Originally Posted by
cnmath16 If someone could check my answer that would be AWESOME
Q. 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 thse engines?
My answer*:
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.
Is this CORRECT?