Example of calculating conditional probabilities We fill a bag with three fruits. Every fruit we put in the bag has the same chance to be an apple, a banana, or a pear. One randomly pulls a fruit out of the bag, it's an apple. What is the probability that the others two fruits in the bag to be apples as well?
I would say that we don't care about the result of the first draw and the probability that the other two are apples is $(\frac 13)^2$
 A: The three fruit are placed into the bag and then one of the three is randomly pulled out.
We may assume without loss of generality that the first fruit placed in the bag was indeed the same fruit as what was pulled out.  We are told that the fruit pulled out was an apple.  This leaves just the question of what the second and third fruit were.  They were each independently an apple with probability $\frac{1}{3}$ for a final probability of $$\frac{1}{9}$$
If this is unconvincing, go ahead and explicitly refer to all $3^4=81$ possible events, keeping track of not only the outcome of the first fruit but also which of the fruits was the one randomly selected and approach directly with definition of conditional probability.
You will find that of these $81$ equally likely different possible events, $3\times 1\times 3\times 3$ correspond to the fruit being drawn being an apple (pick which numbered fruit was the one that was drawn (3 options) set it as an apple (1 option) and pick the type of fruit for each of the remaining (9 options)).  Of these, $3\times 1\times 1\times 1$ correspond to all fruit being apples.  This gives a probability of:
$$\frac{3\times 1\times 1\times 1}{3\times 1\times 3\times 3} = \frac{1}{9}$$

The wording of the question is important.  What we are conditioning on is not the event that there is at least one apple in the bag... what we are conditioning on is that when randomly pulling a single fruit from the bag that it was an apple.  The exact text from the OP is "One randomly pulls a fruit out of the bag, it's an apple."  This is different than "Timmy only likes to eat apples and so he searches the bag for an apple and intentionally pulls one out."  In our actual problem, there was a chance to fail to have procured an apple even in the scenario that there were apples available.  Timmy searching the bag there is no chance to fail to procure an apple except in the scenario that there were no apples in the bag at all.
Compare this to the Boy-Girl Problem and very specifically to the First Question in the boy-girl problem where we are told that the eldest child is a girl.
Compare this to the boy-girl problem where a girl answers the door when the doorbell is rung but we don't know whether it was the elder or the younger child.  Here, even though we don't have a way of knowing the child's relative age... we are still able to distinguish her from her sibling by the very fact that it was she and not her sibling who answered the door.  This is the same scenario as we are in here with this problem.  The fruit we happened to have pulled was an apple.  We did not seek out an apple intentionally.
This is a different question than the Second Question where we are told at least one of the children are girls but not which and are given no information or way to distinguish which.
A: I apologize to both @JMoravitz and @Cerice for causing confusion. I originally had an answer of $\frac{1}{19}$ because I reckoned that there were $27$ possible bags of fruit and only $8$ of them don't have an apple which leaves $19$ possible bags of fruit of which only one contains three apples.
This reasoning is incorrect because although there is only one possible bag of fruit with three apples there are three possible ways to observe it. The apple pulled from the bag could have been the first, second or third apple from the bag with three apples.
If the apple drawn was the first fruit placed then there are $3\times 3=9$ choices for the other two.
Likewise, if the apple drawn was the second fruit placed there are $9$ choices for the other two and if the apple drawn was the third fruit placed there are $9$ choices so there are $27$ possible ways to observe a single apple selected at random and $3$ of these contain all apples. Therefore, the probability of observing two more apples is $\frac{3}{27}=\frac{1}{9}$.
I only posted this to clarify any confusion that I may have caused. Please accept JMoravitz's answer since he suffered from no such malady.
