I'm going to go ahead and take this to mean that the TSA just doubled my chances of death. Seems appropriate in the current climate of hysteria, and makes an excellent sound bite.
It doesn't make a very good soundbite, but it plays well to the climate of hysteria!
At the (incredibly optimistic) risk rate of 1 in 30 million, that's about 20 people given cancer, each year, if this policy goes forward and these devices become standard procedure.
Are you willing to kill 20 people per year to keep safe?
I wonder how long the cancers will take to start showing up.
Unfortunately, the answer for most is yes. That's more or less the definition of war.
Edit: Yikes! I'm really getting downvoted here, but no one's saying why. I didn't say I agreed with it! I just said that most people would probably say yes.
If terrorists never attempt another plane attack they can count every cancer death that results from the scanners as part of their death toll. And they never have to even waste a resource or another moment of planning.
In fact if I were a terrorist I'd be looking at not only the effect of the attack itself, but the downstream residuals. For example, attacking the water supply would have large casualties, but little downstream residual except tightening security around water supplies.
But attacking elementary schools would result in less upfront damage, but if you could force elementary school kids to walk through these types of scanners everyday of their elementary school lives -- well chalk up more deaths for the terrorists.
I really really hope the government knows something we don't, otherwise that tired old saying "then the terrorists have won" is really beginning to apply.
1 - (1-p)(1-p) = 1 - (1-2p+p^2) = 2p - p^2 ~ 2p
(where the last step is because p^2 << p because p << 1; the basic idea is that it's "certainty minus the probability that neither occur")
We can safely assume that the chance of their occurrence together is going to be very small and so the chance of one or the other is approximately the sum of their individual probability.
P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y) ~ P(X) + P(Y)