Box has 10 balls, 6 black, 4 white. Three drawn, color not recorded. What is the probability the fourth ball is white? In this question we assume the $10$ balls are equally likely to be drawn from the box.
What I did was to partition this and say:
$P(\text{Fourth Ball is white}) = P(4^{th} \text{ is } W|3 \text{ are } W) + P(4^{th} \text{ is } W|2 \text{ are } W, 1 B) + P(4^{th} \text{ is } W|1 \text{ is } W , 2 B) + P(4^{th} \text{ is } W|0 \text{ are } W, 3 \text{ are } B)$.
Doing this gave me a number greater than 1. Not sure where I went wrong. 
 A: We described the easy way in the comments. So let's do it the hard way. Let $B_0$ be the event there are $0$ black in the first $3$, $B_1$ be the event there is $1$ black in the first $3$. Define $B_2$ and $B_3$ analogously. Let $W$ be the event the $4$-th is white. Then
$$\Pr(W)=\Pr(B_0\cap W)+\Pr(B_1\cap W)+\Pr(B_2\cap W)+\Pr(B_3\cap W).$$
Now let us calculate the various probabilities.
For example, $\Pr(B_0\cap W)=\Pr(W|B_0)\Pr(B_0)$. We have $\Pr(W|B_0)=\frac{1}{7}$ and $\Pr(B_0)=\frac{4}{10}\cdot\frac{3}{9}\cdot \frac{2}{7}$. Another expression for $\Pr(A_0)$ is $\frac{\binom{6}{0}\binom{4}{3}}{\binom{8}{3}}$.
Let's do another one. We have $\Pr(B_1\cap W)=\Pr(W|B_1)\Pr(B_1)$. We have $\Pr(W|B_1)=\frac{2}{7}$. And $\Pr(B_1)$ can be done in various ways. For instance it is $\frac{\binom{6}{1}\binom{4}{2}}{\binom{10}{3}}$.
Now you can do the remaining two calculations, for $B_2$ and $B_3$, add up, and simplify. After a while you should get $4/10$. And the unpleasantness of the work will show how useful the other viewpoint is. 
