conditional probability dice problem Probability of rolling:
1: 0.3, 2: 0.1, 3: 0.15, 4: 0.15, 5: 0.15, 6: 0.15
So the probability of not rolling number 1 and 2 in 8 rolls is $(0.6)^8$ which is about $0.0168$
What is the probability of numbers 1 and 2 both occurring at least once in 8 rolls?
I do not understand why it isn't $1-0.0168$. Answer is $0.5287$
 A: Let $A$ be the event that a 1 shows in eight rolls.   Let $B$ be the event that a 2 shows at least once in eight rolls.
You have found $\mathsf P[A^\complement\cap B^\complement] = 0.6^8$   The probability that neither number occurs in eight rolls.
You want to find $\mathsf P[A\cap B]$.   The probability that both numbers show (at least once) in eight rolls. 
  These events are not complements.
You should use: $\mathsf P[A\cap B] = \mathsf P[A]+\mathsf P[B]-\mathsf P[A\cup B] \\\qquad\qquad = (1-\mathsf P[A^\complement])+(1-\mathsf P[B^\complement])-(1-\mathsf P[A^\complement\cap B^\complement])$
A: The probability of getting at least one $1$ but no $2$s is
$$P(1,3,4,5,6) = \sum_{k=1}^8 {8 \choose k}(0.3)^k(0.6)^{8-k}.$$
The probability of getting at least one $2$ but no $1$s is
$$P(2,3,4,5,6) = \sum_{k=1}^8 {8 \choose k}(0.1)^k(0.6)^{8-k}.$$
The probability of getting no $1$s or $2$s is
$$P(3,4,5,6) = (0.6)^8.$$
Then, $$1 - P(1,3,4,5,6) - P(2,3,4,5,6) - P(3,4,5,6) = 1 - 0.4713 = 0.5287.$$
A: Shown Using the Methods in the Question
The probability that a $1$ shows up in $8$ rolls is $1-0.7^8$
The probability that a $2$ shows up in $8$ rolls is $1-0.9^8$
The probability that a $1$ or $2$ shows up in $8$ rolls is $1-0.6^8$  

Using Inclusion-Exclusion
Inclusion-Exclusion for two sets says $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Thus, the probability that a $1$ or $2$ shows up in $8$ rolls, $P(A\cup B)$, is the probability that a $1$ shows up in $8$ rolls, $P(A)$, plus the probability that a $2$ shows up in $8$ rolls, $P(B)$, minus the probability that a $1$ and $2$ shows up in $8$ rolls, $P(A\cap B)$.
Therefore, the probability that a $1$ and $2$ shows up in $8$ rolls is
$$
\overbrace{\left(1-0.7^8\right)}^{\large P(A)}+\overbrace{\left(1-0.9^8\right)}^{\large P(B)}-\overbrace{\left(1-0.6^8\right)}^{\large P(A\cup B)}=\overbrace{0.52868094\vphantom{\left(1^1\right)}}^{\large P(A\cap B)}
$$
