Probability of an airplane crash 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?

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 $\displaystyle \frac{1}{2}$.
