Combinatorial and probability question I have a basic combinatorics and probability question (not homework) that I cannot seem to figure out because I have clearly misunderstood something. I apologise if this has been asked before as I did try, but am not aware of the right keywords to use. Nonetheless, it is as follows: 
Suppose I have 5 boxes $(a,b,c,d,e)$ to put 2 balls in, where each box can only take 1 ball at most. Then, suppose I am going to choose 2 boxes out of the 5. I would like to know the probability of both chosen boxes being empty, given that somewhere inside the 5 boxes, there will be 2 balls. So, here is my combinatorial perspective:
There are 5 boxes to put 2 balls in, thus there are $^5C_2 = 10$ possible configuration for this. Then, suppose I choose 2 boxes (say $a,b$). Then, out of the 10 configurations above, only 3 will have box $a$ and $b$ being empty. And this is the case regardless of which 2 boxes I choose, so the probability of 2 boxes being empty is $3\over10$?
From the probability perspective:
We have 5 boxes, and 2 balls, so the probability of a single box being filled is $2\over 5$. So, the probability of a single box being empty is $1 - {2\over 5} = {3\over 5}$. Then, the probability of 2 boxes being empty is ${3 \over 5}^2 = {9 \over 25}$, which is not the same as $3\over10$. 
Where have I gone wrong? It must be misunderstanding of some very basic concepts.
Thank you so much :) 
Jon
 A: Thanks to the comments, the second part of my answer is wrong, due to the fact that putting a ball into a box will then influence the probabilities of the remaining boxes having or not having a ball inside. The probability of the first ball being in a box is 6/10 (e.g. only 6 out of the 10 valid configurations will have box a being empty). However, this then means that out of these valid 6 configurations, only 3 will be valid for a second box (e.g. b) being empty, hence the final probability of 2 boxes being empty is ${6\over10}\times{3\over6}={3\over10}$.
A: Here's the correction for the probability approach.
The first ball is in box $a$ with probability $1/5$. If the first ball is not in box $a$, it must be in one of the others, so given the first wasn't in box $a$, the chance of the second ball being is $1/4$. This gives the probability of box $a$ containing a ball to be $1/5+4/5\times1/4=2/5$ (as correctly stated by the OP), and is empty with probability $3/5$.
Given box $a$ is empty, there are only 4 boxes left for the two balls to be in, so box $b$ is empty with probability $2/4$, giving the probability of two boxes being empty to be $3/5\times2/4=3/10$.
