Let's say I have three 6-sided dice. One is red, one is green, one is yellow. The red dice have 3 skulls on 3 of its faces, the other faces are blank. The green dice have 1 skull each. The yellow dice have 2 skulls each.
What are the chances that I roll all three and get a total of 1 skull?
I figure there's 3 ways to get one skull, the probability of each being:
skull on only the red dice: $(\frac{3}{6}) (1-\frac{1}{6}) (1-\frac{2}{6})=0.278$
skull on only the green dice: $(1-\frac{3}{6}) (\frac{1}{6}) (1-\frac{2}{6})=0.056$
skull on only the yellow dice: $(1-\frac{3}{6}) (1-\frac{1}{6}) (\frac{2}{6})=0.138$
How do I consolidate these to find the chance of getting only 1 skull with a 3-dice roll?
In this case, you have calculated the probability of three outcomes 'skull only on red', 'skull only on green' and 'skull only on yellow'; call them $A$, $B$ and $C$. If any one of these events occurs, then neither of the other two can as well. In this case we say that $A$, $B$ and $C$ are mutually exclusive.
Now, what you are interested in is the chance that any of $A$, $B$ and $C$ occur. For mutually exclusive events, $P(A, B,$ or $C) = P(A) + P(B) + P(C)$.
This makes intuitive sense; imagine a simpler scenario with one roll of one die. If we know that the chance of rolling a $5$ is $\frac{1}{6}$ and that of rolling a $2$ is $\frac{1}{6}$, then intuitively the chance of rolling either a $5$ or a $2$ is $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$. Indeed, one cannot roll both a $2$ and a $5$ in one roll; these events are mutually exclusive.
Extra credit: To see why $A$ and $B$ must be mutually exclusive, consider again the example of one roll of one die. Now consider the event that you roll either $1$ or $2$, call it $A$. Let $B$ denote the event that you roll a $2$ or a $3$. Now, intuitively, $P(A) = P(B) = \frac{1}{3}$. However $A$ and $B$ are clearly not mutually exclusive! Indeed, $P(A$ or $B) = \frac{1}{2} \ne \frac{2}{3} = P(A) + P(B)$. The problem is that in adding those two probabilities, we counted the chance that a $2$ comes up twice.
Indeed, the correct equation is $P(A$ or $B) = P(A) + P(B) - P(A $ and $B).$ Indeed, if both $A$ and $B$ occur in our single roll, this means that $2$ must have been rolled. So $P(A) + P(B) - P(A $ and $B) = \frac{1}{3} + \frac{1}{3} - \frac{1}{6} = \frac{1}{2}$. (Note that this new equation is consistent with the first situation, because if $A$ and $B$ are mutually exclusive, then $P(A $ and $B) = 0$.)
All of these sorts of things can be found in more general terms in any introduction to probability.