Probability error - binary message Please, I need some support (not solution) and input to know if the way is right to go on. 
Consider the communication of binary messages in a transmission medium. Any message sent is selected from two possible symbols, $0$ or $1$. Each symbol occurs with equal probability. It is also known that any numerical value sent on that channel is subjected to distortion. If a value $x$ is transmitted, the $y$ value is received at its destination, described by $y = x + n$, where n represents a random variable that additive noise is independent of $x$. The noise has a normal distribution with parameters $σ^2 = 4$ and $\mu = 0$.


*

*Suppose the transmitter encodes the symbol $0$ with the value $x = -2$ and $1$ with the symbol value $x = 2$. At the destination, the received message is decoded according to the following rules:

*not yet ....


*

*If $y ≥ 0$, one concludes that the symbol $1$ was sent.  

*If $y <0$, conclude the symbol $0$ was sent.



Q: Determine the probability of error for this schema encoding / decoding.
So, I know that....
Bit error probability: $P(x=0|y=1) ~\&~ P(x=1|y=0)$. The probabilities of transmitting each signal $(0,1)$ are equal (i.e., $1/2$).
$y_t = x_t + n_t$ where $x_t$ is a free signal of noise.
Assuming that the noise has a Gaussian distribution. 
 A: Let $E$ be the event that you make an error in reading the message.
$$P\left\{E\right\}=P\left\{\left(E\cap\left(x=-2\right)\right)\cup\left(E\cap\left(x=2\right)\right)\right\}$$
The events $\left(E\cap\left(x=-2\right)\right)$ and $\left(E\cap\left(x=2\right)\right)$ are mutually exclusive because $x$ may only take on a single value. Now we have that
$$P\left\{E\right\}=P\left\{E\cap\left(x=-2\right)\right\}+P\left\{E\cap\left(x=2\right)\right\}$$
Conditional probability tells us that $$P\left\{E\right\}=P\left\{E\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{E\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$
You should be able to rewrite the last probability statement in terms of the observed values $y$ rather than the event $E$, and by extension the probability statement may be written in terms of only $x$ and $n$. That should lead to a simple answer.
EDIT: At OP's request, I'm adding more information.
When $x=-2$, an error is made if $y\geq0$. When $x=2$, an error is if $y<0$. The probability statement becomes $$P\left\{E\right\}=P\left\{y\geq0\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{y<0\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$
We know that $y=x+n$, so change the $y$'s accordingly.
$$P\left\{E\right\}=P\left\{x+n\geq0\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{x+n<0\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$
When $x=-2$, $x+n\geq0$ is exactly the same as $n\geq2$. When $x=2$, $x+n<0$ is exactly the same as $n<-2$.
$$P\left\{E\right\}=P\left\{n\geq2\mid\left(x=-2\right)\right\}P\left\{x=-2\right\}+P\left\{n<-2\mid\left(x=2\right)\right\}P\left\{x=2\right\}$$
We also know that $x$ an $n$ are independent, so the conditional probabilities become marginals.
$$P\left\{E\right\}=P\left\{n\geq2\right\}P\left\{x=-2\right\}+P\left\{n<-2\right\}P\left\{x=2\right\}$$
You know that $n\sim N(0,4)$ and $x$ is equally likely to be either message.
\begin{align}
P\left\{E\right\}&=P\left\{n\geq2\right\}P\left\{x=-2\right\}+P\left\{n<-2\right\}P\left\{x=2\right\}\\
&=P\left\{\frac{n-0}{2}\geq\frac{2-0}{2}\right\}P\left\{x=-2\right\}+P\left\{\frac{n-0}{2}<\frac{-2-0}{2}\right\}P\left\{x=2\right\}\\
&=\left(1-\Phi(1)\right)\times\frac{1}{2}+\Phi(-1)\times\frac{1}{2}\\
\end{align}
Using the $68-95-99.7$ rule, the probability is $\approx0.16/2+0.16/2=0.16$.
