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Re:Chance of Dying From Backscatter Radiation Same as Chance... by noonespecial on Thursday November 18, 2010 @03:34PM (#1919134)
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.
Re:Chance of Dying From Backscatter Radiation Same as Chance... by eleusive on Thursday November 18, 2010 @04:18PM (#1919292)
I'll take it to mean that (given the 1 in 30 million estimate and that there are well over 250 million outgoing airplane seats filled per year by U.S. airlines), the expected number of scans which will cause cancer is nearly 10. That's 10 people who's lives may be put in very real danger.

It doesn't make a very good soundbite, but it plays well to the climate of hysteria!

Re:Chance of Dying From Backscatter Radiation Same as Chance... by roc on Thursday November 18, 2010 @04:33PM (#1919350)
According to the Bureau of Transportation Statistics, there were 617,977,711 passengers on domestic flights in 2009 [1].

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.

[1] http://www.transtats.bts.gov/Data_Elements.aspx?Data=1

Re:Chance of Dying From Backscatter Radiation Same as Chance... by drawkbox on Thursday November 18, 2010 @04:47PM (#1919397)
The TSA kills 20 people per year. Keep the hysteria simple.

Are you willing to kill 20 people per year to keep safe?

Re:Chance of Dying From Backscatter Radiation Same as Chance... by KiwiNige on Thursday November 18, 2010 @05:22PM (#1919528)
It's the TSA agents standing by the machines all day that are first in the fireing line.

I wonder how long the cancers will take to start showing up.

Re:Chance of Dying From Backscatter Radiation Same as Chance... by OneWhoFrogs on Thursday November 18, 2010 @04:59PM (#1919439)
> Are you willing to kill 20 people per year to keep safe?

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.

Re:Chance of Dying From Backscatter Radiation Same as Chance... by msquared on Thursday November 18, 2010 @05:50PM (#1919632)
Uhh, wouldn't these things not have even caught the underwear bomber? Soo.. would you kill 20 people a year for the false sense of security and safety? If so.. well.. you're a douche nozzle, no offense.
Re:Chance of Dying From Backscatter Radiation Same as Chance... by KiwiNige on Thursday November 18, 2010 @05:24PM (#1919536)
I suppose it depends on how many attacks are intercepted by the scanners.
Re:Chance of Dying From Backscatter Radiation Same as Chance... by kenjackson on Thursday November 18, 2010 @05:57PM (#1919653)
Which also depends on how many attampted attacks there are.

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.

Re:Chance of Dying From Backscatter Radiation Same as Chance... by kwantam on Thursday November 18, 2010 @05:28PM (#1919555)
VVVVVV Ahh, ayup, you're correct. Brainfart.
Re:Chance of Dying From Backscatter Radiation Same as Chance... by andrewcooke on Thursday November 18, 2010 @05:51PM (#1919635)
Sorry for being pedantic, but no - these are probabilities not random variates. If the probability of either alone is p then the probability of one or the other or both is:

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")

Re:Chance of Dying From Backscatter Radiation Same as Chance... by joe_the_user on Thursday November 18, 2010 @06:05PM (#1919675)
Not close to true for two small, independent events.

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)