Airplane Engine - Probability Question!

• Feb 4th 2009, 04:57 PM
cnmath16
Airplane Engine - Probability Question!
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?

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?
• Feb 4th 2009, 06:33 PM
mr fantastic
Quote:

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?

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?

Looks OK.

Something to think about: Let the probability of engine failure be p. Find the values of p such that the two-engine plane is safer.