[RocketManJames]

I reasoned that there need only be two types of balls, Blue and Non-Blue (let's call it W), and that (B c 5) / [(W+B) c 5] = 1/2.

This reduced to
[B*(B-1)*...*(B-4)] / [(B+W)*(B+W-1)*...*(B+W-4)] = 1/2.

I did not see a straightforward way to get a B and W that satisfies the above, so I had to iterate using a computer.

The result I got was B=9, W=1.