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rthmjohn

Chances of dying in a plane crash

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They say the chances of dying in a plane crash are 1 in 9 million. Am I correct in thinking that these chances increase the more you fly in airplanes? If that is true then how do you calculate your chances of dying in a plane crash after every new flight you board? That is, what is the formula for calculating this?

 

Thanks,

John

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How many flights are there every year globally?

How many crashes ar there every year globally?

What is the ratio of safe flights to crashed flights annually?

 

Then... How often do you fly?

 

Compare that crash ratio to your own frequency of flight, and you have the probability of yourself being in a crash.

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The chances of dying in the first flight you go on it 1/9,000,000.

 

If you go on a second flight, the chance of dying on that flight is also 1/9,000,000, but as you have now been on two flights, and it only takes one to kill you, you say (all are probabilities):

dead = ( flight1 crash & flight2 fine ) or (flight1 fine & flight2 crash ) or ( flight1&2 crash )

mathematically you say that as [p(x) is the probability of x]:

p(dead) = ( 1/9m * 8,999,999/9m ) + ( 8,999,999/9m * 1/9m ) + ( 1/9m * 1/9m ) = 0.0000002

 

However then adding a 3rd flight gets a bit more complex, and the expression will get even longer. Luckily there is a much better way of doing this! Instead of all that above, we could simply have said:

p(dead) = 1 - p(live)

[that is to say, probability of being alive or dead is 1 (it is certain you will either be alive or dead, no other option exists, at least not within the confines of your question), so the probability of being dead is one subtract the probability of being alive. e.g. if it is 1% likely you die, then it is 99% likely you live]

= 1 - ( flight1 fine & flight2 fine )

= 1 - ( 8,999,999/9m * 8,999,999/9m )

= 0.0000002

 

Getting back to your question (I word this the long way to help you understand better) we now have a much better formula. If you went on 3 flights now we can say:

p(dead) = 1 - p(live)

= 1 - ( flight1&2&3 fine )

and so on for more and more flights.

 

To express this mathematically in a slightly different way, using an exponential, instead of lots of 'and' probabilities (i.e. multiplications):

 

p(dead) = 1 - [ p(one flight fine) ]number of flights

 

so for x flights:

 

p(dead) = 1 - (8,999,999/9,000,000)x

 

So for 100 flights:

p(dead) = 1 - (8,999,999/9,000,000)100

= 0.00001

 

Interestingly, assuming your original statistic to be correct, if a pilot flies every 2nd day of his career (~150days/yr) which lasts 40yrs, the probability of him being in one crash (exponential used it 150*40) is 0.0006.

 

Hope that all helped!

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When I read the thread title I instantly thought that the chances were very high! Not quite, but near to 100% I would guess. i.e. If you are involed in a plane crash then you are likely to die.

 

Does anyone know the the chances of surviving a plane crash?

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The answer is actually very complicated, the chances of survival, like all things, depends on the details.

 

For example, if your plane is flown by Captain Sullenberger, survival is 100% so far. No one has yet died in any of his crashes.

 

http://abclocal.go.com/wabc/story?section=news/local&id=6606410

 

Is the crash falling 30,000 feet or simply the plane running off the end of the runway? Exactly what type of crash are we talking about?

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