- #1
Addez123
- 199
- 21
- Homework Statement
- Probability for an engine to fail is p.
A plane can fly using only half of their engines, for what p is it safer to use a two engine plane instead of a four engine one?
- Relevant Equations
- Binominal, Poisson, multinomial, Normal -distribution formulas.
Given we only have one number I assume we are to use Poisson distribution.
Probability for a plane with two engines to fail require both engines to fail:
$$P_2 = P_o(2) =p^2/{2!} * e^{-p}$$
Probability of a four engine plane to fail requires 3 or 4 engines to fail:
$$P_4 = P_o(3) + P_o(4) = e^{-p}(p^3/{3!} + p^4/{4!} )$$
This leads the the equation $$P_2 < P_4$$
$$p^2/2! * e^{-p} < e^{-p}(p^3/{3!} + p^4/{4!} )$$
$$p^2 < p^3/3 + p^4/12$$
$$1 < p/3 +p^2/12$$
$$12 < 4p + p^2$$
Which we use PQ formula to calculate the points from:
$$p^2 + 4p - 12 = 0$$
The two points are p = -2, p = 2.
The answer is 1/3 < p < 1.
I probabily did everything wrong but some hints as to where I first did wrong would be helpful.
Probability for a plane with two engines to fail require both engines to fail:
$$P_2 = P_o(2) =p^2/{2!} * e^{-p}$$
Probability of a four engine plane to fail requires 3 or 4 engines to fail:
$$P_4 = P_o(3) + P_o(4) = e^{-p}(p^3/{3!} + p^4/{4!} )$$
This leads the the equation $$P_2 < P_4$$
$$p^2/2! * e^{-p} < e^{-p}(p^3/{3!} + p^4/{4!} )$$
$$p^2 < p^3/3 + p^4/12$$
$$1 < p/3 +p^2/12$$
$$12 < 4p + p^2$$
Which we use PQ formula to calculate the points from:
$$p^2 + 4p - 12 = 0$$
The two points are p = -2, p = 2.
The answer is 1/3 < p < 1.
I probabily did everything wrong but some hints as to where I first did wrong would be helpful.