Conditional probability with "at least" We split 8 colored (and distinguishable from each other[each ball is unique]) balls to 4 kids, 2 balls for each kid.
There are 2 blue balls, 2 red balls, 2 yellow balls, 2 green balls [still each ball is unique]  

A) It is known that Amy got at least 1 red ball, what is the
  probability that also John got at least 1 red ball?
B) It is known that Amy got balls of different colors, what is the
  probability that also John got balls of different colors?

What I have done is as follows:
A) let A be "Amy got at least 1 red ball"(B the same for John) hence $P(A)=1-{6\over8}*{5\over7}={13\over28}$ (all the cases subtract the case in which the first ball isn't red and also the 2nd isn't red)
So $P(A|B)={{{(2*{2\over8}*{6\over7})}^2}\over{13\over28}}=0.395$
This is because there is $2\over8$ chance to get red ball first and then $6\over7$ chance of getting non red ball, this chance is multiplied by 2 since we could do it the other way around(1st non red 2nd yes red) and then I square it all since the same chance applies to B(John).
I feel like I may have done it too complicated? is it even the right answer? not sure.
B)first ball can be any color $8\over8$ the 2nd ball has to be different than the first so $6\over7$ chance of that.
But now when I tried to find the new(A is now different color balls so is B) $P(A|B)$ I got something weird like that: ${({6\over7}*{6\over7})\over{6\over7}}={6\over7}$ which I really dont feel good about

We solve the first problem. Let $A$ be the event Amy got at least one red, and $B$ the event John got at least one red. We are asked to find $\Pr(B\mid A)$, which by definition is equal to $\Pr(A\cap B)/\Pr(A)$.
We compute the two required probabilities. You found $\Pr(A)$ correctly. Now we need $\Pr(A\cap B)$. This is the probability Amy and John each got one red.
Imagine that Amy drew a ball, then another, then John drew a ball, then another. The probability Amy got exactly one red is $\frac{2}{8}\cdot \frac{6}{7}+\frac{6}{8}\cdot\frac{2}{7}$, that is, $\frac{24}{56}$.
Given that Amy got exactly one red, the probability John got a red is $\frac{1}{6}\cdot \frac{5}{5}+\frac{5}{6}\cdot \frac{1}{5}$, that is, $\frac{10}{30}$. 
Thus $\Pr(A\cap B)=\frac{24}{56}\cdot \frac{10}{30}$. Now we can find $\Pr(B\mid A)$.
It might be a little smoother to use binomial coefficients. For example, the probability that Amy gets exactly one red is $\frac{\binom{2}{1}\binom{6}{1}}{\binom{8}{2}}$.
The second problem is in a sense somewhat easier than the first. Change the meanings of $A$ and $B$ in the obvious way. You can find $\Pr(B\mid A)$ directly, without finding $\Pr(A)$ and $\Pr(A\cap B)$.
