What values of p is the probability of landing in a 4 engined plane higher than a 2 engine plane?

Suppose an airplane engine will fail(when the plane is in flight) with probability 1-p. Failure of engines are independent events. Suppose, that the plane will land successfully if at least half of its engines work. For what values of p is a 4 engine plane preferred to a 2 engined plane (for what values of p is the probability of a successful landing higher with a 4 engine plane?)|||one engined not counted


one engine min required


P of failure of two of two =1/2 x 1/P of 3 of 2 fail is 1/2^3





ETOPS ( extended twin Operations) for over ocean requirement also known as :


Engines Turn or People Swim


the actual measure of safety is of the chance that any one engine will fail must be very small


each engine design has its own problems





the space shuttle was to be safe with a P failure ,1,000,000 ( or so)


actually TWO complete failures in less than 150 flights, unacceptable|||we shall solve using the binomial distribution

with which you must be very familiar to get such a q,

and shall thus simplify binomial coefficients etc for compact solution.



2-engine plane:

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P[2-engine plane lands ok] = (1- p)^2 + 2p(1- p) = (1-p)(1+p)



4- engine plane:

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P[4-engine plane lands ok] = (1- p)^4 + 4p*(1- p)^3 + 6p^2*(1- p)^2

= (1-p)虏(1+p虏 -2p + 4p -4p虏 +6p虏)

= (1-p)虏(1+ 2p + 3p虏)

= (1-p)(1+p +p虏 -3p^3)



The 4 -engine plane is preferred if

P[4-engine plane lands ok] %26gt; P[2-engine plane lands ok], ie

(1-p)(1+p +p虏 -3p^3) %26gt; (1-p)(1+p)

=%26gt; 1+p +p虏 -3p^3 %26gt; 1+p

=%26gt; p虏-3p^3%26gt; 0

=%26gt; p虏(1-3p) %26gt; 0

ans:p %26lt; 1/3

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