I'm not too great with probability, but I think you're right with \(\frac{1}{2}\times \frac{3}{7}\).
You can look at it like what are the chances of picking the first ball as red and the second ball as red.
First we need to pick one red ball. So you look at how many ways can you pick a red versus how many picks there are. That is \(\frac{4}{8}=\frac{1}{2}\).
So now we have looked at picking the first red ball, and we move on to picking the second. There are 3 red balls out of 7 total, so \(\frac{3}{7}\).
All in all, there are 4 out of 8 ways a red ball would be picked first; then, out of those 4 ways, only 3 ways out of 7 will give you a red ball.
Out of all the ways to pick that exist, I think \(\frac{4}{8}\times\frac{3}{7}\) is the answer.