Dice and Balls Probability Question: Where Did I Go Wrong? The Question
We roll a die. If we obtain a $6$, we choose a ball from box A where three balls are white and two are black. If the die is not a six, we choose a ball from box B where two balls are white and four are black. What is the probability of selecting a white ball?
My Analysis
First, we find the sample space. We either roll a $6$ or we don't. If we roll a $6$ (exactly $1$ way to do this), there are $5$ options. If we don't roll a $6$ (there are 5 ways to do this), we have $6$ options. Therefore, there are $1 \times 5 + 6 \times 5 = 35$ outcomes to this experiment. 
Now we find the event space. If we roll a $6$, there are $3$ ways to select a white ball, leading to $3$ total outcomes. If we don't roll a $6$, there are $2$ white balls to choose from, leading to $5 \times 2 = 10$ outcomes. By rule of sum, there are $3 + 10 = 13$ outcomes that end in a white ball being drawn.
Therefore, the probability is $\frac{13}{35}$.
My Problem
The book gives the answer $\frac{17}{45}$ which is actually quite close to my answer, but not exactly equal. So, what's the problem with my analysis, stackexchange?
 A: The problem with your solution is that you assumes each outcome is equally likely.
In reality, the outcome "The first ball is selected from box $A$" is different from the outcome "The first ball is selected from box $B$".
Or think this way, "The first ball is selected from box $A$ given $6$ is rolled" is different from the outcome "The first ball is selected from box $B$ given $1$ is rolled".
Essentially, every ball in box $A$ can be selected $6\over 5$ times as likely as every ball in box $B$ can be selected.
So the event $6$ is rolled and white ball $X$ is selected from box $A$ is $6\over 5$ times as likely as the event $5$ is rolled and white ball $Y$ is selected from box $B$.
BTW: the correct way: ${1\over 6}\cdot{3\over5} + {5\over6}\cdot {2\over6}$
A: The 35 events are not equally likely. 
5 of them will occur 1/6 of the time (so each have a 1/30 chance of happening; not a 1/35)
30 of them will occur 5/6 of the time (so each have a 1/36 chance of happening; not 1/35)
So the ways to get a white are $3$ with a probability of $1/30$ and $10$ with probability $1/36$.  So I get $1/10 +5/18=9/90+25/90=34/90=17/45$.
A: $\begin{array}{c} 6 & \circ&\circ&\circ&\bullet&\bullet&\times
\\ 5 & \circ&\circ&\bullet&\bullet&\bullet&\bullet
\\ 4 & \circ&\circ&\bullet&\bullet&\bullet&\bullet
\\ 3 & \circ&\circ&\bullet&\bullet&\bullet&\bullet
\\ 2 & \circ&\circ&\bullet&\bullet&\bullet&\bullet
\\ 1 & \circ&\circ&\bullet&\bullet&\bullet&\bullet
\end{array}$
Using your method we the outcomes would be equally likely only if there were thirty six of them; a selection from six balls for each result of the die.   However, as there are only 35 outcomes, then to obtain the proper weights, we need to treat the forbidden ($\times$) outcome as $3/5$ in the white event, and $2/5$ in the black event.   So the probability for selecting a white ball is: $$\dfrac{13.6}{36}=\dfrac{17}{45}$$

More rigorously, the probability for drawing a white ball is calculated as: $$\tfrac 16\cdot\tfrac 35+\tfrac 56\cdot\tfrac 26 = \tfrac {3.6}{36}+\tfrac{10}{36}=\tfrac{17}{45}$$
A: I find drawing a tree diagram is always helpful:

We can see the probability of drawing a white ball from box A is $1/10$, and the probability of drawing a white ball from box B is $5/18$, so the probability of drawing a white ball from either box is
$$\frac{1}{10}+\frac{5}{18}=0.377\cdots=\frac{17}{45}$$
You could approach your error from a formulaic stance: you are using the formula $P(E)=n(E)/n(U)$, but in actuality this formula doesn’t apply to combined events. Your event $E$ is a combined event $A\cap w_A$ (rolling box A and then drawing a white ball) united with the mutually exclusive event $B\cap w_B$ (rolling box B and then drawing a white ball), so you need to calculate
$$\begin{align}
P\left[ (A\cap w_A) \cup (B\cap w_B)\right] &= P(A\cap w_A) + P(B\cap w_B) \\
&= P(A)\,P(w_A) + P(B)\,P(w_B) \\
&= \frac16\frac35+\frac56\frac26 \\
&= \frac{17}{45} \\
\end{align}$$
As you can see, once you break down the probability in terms of non-combined events, you can use the counting formula.
