Expert: Steve Holleran - 9/5/2008

Question
Assume that airplane engines operate independently of each other and that at least half of the engines on a plane must operate for the plane to continue flying. A particular airplane engine fails with a probability of 1/7. Which is safer, an airplane with 2 of these engines or an airplane with 4 of these engines?

Answer
Hi Arti,

It seems to me that this is the setup:

Plane w/ 2 engines: You want 1 or 2 of these engines to not fail.  That is the same as 1 - P(2 failures).  If P(failure) = 1/7 for each engine,
then for 2 failures you want both engines to fail, which would be

P(2 failures) = 1/7 * 1/7 = 1/49  

Then 1 - 1/49 = 48/49 = 0.9796 = P(plane is safe)


Plane w/4 engines:

If half or more of the engines have to be good, then that means
2 or 3 or 4 good engines.  But again, its easier to calculate

P(2 or 3 or 4 good) = 1 - P(0 or 1 good).

P(0 good)= P(4 failures) = (1/7)^4 = 1/2401

P(1 good) = 4C1 * (1/7)^3 * (6/7)^1 = 4 * 6/2401 = 24/2401

Then 1 - P(0 good or 1 good) = 1 - (1/2401 + 24/2401)

                            = 1 - (25/2401) = 2376/2401 =

                            = 0.9896

So it looks like the probability for the 4 engine plane is slightly higher.

Sorry this took so long

Steve