# Why is $E[T_A] = \sum_{y \in X} P[X_1=y|Xo=x]E[T_A|X_0 =x, X_1 =y]$

I am studying a proof of a theorem in stohastic processes and there is this part thas confuses me :

$$E[T_A] = \sum_{y \in X} P[X_1=y|Xo=x]E[T_A|X_0 =x, X_1 =y]$$

where $$T_A = inf\{k \geq 0: X_k \in A\}$$

I do not get it because how come we have into the sum $$E[T_A]$$?
Since, $$E[X] = \sum px$$ where x is every possible value of random variable $$X$$.
So , we are trying to say something like :
The fact that Markov chain $$\{X_n\}_{n \ in No}$$ is hitting $$A$$
is the same with the fact $$\{X_0=1, X_1=y\}\wedge \{X_n\in A \}$$ for all possible $$(x,y)$$ - of course , that's why we have a sum - . And because we are interested in hitting time ... in a similar way we can say that the fact of $$\{X_n\}_{n \ in No}$$ is hitting $$A$$ for the first time is equivalent to :
$$\{X_0=1, X_1=y\}\wedge \{T_A =n\}$$ .
So it would make more sense to me to write something like:
for $$T_A \in N_0$$
$$E[T_A] = \sum_{y \in X} \sum_{n\in N_0}n P[T_A=n|Xo=x,X_1 = y]$$