Conditional Probability - Two methods There are two methods, A and B, to finish a work.


*

*Method A succeeds with probability $1/3$, but if it fails one tries method B with probability $3/4$ or method A again with probability $1/4$.

*Method B succeeds with probability $1/4$, but if it fails one tries method A with probability $2/3$ or method B again with probability $1/3$.

*Successive choices are made according to the scheme above until success.


What is the probability that success will be made by method A if

a) one starts with method A ?.

b) one starts with method B ?.
I thought the problem was dealing with a markov chain, but i think it's more related to conditional probability now. Still stuck!
Any help will be greatly appreciated. Thank you very much!
 A: Let $a$ be the probability one will (ultimately) succeed with Method $A$, given that one starts with Method $A$, and let $b$ be the probability one will succeed with Method $B$, given that one starts with Method $B$.
We assume without proof that if we start with $A$, then with probability $1$ we ultimately succeed, and the same holds if we starts with $B$. We have then 
$$a=\frac{1}{3}+ \frac{2}{3}\left(\frac{1}{4}a +\frac{3}{4}(1-b)\right),$$
and
$$b=\frac{1}{4}  +\frac{3}{4}\left(\frac{2}{3}(1-a) +\frac{1}{3}b\right).$$
Solve for $a$ and $b$. We get $a=\frac{2}{3}$ and $b=\frac{5}{9}$. 
Remark: It is a Markov chain problem, but does not require us to deploy the machinery. The two key equations come from conditional probability considerations. For the first equation, suppose we start with Method $A$. There is immediate success with probability $\frac{1}{3}$. With probability $\frac{2}{3}$ we continue. If (probability $\frac{1}{4}$) we use Method $A$, then the probability that (ultimate) success comes with Method $A$ is $a$. If (probability $\frac{3}{4}$) we switch to Method $B$, then the probability success still comes from Method $A$ is $1-b$.  
