"An airplane needs at least half of its engines working to stay airborne. If an engine independently functions with probability p, for what values of p is an airplane with three engines safer than one with five engines?"
Ok, so I'm going to try to work through this. I'm pretty much thinking aloud (atype? is there a word for that?) so please forgive my rambling or tangents.
If we say that an airplane is only as safe as it's likelihood of remaining airborne (which I don't think is an unfair assumption) then I need to find values for p such that the probability for remaining airborne is better with only three engines rather then five engines.
So the chance that the airplane is airborne (for a 3 engine plane) at any point in time is:
Since we have 3 engines we need exactly 1.5 engines running at any point in time. Since this is a discrete problem and there is no concept of a engine half running, we need at least 2 engines running.
So, P(X) = p, where X = 1, 2, 3 (for a 3 engine plane)
So the chances that we're in the air at any given point in time is:
The chances that Engines 1 and 2 are running, the chances that engines 1 and 3 are running and the chances that engines 2 and 3 are running.
P(1) * P(2) + P(1) * P(3) + P(2)* P(3) = p^2 + p^2 + p^2 = 3p^2
I'm not really sure what significance 3p^2 has, but I'm going to try it for the five engine plane now:
So again, P(X) = p, where X = 1,2,3,4,5 (for a 5 engine plane)
We need exactly 2.5 engines running (or in this case 3 engines). Meaning our possibilities are:
P(1) * P(2) * P(3) + P(1) * P(2) * P(4) + P(1) * P(2) * P(5) + P(1) * P(3) * P(4) + P(1) * P(3) * P(5) + P(1) * P(4) * P(5) + P(2) * P(3) * P(4) + P(2) * P(3) * P(5) + P(2) * P(4) * P(5) + P(3) * P(4) * P(5) = 10p^3
So now I guess I need to find values for p where 10p^3 < 3p^2.
10p^3 < 3p^2 = 10p < 3 = p < 3/10.
So any value for p less then 30%, we're safer with 3 engines then with 5 engines? what is the significance of the equations I came up with (10p^3 and 3p^2)?