probability of failing

[kiroro22]

Hi,there
first of all, here is the question

A four-engined aircraft can fly safely on two engines; a three-engined aircraft can fly safely on one. On a particular trip, the probability of any engine failing is p, independently of all other engines and of the aircraft type.

which type of aircraft has the highest probability of being able to make this journey safely?


My intuitive sense tells me that three-engined aircraft will have a higher probability of being able to make this journey safely because it can fly without 2 engines.

However, the first one has four engines but it's safe when it has two engines so my thought is this has a higher chance to fail the journey than the first one.

My question is how to solve this kind of question formally using a proper probability theorem?

If I set the failing is p then not failing is 1-p....
do I start with this?

well,any help would be good for me

thank you so much in advance.

thanks

 

[tkhunny]

Rather than 1-p, why not make it q = 1-p.

Use the Binomial Theorem and expand both (p+q)^3 and (p+q)^4.

Throw away the q^3 term from the first and the q^3 and q^4 terms from the second. Tell me why I did that?

Rewrite what is left with the dubious and hopeful expression: p^2 +(3)pq + (3)q^2 > (p)p^2 + (4p)pq + (6p)q^2

It's probably an excessive criterion, but one measure would be a term-by-term coefficient comparison.

p^2 -- Pretty obviously, p < 1
pq -- When is 4p < 3? p < 3/4 -- Since p is the probability of failure, I sincerely hope p < 3/4
q^2 -- When is 6p < 3? p < 1/2 -- Again, I hope so!!

Well, then...What is to be concluded?

 

[soroban]

Hello, kiroro22!

I'll explain this in bite-sized pieces.

A four-engined aircraft can fly safely on two engines; a three-engined aircraft can fly safely on one.
On a particular trip, the probability of any engine failing is \(\displaystyle p\), independently of all other engines and of the aircraft type.

Which type of aircraft has the highest probability of being able to make this journey safely?

\(\displaystyle \text{The 4-engine plane will crash if 3 or 4 engines fail.}\)
\(\displaystyle P(\text{3 fail}) \:=\:{4\choose3}p^3(1-p) \quad\text{or} \quad P(\text{4 fail}) \:=\^4\)
\(\displaystyle P(\text{Four-engine, crash}) \:=\:4p^3(1-p) + p^4 \:=\:4p^3 - 3p^4\)

. . \(\displaystyle P(\text{Four-engine, safe}) \:=\:1-(4p^3 - 3p^4) \;=\;1 - 4p^3 + 3p^4\)



\(\displaystyle \text{The 3-engine plane will crash if all 3 engines fail.}\)

\(\displaystyle P(\text{Three-engine, crash}) \:=\^3\)

. . \(\displaystyle P(\text{Three-engine, safe}) \:=\:1-p^3\)



\(\displaystyle \text{Which is greater?}\)

. . \(\displaystyle P(\text{Four-engine, safe}) \,\text{ or }\,P(\text{Three-engine, safe})\)


\(\displaystyle \text{We have: }\;1 - 4p^3 + 3p^4 \;\;^>_< \;\;1 - p^3\)

. . . . . . . . . . . . . .\(\displaystyle 3p^4\;\;^>_<\;\;3p^3\)

\(\displaystyle \text{Divide by pos. }3p^3\!:\;\;\;p\;\;^>_<\;\;1\)


\(\displaystyle \text{Since we know that }p < 1\text{, the inequality is:}\)

. . \(\displaystyle P(\text{Four-engine, safe})\) .< .\(\displaystyle P(\text{Three-engine, safe})\)


\(\displaystyle \text{Therefore, the three-engine plane is safer.}\)

 

[kiroro22]

thanks for your help soroban,

it is so easy to follow. good explanation

thank you very much indeed.