According to this page from the New York Times (warning: free registration required), http://www.nytimes.com/2003/02/06/technology/circuits/06risk.html?8hpib
So let’s say I run a trial in which I launch 113 space shuttle missions. Let’s assume that each mission has a .0069% chance of catastrophic failure and that the chance of failure on one mission is independent from the chance of failure on other missions.
What percentage of 113-mission trials will have exactly two catastrophic failures? Less than two failures? More than two?
I used to know how to answer these sorts of questions, but that class was over 20 years ago…
Ah, crap.
That probability should be 0.69%, obviously.
1/145 is roughly 0.006896552.
I guess it’s a good thing I forgot everything I learned in Probability and Statistics. If I can’t manage turning a decimal into a percentage, I’m obviously unfit to handle probability questions…
The probability of two catastrophic failures is 113!/(2!*111!) * (1/145)[sup]2[/sup] * (144/145)[sup]111[/sup]. The probability of less than two is 113!/(1!*112!) * (1/145)[sup]1[/sup] * (144/145)[sup]112[/sup] + 113!/(0!*113!) * (1/145)[sup]0[/sup] * (144/145)[sup]113[/sup]. The probability of more than two is one minus the probability of two or less–that’s the sum of the probability of two and the probability of less than two.
I’m no expert, but here’s my best thinkage:
Probability of catastrophic mission failure:
1/145
Probability of a mission the astronauts can walk away from:
1 - (1/145) = 144/145
You need to know the probability of 2 catastrophic mission failures, and 111 successful voyages:
(1/145)^2 * (144/145)^111
.0000476 * 0.464 = 0.0000221
Which means that 0.00221% of all 113 mission trials will have two accidents.
Given the 1/145 odds, there is a 45% chance of having zero catastrophic failures, and almost a 55% chance of having exactly 1. Which means that until Saturday, the 1/145 statistic seemed correct… and now it seems likely that the 1/145 statistic should probably be reconsidered.
[QUOTE]
*Originally posted by ultrafilter *
**The probability of two catastrophic failures is 113!/(2!*111!) * (1/145)[sup]2[/sup] * (144/145)[sup]111[/sup].
Hmmm… simulpost.
But perhaps I got my math wrong… why are the factorials needed here?
Because you don’t want to count a failure on missions 2 and 17 as something different from a failure on missions 17 and 2.
Fair enough… thanks for the explanation. That would bring the answer of the OP to about 14%… which is a more reasonable figure, and not as challenging to the 1/145 stat.
Well, someone obviously should check my arithmetic, but the numbers I get are:
Probability of less than two crashes in 113 missions: 81.6%
Probability of exactly two crashes in 113 missions: 14.0%
Probability of more than two crashes in 113 missions: 4.4%
I’ve not checked the figures, but I’ve simulated the situation, by generating 1000 random numbers according to the binomial distribution. Basically that like coin flipping, where the result of each flip is 0 or 1, with a fixed probability of each (at .0069), with the successes (or failures), independent, as you wanted. Here are the percentages for less than 2, exactly 2 and more than 2 failures respectively:
<2 - 82.6%
=2 - 14.3%
>2 - 3.1%
These aren’t exactly the same as your figures chukdung, but they are fairly close, suggesting that any error’s you are making (if any) aren’t affecting your results very much. I’ll do a larger sample size, to confirm…
This time, I’ve done 100,000 numbers, to get the following percentages:
<2 - 81.624%
= 2 - 14.024%
> 2 - 4.352%
These match almost exactly with your figures, so that’s a very good confirmation that your arithmetic is fine, as my method is independent of yours, but the answer is about the samel, and my method is only an estimate of the true probability (although a very good, one it seems)