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?**