Finding $P(A\cup B \cup C)$ when rolling a set of 3 dice Consider a set of 3 6-sided die divided into 3 groups, with each group having 2 dice and every die being in exactly 2 groups. Let $p_e$ be the probability that, after rolling all the die, there is at least one odd number rolled in at least one of the groups. Calculate $p_e$ using the expression for $P(A \cup B \cup C)$, and using another (easier) method.
My solution so far:
Let $X,Y,Z$ be the three different dies and the three groups are:
$XY, XZ, YZ$
We want to find
$$P(XY \cup XZ \cup YZ) = P(XY)+P(XZ)+P(YZ)-P(XY\cap XZ)-P(XY \cap YZ) - P(XZ \cap YZ) + P(XY\cap XZ \cap YZ)$$
$P(XY) = 1-P(no.odd)=1-1/4=3/4$
$P(XY)=P(YZ)=P(XZ)=3/4$
$P(XY\cap XZ)=(3/4)(3/4)=9/16$
$P(XY\cap XZ)=P(XY\cap YZ)=P(XZ\cap YZ)=9/16$
$ P(XY\cap XZ \cap YZ)=3/4(3/4)(3/4)=27/64$
Therefore:
$$P(XY \cup XZ \cup YZ) = 3/4+3/4+3/4-9/16-9/16-9/16+27/64=63/64$$
Easier Method:
$$P(XY \cup XZ \cup YZ) = 1- P(XY \cup XZ \cup YZ)^c = 1-P(XY^c\cap XZ^c \cap YZ^c)$$
$P(XY^c)=P(no.odd)=1/4=P(XZ^c)=P(YZ^c)$
So we have $1-(1/4)(1/4)(1/4)=1-1/64=63/64$
Am I correct in my solution?
 A: 
"Let $p_c$ be the probability that, after rolling all the die, there is at least one odd number rolled in at least one of the groups."

It never occurred to me that MathSE reviewers would interpret this to mean that the number on each die was to be examined to see if it was even or odd.
I interpreted the problem differently.  That is, I interpret the problem to intend that the Math student is supposed to examine the three sums Die-1 + Die-2, Die-1 + Die-3, and Die-2 + Die-3.  My interpretation is based on the idea that if the problem composer intended the original interpretation instead, why did the problem composer create the three groups, where each group represented examining the two dice in that group.
What would be the point?

First of all, the probability of at least one odd number is equal to $(1)$ minus the probability that you don't have $(3)$ even numbers.
You will have $(3)$ even numbers, if and only if, all $(3)$ dice have the same parity, odd or even.  Without loss of generality, you can assume that the first die has parity even.
Then, you will have all $(3)$ dice with the same parity if and only if the 2nd and 3rd dice are also of even parity.
The probability of this happening is $\displaystyle ~\left[\frac{1}{2}\right]^2 = \frac{1}{4}.$
Therefore, the easy approach is to compute
$$1 - \frac{1}{4} = \frac{3}{4}. \tag1 $$
So, (1) above is the probability that you have at least one odd sum, among the three groups.

For the hard way, I recommend approaching the problem combinatorically, where the probability will be expressed as
$$\frac{N}{D},$$
with $D = 2^3.$  Here, you should resist the idea of the simpler approach of $D = 6^3$, because there is no reason to differentiate each of the $(6)$ numbers that may show on a specific die.  The nature of the problem is such that it is sufficient to only focus on the odd/even parity of each die.
Let $S_1$ denote all of the ways that Die-1,Die-2 can have different parities.
Similarly, define $S_2, S_3$ to cover Die-1,Die-3 and Die-2,Die-3, respectively.
Then, the overall computation is
$$\frac{|S_1 \cup S_2 \cup S_3|}{2^3}.$$

See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Let $T_1$ denote $|S_1| + |S_2| + |S_3|.$
Let $T_2$ denote $|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|.$
Let $T_3$ denote $|S_1 \cap S_2 \cap S_3|.$
Then, in accordance with Inclusion-Exclusion theory, the overall computation will be
$$\frac{T_1 - T_2 + T_3}{2^3}.$$
To enumerate $|S_1|$, you can have Die-1, Die-2 be odd/even or even/odd.  In each case, Die-3 is unconstrained.
Therefore, $|S_1| = 2 + 2 = 4.$
By symmetrical considerations, you have that
$|S_2| = |S_3| = |S_1|.$
Therefore,
$$T_1 = 12. \tag2 $$

To enumerate $S_1 \cap S_2|$ you want the number of ways that Die-2 can have a different parity than both Die-1, Die-3.
Die-2 can be either odd or even.  In each case, the parities of both Die-1 and Die-3 will then be set.
Therefore, $|S_1 \cap S_2| = 2.$
Again, by symmetrical considerations, 
$|S_1 \cap S_2| = |S_1 \cap S_3| = |S_1 \cap S_3|$.
Therefore,
$$T_2 = 3 \times 2 = 6. \tag3 $$

The enumeration of $|S_1 \cap S_2 \cap S_3|$ must be $(0)$.
The reason for this is that if you take any two of the three subsets, (for example) $S_1$ and $S_2$, then the subset given by $S_1 \cap S_2$ demands that Die-1 and Die-3 both have a different parity than Die-2.
This requires that Die-1 and Die-3 have the same parity.
Similar considerations will apply throughout the attempt to enumerate $|S_1 \cap S_2 \cap S_3|$.
Therefore,
$$T_3 = 0. \tag4 $$

So, using Inclusion-Exclusion, the overall probability that at least one of the sums is odd is
$$\frac{12 - 6}{8}.$$
