How do I model the probability that a straw is missing and another is in perfect condition

Question

In a pack of six chocolate drinks every carton is supposed to have a straw, but it is missing with probability $1/3$, with probability $1/3$ it is broken, with probability $1/3$ it is in perfect condition. Let $A$ be the event 'At least one straw is missing and at least one is in perfect condition'.

Question: Find a suitable probability space, formulate $A$ set-theoretically and find $P(A)$.

My ideas:

$\Omega:=\{1,2,3\}^6$ and $\mathcal{F}:=2^{\Omega}$

Where

$1:=$ "Straw missing"

$2:=$ "Straw broken"

$3:=$ "Straw in perfect condition"

$A:=\{\omega \in \Omega:\exists i,j \in \{1,2,3,4,5,6\}$ while $i \neq j$ and $\omega_{i}=1,\omega_{j}=3\}$

I think until this point I am correct. But I am struggling to find $P(A)$:

It may to be easier to look at $A^{c}:=$At most one straw is missing or at most one is in perfect condition. But I cannot come up with anything adequate.

I've though of something:

$P(A)=1-P(A^{c})=1-((\frac{1}{3})^{6}+\frac{6!}{5!1!}(\frac{1}{3})^{6}+\frac{6!}{5!1!}(\frac{1}{3})^{6})=\frac{716}{729}$

Answer 1

It is easier to compute $A^c$, which is that either no straws are missing or none are in perfect condition. There are $2^6$ ways you can have no straws missing, because each must be either broken or perfect. There are also $2^6$ ways you cam have no straws perfect. We have counted all broken in both cases, so $|A^c|=2\cdot 2^6-1=127$. The probability of $A$ is then $1-\frac {127}{3^6}$

Answer 2

If you're not familiar with the inclusion-exclusion principle just define the following events:
1)$A_1:=${$w\in\Omega:$ $\forall i, w_i \neq 1$}
2)$A_3:=${$w\in\Omega:$ $\forall i, w_i \neq 3$}
$P(A_1)=P(A_3)=\frac{2⁶}{3⁶}$, but if you want to find $P(A_1\cup A_3)$ it is:
$P(A_1\cup A_3)= P(A_1)+P(A_3)-P(A_1\cap A_3)=\frac{2⁶}{3⁶}+\frac{2⁶}{3⁶}-\frac{1}{3⁶}=\frac{127}{729}$, and for you remains to see why $A_1\cup A_3$ is actually $A^c$