Engine failure probability The probability of an airplane engine failure during the flight is $p$. Engines fail independently of one another. The plane can fly if at least half of the engines are working.
If we consider two and four engine airplanes, for which $p$ is the two engine airplane more secure than the four engine?
The way I approached this is calculated the probability of each airplane flying.
For the two engine airplane, we can consider the following cases when it functions properly:
$$WW \cup WN \cup NW$$
where $W$ means the engine is working and $N$ not working.
If the probability of engine failure is $p$, then the probability of engine functioning properly is $(1-p)$. So, the probability of a two engine airplane functioning properly is:
$$P(2)=(1-p)^2+2p(1-p)=(1-p)(1+p)$$
Now, if we do the same for the four engine airplane, we consider the cases when it flies normally:
$$ WWWW \cup WWWN \cup WWNW \cup WNWW \cup NWWW \cup WWNN \cup WNWN \cup WNNW \cup NWNW \cup NWWN \cup NNWW $$
So it can fly with probability:
$$P(4)=(1-p)^4+4p(1-p)^3+5p^2(1-p)^2=(1-p)^2(2p^2+2p+1)$$
Now, in order for the two engine airplane to be safer than the four engine, the probability of the two engine working has to be greater than the four engine working:
$$(1-p)(1+p)>(1-p)^2(2p^2+2p+1)$$
Which is correct for: $0<p<1$
So, it means the two engine airplane is safer than a four engine airplane in all cases, but it doesn't make logical sense. Did I go wrong in my calculations?
 A: I see $P(4)$ is calculated incorrectly - it should be $(1−p)^4+4p(1−p)^3+6p^2(1−p)^2$ instead of $(1−p)^4+4p(1−p)^3+5p^2(1−p)^2$. 6 instead of 5 because we have $ {4 \choose 2} $ possibilites of having 2 engines working and 2 engines failing.
A: There actually is no "obvious" intuitive answer to this problem, either plane can be safer or more dangerous depending on the engine reliability.
Given independent failures, the probability of disastrous engine failure is $$4(1-p)p^3+p^4=4p^3-3p^4=p^3(4-3p)$$ for the four-engine and $p^2$ for the two engine plane. (This concords with your calculation of $1-p^2$ for a successful journey for the twin engine, but I didn't check against your formula for four engines. There is an error where 5 should be 6 as already pointed out.)
Four is more dangerous sometimes
Say $p=1/2$ then the four-engine plane has $1/8(4-3/2)=5/16$ chance of disaster, and the two-engine plane has chance $1/4$ of disaster, so the four-engine plane is more dangerous.
Two is more dangerous sometimes
If $p=1/10$ then the four-engine plane has chance $1/1000\times(4-3/10)\approx 4/1000$ of disaster and the twin-engine job has chance $1/100$ of disaster, so the four-engine plane is safer.
The threshold where the four-engine starts to become more dangerous is $p(4-3p)>1$ which has solution $p>1/3$. I can't think of a nice argument to justify this at the moment.
