Conditional Probability Cupcakes This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with conditional probability, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Two boxes each contain $4$ cupcakes.
One box has $3$ chocolate and $1$ vanilla, and the other box has $2$ chocolate and $2$ vanilla.
A box is randomly selected, opened, and a cupcake is randomly selected.
This first cupcake is vanilla.
If one more cupcake is randomly selected from the same box, what is the probability that it will be vanilla ?

My working so far:
If we pick a box at random, and pick a vanilla cupcake,
then there is a $\frac23$ probability we picked the second box.
($2$ chances for having picked vanilla from there, against $1$ from $1$), and
$\frac13$ for the first box.
If we DID pick the first box, the chances of $2nd$ vanilla are $0$, since
there isn't one, so $\frac13 × 0$ for that.
Now I am stuck.
 A: Defining events:
$v_1$: a vanilla cupcake is selected at first
$v_2$: a vanilla cupcake is selected at second
$b_1$: a vanilla cupcake is selected from box 1
$b_2$: a vanilla cupcake is selected from box 2
It is asked about $P(v_2|v_1)=\frac{P(v_1 \cap v_2)}{P(v_1)}$
$P(v_1 \cap v_2)$: Probability, that two vanilla cupcakes are selected.
It is impossible to select two cupcakes from box 1.
Thus $P(v_1 \cap v_2)=p(b_2)\cdot P(v_1)\cdot p(v_2|v_1)=0.5\cdot \frac{2}{4}\cdot \frac{1}{3}=\frac{1}{12}$
$P(v_1)=P(b_1)\cdot P(v_1|b_1)+ P(b_2)\cdot P(v_1|b_2)=\frac{1}{2}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{8}$
Therefore $P(v_2|v_1)=\frac{1}{12}\cdot \frac{8}{3}=\frac{2}{9}$
A: Let's define the following events: the event that first box is chosen is $B_1$, the event that the second box is chosen is $B_2$, the event that the first cake is vanilla is $V_1$, the event that the second cake (from the same box) is vanilla is $V_2$. What you want to find is exactly $\Pr(V_2 | V_1)$.
First, we have, by definition,
\begin{align}
&\Pr(V_2 | V_1) = \frac{\Pr(V_1, V_2)}{\Pr(V_1)}
\end{align}
Next, we find those two probabilities on RHS,
\begin{align}
\Pr(V_1, V_2) &= \Pr(V_1, V_2, B_1) + \Pr(V_1, V_2, B_2) \\
&= \Pr(V_1, V_2 | B_1) \cdot \Pr(B_1) + \Pr(V_1, V_2 | B_2) \cdot \Pr(B_2) \\
&= 0 + \frac{1}{6} \cdot \frac{1}{2} \\
&= \frac{1}{12}
\end{align}
and
\begin{align}
\Pr(V_1) &= \Pr(V_1 | B_1)\cdot\Pr(B_1) + \Pr(V_1 | B_2)\cdot\Pr(B_2) \\
&= \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \\
&= \frac{3}{8}
\end{align}
A: Working under condition that the first cake chosen is vanilla: let $E$ denote the event that the box picked at random contains another vanillacake. 
There are $3$ vanillacakes that can have been chosen with equal probability and in $2$ of these cases the box contains another vanillacake (exactly one). 
This tells us that $P\left(E\right)=\frac{2}{3}$.
If $V$ denotes the event that the second selected cake is a vanillacake.
Then $$P\left(V\right)=P\left(V\mid E\right)P\left(E\right)+P\left(V\mid E^{c}\right)P\left(E^{c}\right)=\frac{1}{3}\frac{2}{3}+0\frac{1}{3}=\frac{2}{9}$$
Here $P(V\mid E)=\frac13$ because under event $E$ exactly $1$ of the $3$ cakes left to be chosen is vanilla.

This is in fact your own approach, wich is okay except that you failed completing it. 
A: You have realised that P(first vanilla came from box 2) = $\frac{2}{3}$, 
and that only the 2nd box can contribute to the indicated probability.
"Now I am stuck."
Just continue t0 $\frac{2}{3} \times$ P(draw another vanilla from box 2) = $\frac{2}{3}\times \frac{1}{3} = \frac{2}{9}$
