Determining the average probability of error I'm trying to solve the following problem:

In a binary PAM system, the input to the detector is
$$y_m = a_m+n_m+i_m$$
where $a_m = \pm1$ is the desired signal, $n_m$ is a zero-mean Gaussian random variable
with variance $\sigma_n^2$ and $i_m$ represents the ISI due to channel distortion. The ISI term
is a random variable which takes the values, $\frac{-1}{2}, 0, \frac{1}{2}$ with probabilities $\frac{1}{4}, \frac{1}{2},\frac{1}{4}$ respectively. Determine the average probability of error as a function of $\sigma_n^2$.

My try: I think the probability of error is $$P_e = \mathbb{P}(y_m&lt;0 |a_m = 1) + \mathbb{P}(y_m\gt 0|a_m = -1)$$According to the definition of conditional probability, we have $$\mathbb{P}(y_m&lt;0 |a_m = 1) = \frac{\mathbb{P}(y_m&lt;0 \cap a_m = 1)}{\mathbb{P}(a_m = 1)} = \frac{\mathbb{P}(n_m+i_m&lt;-1)}{\mathbb{P}(a_m = 1)}$$ By law of total probability $$\mathbb{P}(n_m+i_m&lt;-1) = \mathbb{P}(n_m+i_m&lt;-1\cap i_m = \frac{-1}{2}) + \mathbb{P}(n_m+i_m&lt;-1\cap i_m = 0) + \mathbb{P}(n_m+i_m&lt;-1\cap i_m = \frac{1}{2}) = \mathbb{P}(n_m&lt;\frac{-1}{2}) + \mathbb{P}(n_m&lt;-1) + \mathbb{P}(n_m&lt;\frac{-3}{2})$$
Each term can be written as a function of $\sigma_n^2$ easily. So it seems we need $\mathbb{P}(a_m = \pm1)$ instead of $\mathbb{P}(i_m = \pm\frac{1}{2},0)$ which can't be true because we should certainly use values of $\mathbb{P}(i_m = \pm\frac{1}{2},0)$. What's my mistake here? Also is there any difference between &quot;average probability of error&quot; and &quot;probability of error&quot;?
Note that
$$
\mathbb{P}(n_m+i_m&lt;-1\cap i_m = \frac{-1}{2})
$$
is equal to
$$
\mathbb{P}(n_m &lt; -0.5)\mathbb{P}(i_m = -0.5) = 0.25\mathbb{P}(n_m &lt; -0.5)
$$
since $n_m$ and $i_m$ are independent.
Another way to obtain the solution is to add the the probability of these three disjoint events:

$n_m &lt; -1.5$,
$i_m = 0$ and $-1.5 &lt; n_m &lt; -1$,
$i_m = -0.5$ and $-1 &lt; n_m &lt; -0.5$

In this case, we can add the three probabilities because the events are disjoint. To keep them disjoint, it is important to keep the noise ranges disjoint too; otherwise, you would be adding them multiple times.