jmc

Let $B$ denote drawing a black ball and $W$ denote drawing a white ball.

There are two possible successful orders: $BWBWBWBW$ or $WBWBWBWB.$

There are $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ ways to arrange four $B$'s and four $W$'s, so the probability that a random arrangement is successful is $\frac{2}{70}=\boxed{\frac{1}{35}}$.

OR

We could also compute this based on the probabilities at each step that we draw a ball of the opposite color. If we do that, we get $\frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}=\frac{144}{5040}=\boxed{\frac{1}{35}}$.