2 ms·
Except that if there are 4000 planes in the sky, that does make (assuming the planes are not overlapping, which seems largely safe) the probability 4000 times l
by lutorm 17y ago
Except that if there are 4000 planes in the sky, that does make (assuming the planes are not overlapping, which seems largely safe) the probability 4000 times larger than for one plane.
As long as the planes don't overlap, the flight hours is the correct number to use, because more planes in the sky gives higher probability. It's not the number of meteors that matter, but the number of "potential meteor strikes".
Equivalently, you can say that the probability of a meteor hitting any one plane at a given time is the fractional area taken up by planes, which if there are 4000 planes in the air is 4000*5.82e-13. (Again assuming the planes don't overlap, which since 5.82e-13<<1 is a safe assumption.)
- tomsaffell 17y agoAh, yes - I mis calculated p(strike of any meteor) by a factor of 3500 (the # of planes in the sky, according to the article). I think this should be applied to the probability, not the count of the occasions - but the numbers are small enough that it doesn't make much difference. p(one or meteor strikes) = 1 - ( (1 - p(strike of any meteor))^ (total # meteors) ) (assuming statistical independence, etc). p(strike of any meteor) = 5.82e-13 * 3500 = 2.04E-9 (total # meteors) = 3000 * 365 * 20 (for testing over 20 years) p(one or meteor strikes) = 0.0436 = 4.36%
- lutorm 17y agoYeah, it doesn't make a difference until the planes cover a significant area of the Earth, in which case just multiplying by the numbers of planes will be wrong for the same case that the initial calculation was wrong. Then it would be 1-(1-5.82e-13)^3500 in this case, too.