Question:
There are $3$ targets $A, B, C$. Whenever you "shoot", there is an equal chance of hitting any target. Obviously in this case the probability would be $1/3$. If you were to shoot twice (one shot after the other), what is the probability that targets $A$ and $C$ will be hit? If a target is already is hit, it is still there and there are no changes.
From my understanding, only the sequences $A, C$ and $C, A$ will be valid. Do I list out all possible variations of these two shots and then have the numerator as $2$ and the denominator as all the possible variations OR do I calculate the probability of $A$ hit with the first shot and then calculate the probability of $C$ hit with the second shot and also add the probability of sequence $C, A$ after.
In the first way, there are a total of $3\times3=9$ possible ways to hit $2$ times. As you have pointed out, only "$AC$" and "$CA$" satisfy the conditions. Hence, the probability is $\dfrac{2}{9}$.
In the second way, there is a $\dfrac{1}{3}$ chance of hitting $A$ and then another $\dfrac{1}{3}$ chance of hitting $C$, giving a total of $\dfrac{1}{9}$ and similarly for hitting $C$ then $A$. Hence, the probability is still $\dfrac{2}{9}$.
Either way works, but I'd say the second way is simpler to understand.