Brian Pfleuger said:
Your calculation is not correct, you do not have the proper volumes involved.
The .308 bullet in question is .308 inches wide (diameter) so we will consider a plane that is .308 inches by 36,000 inches (1,000 yards). Any point on this plane will be struck by the bullet as it travels, to be completely accurate, we'd have to consider the full circumference of the bullet, but that's too complicated and would make it even more likely to hit a drop anyway.
Our plane has 36,000*.308=11,088 square inches of area. We will assume a rainfall rate of 1 inch per hour (quite heavy) and that gives us a total volume of 11,088 cubic inches of water on that plane per hour, or 11,088/3,600=3.08 cubic inches per second.
We will assume an average raindrop size of 3mm in diameter, which is probably a reasonable assumption since drops generally split when they exceed 4mm. That is 0.118 inches in diameter, or 0.059 in radius, which gives us a volume of 0.059^3(4/3)*3.141=0.00086 cubic inches per drop.
This means that each second, 3.08/0.00086=3,581 drops will pass through that plane. So, over the bullet's 2 second flight time, 7,162 raindrops will cross its flight path.
The math to calculate the probability of how many drops it will actually hit is quite a bit more complicated, but just from the number of drops, you can reasonably assume that it's going to hit quite a few of them.
Now, multiply that by 5 for the 5 inches per hour that the first analysis did (which is a hurricane).