Statistics with Baye's theorem and partitions

1. Feb 6, 2010

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 [Broken]

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

Q1
http://img19.imageshack.us/img19/7307/25729239.jpg [Broken]

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 [Broken]

Last edited by a moderator: May 4, 2017
2. Feb 6, 2010

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

3. Feb 6, 2010

rock.freak667

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

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

4. Feb 7, 2010

Mulder

You should just have Σall x P(X=x | Y=2) =1

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