Basic probability aeroplane engines problem

[danmanchester]

Hi all, I've got a past paper question that goes something like this:

An aeroplane has four engines. During a certain journey, each engine fails with a probability of 0.1, independantly of the others. The aeroplane can fly when at least two engines are working. Calculate:

A) The probability that the aeroplane will complete the journey.
B) The probability that the aeroplane will complete four journeys with no engine failures. State clearly any assumptions you make.


Ok so I've decided this is a binomial distribution problem; for part A) I have assigned X to be the r.v. "number of engines failing" where X~B(4,0.1).

I'm looking for Pr(X ≤ 2) = 1 - (Pr(X = 3) + Pr(X = 4))
= 1 - (4C3X0.1^3X0.9 + 4C4X0.1^4)
= 1 - 3.7 x 10^-3
= 0.9963 (which as far as answers go, seems relatively plausible?).

As for part B), I have absolutely no ideas! If someone would be so kind as to assist it'd be much appreciated :)

Cheers,

Dan

 

[AlephZero]

As for part B), I have absolutely no ideas!

Start by finding the probability of one journey with no engine failures.

 

[Ray Vickson]

Hi all, I've got a past paper question that goes something like this:

An aeroplane has four engines. During a certain journey, each engine fails with a probability of 0.1, independantly of the others. The aeroplane can fly when at least two engines are working. Calculate:

A) The probability that the aeroplane will complete the journey.
B) The probability that the aeroplane will complete four journeys with no engine failures. State clearly any assumptions you make.


Ok so I've decided this is a binomial distribution problem; for part A) I have assigned X to be the r.v. "number of engines failing" where X~B(4,0.1).

I'm looking for Pr(X ≤ 2) = 1 - (Pr(X = 3) + Pr(X = 4))
= 1 - (4C3X0.1^3X0.9 + 4C4X0.1^4)
= 1 - 3.7 x 10^-3
= 0.9963 (which as far as answers go, seems relatively plausible?).

As for part B), I have absolutely no ideas! If someone would be so kind as to assist it'd be much appreciated :)

Cheers,

Dan

In B): what is the probability that in any single trip none of the engines fail? What would this become for 4 trips?

 

[danmanchester]

Thanks for the replies guys. So, for one journey the probability that no engines fail is given by:

Pr(X = 0) = 4C0X0.9^4 = 0.6561

And so for four journeys, 0.6561^4 = 0.185? This seems a little bit of an unrealistic figure?

 

[AlephZero]

This seems a little bit of an unrealistic figure?

Yes, because the assumption that the engine failure rate per flight os 0.1 is unrealistic. The actual rate for modern jet engines is of the order of 1 failure per 100,000 hours. Even for 10 hour long haul flights, that is a probability of about 0.0001 per flight, not 0.1.

 

[danmanchester]

Hi Aleph, thanks for the clarification!

 

[Ray Vickson]

Thanks for the replies guys. So, for one journey the probability that no engines fail is given by:

Pr(X = 0) = 4C0X0.9^4 = 0.6561

And so for four journeys, 0.6561^4 = 0.185? This seems a little bit of an unrealistic figure?

I echo the remarks of AlephZero: it is the 0.1 figure that is unrealistic. Would you want to fly in a plane that had only a 99.6% probability of not crashing? But, given p = 0.1 the answer you got is understandable: it is like having a 16-engine plane and asking that none of the engines fail.

 

[AlephZero]

it is like having a 16-engine plane and asking that none of the engines fail.

That general principle applies in real life to 2 and 4 engined planes. Other things being equal, you are twice as likely to have an engine failure with 4 engines than with 2. But the consequence of 1 engine failure out of 4 is likely to be less serious than 1 out of 2, so it's not a clear-cut decision whether 2 or 4 engines is "best" for all plane designs.

Also, the question assumes the engine failures are independent of each other, and that is not necessarily true either. Running out of fuel, or flying into a flock of birds big enough to cause engine damage, are two examples where it is false.