Thread: 3 Equations simple (?) system with probabilities

I have a small problem that delays my project , it seems I am stuck here loads of time. It is probably very easy but I cant see it right now , I am very anxious about this, please take a look:
P0 = (4/5)P0 + (3/4)P1
P1 = (1/5)PO + (1/4)P1
also I know this: P1+P0=1
The solution is P0= 15/19 and P1= 4/19.
I have a very similar problem and I don't know how to continue from this point on, and delays my project!! How am i supposed to solve this ? For more probabilities like P1, P2, P3 what can we do ? Is there some general algorithm for this with steps ?
Thanks a lot people

I'm not sure what your question is. You have two independent equations in two unknowns (note: the equation P0= (4/5)P0 +(3/4)P1 is the same as P1 = (1/5)P0+(1/4)P1, so only one of those equations is important, and the other independ equation is P1+P0=1). The solution for P0 and P1 is as you stated: P0=15/1 and P1=4/19. So - what's your question?