I'm in Canada, and we've been using chip cards for a few years now. I just called my bank 45 minutes ago after noticing a fraudulent charge on my credit card from August 30th. Since I bought a bunch of stuff at Home Depot in May/June, I'm assuming they managed to clone my card from the stolen data. The charge was only $4.56, at a gas station halfway across the country, so I would guess that someone was testing the clone to see if it was a valid card number (maybe testing one number from a batch of 100s or 1000s, to see if the numbers were legit.)
Just so we're clear, I'm not saying the fraudulent purchase itself was made using the chip. I only ever use chip + pin when making purchases, but I suppose a cloned card could use NFC (eg: PayWay) for a purchase that small, or even just the magstripe, neither of which requires them to have compromised my pin. My point is that I thought I was being safe using chip + pin, but still got hit regardless. Fortunately, banks seem to be good about this sort of thing, and my new card is on its way.
Google says:
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.
Trying to be happy is like trying to build a machine for which the only specification is that it should run noiselessly.