Recent content by scriby

Could you please tell me how to do that? Because I think that's the point I don't really understand (how to add the probabilities)

1/3 ? I can't follow your idea here. Making the states d and e absorbing means it is impossible to return once reaching those states right? This would mean their probabailities are always = 1 ? Why is p(a)=2/3 , shouldn't it be only 1/3 ? and why is p(c) = 2/3 , we only got 2 options to...

is this right? probability of hitting d (starting in g): 1+1+1/3 probability of hitting e (starting in g): 1+1+1/3+1/2

I guess 1 in all cases. I guess undirected (this is not specifically mentioned in the task but the image clearly shows an undirected graph since the arrows are missing) thx for your replies

yes of course...I got a little confused since c has two choices, but yes starting at g there's only one choice and that is c

Hi SammyS , thanks for your reply g has 1 choice c has 2 choices a has 3 choices d has 1 choice b has 3 choices f has 1 choice e also has 1 choice , but I think since it should not be relevant for the solution we can put it =0 ??

http://img844.imageshack.us/img844/8333/1111jx.png Homework Statement With which probability, starting in g, node d gets hit before node e?Homework Equations The Attempt at a Solution I think the probability of hitting each node starting in g is the following: p(g) = 1 p(c) = 1/2 p(a) =...