A problem with too many counting required 
We have $10$ bags of candies each with $10$ colorful candies    It is
  known that at least $1$ of the bags contains exactly $5$ red
  candies
We get $1$ candy from each bag what is the probability that we get
  an even($0$ is considered even) number of red candies?

I assumed that each bag that doesn't contain exactly $5$ red candies contain 1 candy of each color(1 red, 1 blue, 1 green...) because I didn't think its possible to solve it without it, but even with this assumption I got stuck pretty quickly.
My question is do we need that assumption at all(cause I'm pretty sure I'm supposed to work without it)?
And more importantly how to solve such a problem with so many cases to consider (0,2,4,6,8,10 -> each of these cases could be a question of its own) 
 A: Somewhat unexpectedly, it turns out that you don't need any information about the distribution of colors in the remaining bags.  This, of course, is a property of symmetry...if you replace the $5$ in your problem with anything else, then the question can not be answered without additional information.
Suppose we know that bag $\#1$ contains exactly $5$. Of course the probability is exactly $.5$ that we get a red from that bag.  What about the $9$ others?  Well, let $p_E$ be the probability that we get an even number (from the remaining $9$), and $p_O$ the probability that we get an odd.  Of course $p_E+p_O=1$  so your answer is $$\frac 12\times p_E+\frac 12\times p_O=\frac 12\times (p_E+p_O)=\frac 12$$
A: The problem as stated is not well posed, as we don't know what is the probability that a bag contain exactly $n$ candies. We could have for example that each candy has probability $\tfrac{1}{p}$ of being red for some $p$, or that the probability for a bag to contain exactly $n$ red candies is exactly $\tfrac{1}{10}$, and those two cases would give drastically different answers.
