How to calculate the complement of rolling dies? I'm having a hard time trying to picturing the complement rule on this one. Can someone please enlighten me?
You roll 10 times of a fair 12 sided die. Find the probability of rolling two or more 12's in 10 rolls?
 A: As @lulu said:
$$P(\text{rolling two or more 12's in 10 rolls}) = 1-P(\leqslant 1 \text{ roll of a 12}) = 1-P(1)-P(0)=$$
$$=1-\binom{12}{1}(\frac{1}{12})^{1}(\frac{11}{12})^{11}-\binom{12}{0}(\frac{1}{12})^{0}(\frac{11}{12})^{12}=1-(\frac{11}{12})^{11}-(\frac{11}{12})^{12}$$
A: I will unpack this intuitively but also with formal notation in case you are interested...
The probability experiment we're facing here is the following: a twelve sided die is rolled 10 times. You are asked to find the probability that 2 or more of the 10 rolls is a twelve. In order to imagine the complement, it is helpful to frame this in terms of sets. If you are lacking an understanding of set theory, I highly recommend acquainting yourself with it if you are undertaking any serious exploration of probability theory. 
Let $S$ be the sample space, that is, the set of all possible outcomes after rolling the die 10 times. You can imagine each outcome in $S$ as a sequence of 10 numbers representing the result of the rolls after performing the experiment. Stated formally,
$S = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}):x_i\in\{1,2,...,12\}$ for $i = 1,2,...,10 \}$
Now, imagine the event $E$ is a subset of $S$ consisting of all outcomes in which 2 or more of those 10 rolls is the number 12 - in any order. Stated formally,
$E\subseteq S$ s.t. $E = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}): (\bigcup\limits_{i=1}^{10} x_i=12)$ $\wedge$ $(\bigcup\limits_{i=j}^{10} x_j=12)$ whenever $i \neq j \}$
With this definition in mind, we want the probability with which the result of the experiment will be a member of $E$, or in other words, we wish to know $P(E)$.
By the complement rule, we know $1-P(E^c)=P(E)$ where the set $E^c$ is the complement of $E$. Hence, as you have discovered, if we can determine $P(E^c)$, then we can solve for $P(E)$. So then, how do we define $E^c$?
By definition of complement, $E^c$ is the set of all members of $S$ that are not members of $E$. In other words, $E^c$ consists of all outcomes in which it is NOT the case that 2 or more of those 10 rolls is the number 12. Well, that is only the case if 1 or less of the 10 rolls is the number 12. Since we cannot have fewer than 0 rolls being the number 12, we know $E^c$ consists of all outcomes in which either no rolls are 12 or only 1 of 10 rolls is 12. This can be stated formally below...
Given $S$ is the sample space, or our "universal set," we can define $E^c$ compactly as follows: 
$E^c = \{y:y \not \in E\}$ where $y=(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10})$
By definition of negation, we have
$E^c = \{y:\neg(y \in E)\}$ 
and by substitution with definition of $E$ we have
$E^c = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}):\neg((\bigcup\limits_{i=1}^{10} x_i=12)$ $\wedge$ $(\bigcup\limits_{i=j}^{10} x_j=12))$ whenever $i \neq j\}$
By DeMorgan's Law
$E^c = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}):\neg(\bigcup\limits_{i=1}^{10} x_i=12)$ $\vee$ $\neg(\bigcup\limits_{i=j}^{10} x_j=12)$ whenever $i \neq j\}$
and by DeMorgan's Law again
$E^c = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}):\bigcap\limits_{i=1}^{10} \neg (x_i=12)$ $\vee$ $\bigcap\limits_{i=j}^{10} \neg (x_j=12)$ whenever $i \neq j\}$
and by definition of not equal
$E^c = \{(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}):\bigcap\limits_{i=1}^{10} x_i \neq 12$ $\vee$ $\bigcap\limits_{i=j}^{10} x_j \neq 12$ whenever $i \neq j\}$
So, the compliment of the probability in which 2 or more of the 10 rolls is the number 12 is... the probability in which either no rolls or only 1 of the 10 rolls is the number 12. If you can solve for the latter, then you can solve for the former via the complement rule.
