What is the Probability of Engine Failure for a Plane with Four Engines?

[Addez123]

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.

 

[Delta2]

I think your mistake is that you use Poisson distribution, while I think the problem is suitable for binomial distribution..
I did it with binomial distribution and I get the suggested answer, that is $\frac{1}{3}<p<1$.

Hint: In using binomial distribution n is the number of engines of the plane. k is the number of the engines that fail, $$Pr(n,k,p)=\begin{pmatrix}n \\k \end{pmatrix}p^k(1-p)^{n-k}$$

 

[Addez123]

Re-did it, now it works out!
Thanks!