Probability of three independent events Let events $X,Y,Z$ be mutually independent events such that $\mathbb{P}(X) = 0.4, \mathbb{P}(Y) = 0.3, \mathbb{P}(Z) = 0.2$. Find $\mathbb{P}((X \cup Y) \setminus Z)$.
Solution: looking at the Venn diagram below

I can rewrite
$$\mathbb{P}((X \cup Y) \setminus Z)$$
as
$$\mathbb{P}(X) + \mathbb{P}(Y) - \mathbb{P}(X \cap Z) - \mathbb{P}(Y \cap Z) + \mathbb{P}(X \cap Y \cap Z) \overset{\text{independence } X,Y,Z}{=} $$
$$\mathbb{P}(X) + \mathbb{P}(Y) - \mathbb{P}(X)\mathbb{P}(Z)- \mathbb{P}(Y)\mathbb{P}(Z) + \mathbb{P}(X)\mathbb{P}(Y)\mathbb{P}(Z) =$$
$$ 0.4+ 0.3 - 0.4\cdot 0.2 - 0.3\cdot 0.2 + 0.4 \cdot 0.3 \cdot 0.2 = 0.584$$
Am I correct?
 A: Rereading.  Your answer appears wrong.  You have counted the region of $X\cap Y\cap Z^c$ once when talking about $\Pr(X)$ and again when talking about $\Pr(Y)$ but this was not later subtracted to correctly account for the overcount like you should have done when mimicking inclusion-exclusion.  Similarly, the region for $X\cap Y\cap Z$ was counted once with $\Pr(X)$ again with $\Pr(Y)$, was discounted once with $\Pr(X\cap Z)$ and discounted again for $\Pr(Y\cap Z)$, but then you added it back in for $\Pr(X\cap Y\cap Z)$ making it so we counted it a total of one time when we intended to have counted it zero times.
When running inclusion-exclusion or mimicking inclusion-exclusion you need to ensure that each region is included in a net total of one occurrences each when we wanted to include it once, or zero times if we intended zero.
Even faster:
$X,Y,Z$ mutually independent also implies $X,Y,Z^c$ are mutually independent.
$(X\cup Y)\setminus Z$ is equivalent to $(X\cup Y)\cap Z^c$
Distributing:  $(X\cap Z^c)\cup (Y\cap Z^c)$
Expanding: $\Pr((X\cap Z^c)\cup (Y\cap Z^c)) = \Pr(X\cap Z^c)+\Pr(Y\cap Z^c) - \Pr(X\cap Y\cap Z^c)$
$$0.4\times 0.8 + 0.3\times 0.8 - 0.4\times 0.3\times 0.8$$

Alternatively, adding each region individually:
$$0.4\times 0.7\times 0.8 + 0.4\times 0.3\times 0.8+0.6\times 0.3\times 0.8$$
