Results 1 to 2 of 2

Math Help - Airplane Engine - Probability Question!

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    53

    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?

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by cnmath16 View Post
    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?
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 4th 2011, 07:09 AM
  2. vectors Airplane question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 1st 2009, 11:23 AM
  3. Vector airplane question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 19th 2009, 05:04 PM
  4. Mathematical search engine
    Posted in the Math Software Forum
    Replies: 1
    Last Post: June 15th 2008, 04:01 AM

Search Tags


/mathhelpforum @mathhelpforum