Statistics expert.. help please?

Binomial Random Variable Question..





A commercial airplane has four engines; each has an independent reliability (probability of working after a flight) of 92%. For an airplane to land safely, at least two engines must be in working condition





a) What is the probability that an airplane lands safely








b) If we want to increase the probability that an airplane lands safely to 99.9% but keep the number of engines the same, what should be the minimum reliability of the each engine?











c) If we want to increase the probability that an airplane lands safely to 99.9% but we can not change the reliability of each engine, what should be the minimum number of engines?











d) If it costs about 10 millions dollars to improve the reliability of each engine by 1% and it costs about 25 millions dollars to add an engine to the airplane effectively, and if we want to increase the probability that an airplane lands safely to 99.9%, what would be a more cost-effective approach (increases the reliabilit

Statistics expert.. help please?
a) What is the probability that an airplane lands safely





P (exactly 4 engines working) = (4!)/((4!)(0!)) * (.92)^4 * (.08)^0 = .7164


P (exactly 3 engines working) = (4!)/((3!)(1!)) * (.92)^3 * (.08)^1 = .2492


P (exactly 2 engines working) = (4!)/((2!)(2!)) * (.92)^2 * (.08)^2 = .0325





P ( at least 2 engines working) = .7164 + .2492 + .0325 = .9981





ANS: 99.81%





b) If we want to increase the probability that an airplane lands safely to 99.9% but keep the number of engines the same, what should be the minimum reliability of the each engine?





similar to above we can write





P (4 engines working) = (x)^4 = x^4


P (3 engines working) = 4 * x^3 * (1-x) = 4x^3 - 4x^4


P (2 engines working) = 6 * x^2 * (1-x)^2 = 6x^2 (1 - 2x +x^2) = 6x^4 -12x^3 + 6x^2





P (at least 2 working) = x^4 + (-4x^4 + 4x^3) + (6x^4 -12x^3 + 6x^2) = 3x^4 -8x^3 + 6x^2





.999 = 3x^4 -8x^3 + 6x^2





x = .936





ANS: 93.6%





c) If we want to increase the probability that an airplane lands safely to 99.9% but we can not change the reliability of each engine, what should be the minimum number of engines?





since we were close before I would try 5 engines





P (exactly 5 working) = (5!)/((5!)(0!)) * (.92)^5 * (.08)^0 = .6591


P (exactly 4 working) = (5!)/((4!)(1!)) * (.92)^4 * (.08)^1 = .2866


P (exactly 3 working) = (5!)/((3!)(2!)) * (.92)^3 * (.08)^2 = .0498


P (exactly 2 working) = (5!)/((2!)(3!)) * (.92)^2 * (.08)^3 = .0043





P(at least 2) = .9998





d) If it costs about 10 millions dollars to improve the reliability of each engine by 1% and it costs about 25 millions dollars to add an engine to the airplane effectively, and if we want to increase the probability that an airplane lands safely to 99.9%, what would be a more cost-effective approach





adding one engine gets you there for $25 mill.ion


increasing to 94% also gets you there for $20 million





so more effective to improve engines
Reply:a) BINOMIAL DISTRIBUTION


n = Trials [4]


r = occurrences [2, 3, 4]


p = probability of occurrence [0.92] (92%)








Probability Safe Landing [r = 2, 3, 4] = n!/[r!*(n-r)!] * p^r * (1-p)^(n-r) = 4!/[2! *(4-2)!] *(0.92)^2 * (1-0.92)^(4-2) +. . .+ 4!/[4! *(4-4)!] *(0.92)^4 * (1-0.92)^(4-4)








b) BINOMIAL DISTRIBUTION





Probability Safe Landing [r = 2, 3, 4] = 0.999 =


n!/[r!*(n-r)!] * p^r * (1-p)^(n-r)





Utilize Power Series Expansion for the "first few terms"


http://www.ucl.ac.uk/Mathematics/geomath...





Compute iteratively, solve for "p".








(c) Compute Probability Safe Landing [r = 3 or r = 3, 4] = n!/[r!*(n-r)!] * p^r * (1-p)^(n-r) = 4!/[2! *(4-2)!] *(0.92)^2 * (1-0.92)^(4-2) +. . .+ 4!/[4! *(4-4)!] *(0.92)^4 * (1-0.92)^(4-4)





(d) Compare costs of (b) and (c)