Does being involved in a plane crash reduce the probability of being involved in a second plane crash?

Question

I know that the two events are probably independent (unless you are the pilot), and therefore P (A!B) should be P(A), but isn't the probablity of being involved in two plane crashes lower than being involved in one plane crash?

Answer 1

being involved in a plane crash stays the same
but being involved in two decreases

Answer 2

Well, the probability of being involved in two plane crashes is less than that of being involved in one. Think about the odds of drawing an ace from two different unsorted decks of cards at the same time, and you'll understand that idea. However, the second time you get on a plane, the odds of you getting in a crash are the same. That's like saying that the odds of pulling an ace from the second deck is the same as pulling an ace from the first, even if you happen to have pulled an ace from the first.

In short, being in a plane crash once does not reduce the probability of being in a second plane crash. But the probability of being in two plane crashes is much lower than being in only one crash, because you must crash twice. To give you a simpler example: if you flip a coin, the probability that you will get heads is 0.5. The second time you flip, the probability is the same. However, the odds of getting heads twice in a row is (0.5)^2. If the probability of crashing in a plane is P, the probability of crashing twice in a row is P^2. If P is small, the odds against getting in two consecutive plane crashes may be very high! I don't know quite how to handle nonconsecutive plane crashes, but I suspect the odds against getting in multiple nonconsecutive plane crashes are high as well.

I see that Unstable has hit the nail on the head. You can only calculate the probability of both events happening before both of them happen. But if one happens first, the overall probability calculation changes. After one crash, you are still just as likely to get in another. However, before any crash happens, the probability of you getting into two crashes is very low.

Answer 3

No it doesn't. Just because you have been involved in one plane crash does not diminish your chances of being involved in another one. The probability is affected by how often you fly and at what time of the year and from where you fly: an actuary would be able to work out the probabilities to a greater degree of accuracy than the usual person.

The chances of there being two people who have been involved in two plane crashes are highly improbable and the chances increase for three people who have been involved in two plane crashes etc.

Answer 4

The probability of being involved in two plane crashes is the same as being involved in none (Provided you are alive in the first and continue to go by plane as you were going before the first). The probability of being involved in just one is the maximum. So it's certainly lower than being involved in just one.

Answer 5

This is true, being involved in 2 air crashes is much smaller than being involved in 1. But.. if you are already been involved in 1 aircrash, then the probability of being involved in the second one is the same, as if you were not involved in the first one. This is because by getting invoved in 1 air crash you already delete this crash from your equation, since its already have been, and has no more probability effect.. Probability is only about future and present.

Answer 6

No it doesn't reduce the probability. As you have identified they are independent.

Look at it this way, if an event has already occurred (it's probability is no longer uncertain and it's value become 1).

Your case for 2 consecutive crash calculation being lower only holds true if both events have not occurred yet. But if one of them has already occured you cannot then compute it any further as part of your probabilty determinations.

Look at it from a simpler situation like throwing a single die, the probability of throwing 2 consecutive 6 one after the other is 1 out of 36. But if you have already gotten a 6 for your first throw, the chance of you getting another 6 is no longer 1 out of 36, but 1 out of 6 instead.

Answer 7

the chances of being involved in a plane crash remains the same no matter if you've already been one.

it's just like picking a black chip from a bag with an equal number of black and white chips.
this first time is 50%
the second time will also be 50% (provided that you put the chip back of course).

but the chances of being involved in 2 plane crashes is different than being involved in 1.
just like the chances of getting two black chips = 0.5 x 0.5 = 0.25.

see the difference?

Answer 8

The probability of being in two plane crashes, which is the P(a!b).
Thats two factorials, so whatever the odds of getting into either accident is exponentially larger than just P(a)

If you havent learned your lesson - to never fly on an airplane again and do so, some god out there might smack your plane down.

Answer 9

actually mathematical the probability is increasing...
the probability theorem says that PROBABILITY=(the number of possible cases)/(the number of all cases)

(the number of possible cases) is the number of all the planes you are flying in your entire life
(the number of all cases)
is the number of all planes in the earth

let's take an example:
there are 100 planes in the world (it's an example) and you fly in your life with 10 planes; if you are involved in one plane crash the number of planes in the world becomes from 100 to 99.

the probability before the plane crash is 10/100=0.100000000000
the probability after the plane crash is 10/99= 0.1010101010101

0.10000000... < 0.101010101010... that means the probability is increasing



well u can still travel with planes the probability is increasing very poor :)))) and after all the probability is increasing for everybody after your plane crash :) (if you are still alive after the crash)

Answer 10

Probability of being involved in two crashes is less but probability of being involved in second crash given the involvement in first crash remains the same.