I would like to understand the way to calculate this particular problem.

I understand that each flight still has a 2.6 in 1M chance of accident. But a person who takes more flights would seem to be at an overall at higher risk.

  • $\begingroup$ Try a geometric distribution model. $\endgroup$ – Zhanxiong Jun 26 '15 at 13:17
  • $\begingroup$ Strongly related: stats.stackexchange.com/questions/137711/… $\endgroup$ – Tim Jun 26 '15 at 13:47
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    $\begingroup$ Is this a homework question? If it is, please use the self-study tag. $\endgroup$ – Rob Hall Jun 26 '15 at 14:01
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    $\begingroup$ Not a homework question and I don't want to do the math - was hoping for an answer and logical explanation of method. My spouse and I just talk about crazy things and this was the discussion in bed this morning. :) $\endgroup$ – Mary Camacho Jun 26 '15 at 17:50
  • $\begingroup$ Your post is not clear. Are you asking for the probability of at least one accident in 1000 flights? (If so, the easy strategy is to work out the probability of the complementary event -- no accidents in 1000 flights - assuming independence). $\endgroup$ – Glen_b May 22 '16 at 2:00

You need to formulate a sequence of conditional probabilities which can then be multiplied by each other. Single flight survival Pr(S) = 1=Pr(crash). So id p= Pr(crash), the probability of surviving all of 1000 flights is the probability of surviving a single flight raised to the 1000th power = (1-p)^1000 ...or with R: (And think of the frequent flyer miles you would have!)

(1- 2.6/1e6)^1000
#[1] 0.9974034

So the probability of having an adverse event prior to flight #1001 would be:

1 - .Last.value
#[1] 0.002596626

Technically these are not "odds" but rather probabilities. People often conflate those terms. In this case the distinction is unimportant.

  • $\begingroup$ I like this approach. Simple and clean :) $\endgroup$ – Kitter Catter May 21 '16 at 16:20

Under a simple Bernoulli experiment model setting, we may view each flight travel as an "experiment" with "success" rate $p = 2.6/1,000,000$ and "failure" rate $q = 1 - p$. Let $X$ denote the random variable that the number of flights a particular person travelled to suffer a first accident. Clearly $X \sim \text{Geometric}(p)$, and the odds you are asking is the probability $$\mathbb{P}(X \leq 1000) \tag{1}$$

This probability can be calculated easily in R, by using pgeom(999, prob = p), results in $0.002596626$.

It is easily seen from $(1)$ that the more flights a person takes, the higher risk he/she suffers, as $\mathbb{P}(X \leq n)$ is an increasing function of $n$.

The above analysis of course depends on the Bernoulli model assumption. Stories could be different if you use other probabilistic models.


I'm going to give you a simple, almost math-free answer that is hopefully sheds a different light than previous answers.

Each time we fly, we have a small probability of accident 0.0000026. This means, on average, if each trip we take is independent of the other (eg. whatever happens in your first flight cannot affect the chances of accident on your second and so forth), then if we took 10,000,000 flights, we would on average experience 26 accidents. But what if we only took 1000 flights? What's the probability of at least one accident?

Instead of solving for the chance of at least one accident, which will include way too many cases (proability of one accident, proability of two accidents,...,proability of one thousand accidents), let's solve for the probability there isn't an accident at all (known as the complement). If there is a 0.0000026 chance of accident in one flight, then there is a $$1-0.0000026 = 0.9999974$$ chance of having no accidents. Since there are 1000 flights, we need to take this probability to the power of 1000, or rather, the chance of having absolutely no accident whatsoever on 1000 flights is $0.9999974^{1000} = 0.9974033737$.

What we originally wanted was the probability of at least one accident. What we solved for was however to have no accidents, the exact opposite of this. To get the answers we want, then, we simply again take the complement of this probability $1-0.9974033737=0.0025966263$. In words: If we fly 1000 times, there is a $.3\%$ chance we will face at least one accident.

If some of these concepts (like exponentiating the probability or taking the complement) seem unfamiliar, I recommend reading the first chapters of "A first course in Probability" by Sheldon Ross.


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