). So X can only be equal to 3 thus I am not sure how to write out the distribution table.http://img222.imageshack.us/img222/9981/95680883.jpg

 

[Mulder]

for Q1, there's no reason not to include (2,2) as well (twice)

for Q3, three events would be independent if p(A and B and C) = p(A)p(B)p(C), and the three possible pairs are independent (e.g A and C), just a natural extension from two events

 

[rock.freak667]

for Q1, there's no reason not to include (2,2) as well (twice)

Right, but wouldn't then my distribution not all add up to one either way? Σ_{all x} P(X=x)≠1

for Q3, three events would be independent if p(A and B and C) = p(A)p(B)p(C), and the three possible pairs are independent (e.g A and C), just a natural extension from two events

This I know, but I don't know what numbers I'd write down for P(A) and so on.

 

[Mulder]

Right, but wouldn't then my distribution not all add up to one either way? Σ_{all x} P(X=x)≠1

You should just have Σ_{all x} P(X=x | Y=2) =1

This I know, but I don't know what numbers I'd write down for P(A) and so on.

I drew the square with corners at 0, (0,1), (1,1), (1,0) ao P(A) = P(B) =1/2 etc.