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Some time ago I read a problem about probability of an airplane crash. I can't recall it correctly but I ll try to write it here. Please tell me if it doesn't make sense and correct me.

Any engine of an airplane has 50% chance of failure during travel and if half of the engines survive during travel then airplane will reach its destination. Which plane voyage would be safer, four engine one or two engine one?

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    $\begingroup$ Damn. I surely wouldn't want to be in that plane! $\endgroup$ Mar 6 '14 at 21:57
  • $\begingroup$ The problem is well defined. Is that what you are asking, or about how to solve it? $\endgroup$ Mar 6 '14 at 21:58
  • $\begingroup$ I admired solution of the problem indeed, because it also gave probability of other conditions such as if failure of any engine has > 50% chance graphically. But it's just a fuzzy memory. $\endgroup$
    – kenn
    Mar 6 '14 at 22:04
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The four-engine plane will crash if more than half of its engines fail during travel. This happens when $3$ of the engines fail or all $4$ fail.

The probability of $3$ engines failing is $$\dbinom{4}{3} \left(\frac{1}{2}\right)^4 = \frac{1}{4}$$

and the probability of $4$ engines failing is $$\dbinom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$

so the probability of a four-engine plane crashing is $\displaystyle \frac{1}{4} + \frac{1}{16} = \frac{5}{16}$.

The two-engine plane will crash if more than half of its engines fail during travel. This happens only when both fail.

The probability of $2$ engines failing is $$\dbinom{2}{2} \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

so the probability of a two-engine plane crashing is $\displaystyle \frac{1}{4}$.

So we get the ironic result that the two-engine plane is safer than the four-engine one.


For the general case, if we're worried about $n+1$ to $2n$ engines failing, the probability would be $$\sum_{i=n+1}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n}$$

Let's call this sum $S$. As $n$ approaches $\infty$, we note that $$\sum_{i=0}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} = 1 = \sum_{i=0}^{n-1} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} + \sum_{i=n+1}^{2n} \dbinom{2n}{i} \left(\frac{1}{2}\right)^{2n} + \dbinom{2n}{n} \left(\frac{1}{2}\right)^{2n}$$

$$ = 2S + \dbinom{2n}{n} \left(\frac{1}{2}\right)^{2n} \approx 2S$$

and we can see that $S$ approaches $\displaystyle \frac{1}{2}$.

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  • $\begingroup$ One might ask what's the limit as $n\to\infty$, where $2n$ is the number of engines... $\endgroup$ Mar 6 '14 at 22:08
  • $\begingroup$ Yes, that's right! I remember it like that. Thank you for the answer. $\endgroup$
    – kenn
    Mar 6 '14 at 22:08
  • $\begingroup$ @user2345215 as n goes to infinity, the distribution gets more exact with the probability approaching $1/2$ $\endgroup$
    – qwr
    Mar 6 '14 at 22:10
  • $\begingroup$ @qwr: I see it, because you sum up almost half of the binomial coefficients, I was just suggesting to add that information to the response :) $\endgroup$ Mar 6 '14 at 22:29
  • $\begingroup$ @2012ssohn I just have time now to inspect your solution. Is there any other exponential formula like e^x that we can apply in this situation? I think I saw something like that in the solution of the problem but I don't remember exactly. Anyway, your solution is simple and instructive. Thank you. $\endgroup$
    – kenn
    Mar 7 '14 at 9:00

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