- To make the math simple, let's first assume CMEs can be fired in any direction.
- For a CME to hit the Earth, it has to occupy the same space as us at the same time.
- The Earth is approx 1 AU from the sun at any given time; so to hit the Earth, the CME has to hit a particular spot on a sphere of space 1 AU in radius.
- So the probability of a given CME hitting Earth is approximately equivalent to the ratio of half the Earth's surface area (since only half faces the Sun at a time) to the surface area of a sphere with a radius of 1 AU.
- 1 AU = 149,597,871 km
- Surface area of a sphere is 4*pi*r^2, so our orbital sphere has an area of approx 2.8 x 10^17 km^2.
- Surface area of the Earth = 510,072,000 km^2, or 5.1 x 10^8 km^2
Therefore the probability of being hit by a given CME is (2.8 x 10^17) / (5.1 x 10^8) = 5.5 x 10^-8, or a 0.0000055% chance.
Now the number of CMEs per year is actually higher than I expected, which I suppose explains why we do in fact get hit between 0 - 70 times per year. However the number of annual large CMEs is quite low, with none of the sites I visited actually agreeing on the number (most seemed to agree it's less than 5 per year in a solar maximum.) Let's say there are 5 per year. That only brings the chance of being hit by one of them up to 0.000028% per year. So if I live to be 100, the chances I'll see one in my lifetime are only 0.0028%.
caveat: These calculations ignore CME cross-section (essentially width and height) and duration (essentially length), since I couldn't find any accurate information on those. If you find those, you can factor them into these calculations by multiplying by the cross-section, multiplying by the % duration that the CME's strength is high, and multipyling by the Earth's average orbital velocity. That will modify the probility to take into account the volume of space the Earth occupies while the CME is traversing the edge of our 1 AU sphere, and how much of the surface of the sphere is touched by the CME.