# Two and four engine probability problem

Austen, a rocket designer, has come to Chris, a reliability expert, with the following problem:

“The vehicle is designed. We can use two large engines or four small engines and get the same thrust and the same weight. However, we know that the engines are subject to catastrophic failure, and we have designed the vehicle so that we will still get into orbit if half of the engines fail. Now, if you tell me the probability of an engine failing in the time required to get into orbit, I can decide whether to use two or four.”

Chris replied, “We have analyzed the test data on the engines and have found that the large and small engines have the same probability of failing in a given time. I can assure you that it makes no difference whether you use two or four engines. However, this failure probability is classified top secret and I cannot give it to anyone.”

Austen said, “Never mind. From what you have just told me, I can calculate by myself the failure probabilities of an engine and of the rocket.”

a. What is the failure probability for a single engine?

b. What is the failure probability for the rocket?

• Why are those your answers? – Mike Pierce Mar 7 '16 at 1:03
• It is more useful that you show us your work and logic, not a numerical answer. Formatting tips here. – Em. Mar 7 '16 at 1:04
• I assumed independence and each engine should have an equal probability of failing so 0.5? Since the rocket fails when an engine fails doesn't it also have a probability of 0.5? – user3777772 Mar 7 '16 at 1:37

Let the probability of failure of an engine be $p$.

Then the probability the two-engine rocket fails is $p^2$ (both engines must fail). Here we are assuming independence. This may not be a reasonable assumption.

The probability the four-engine rocket fails is $\binom{4}{3}p^3(1-p)+p^4$. Our expression is the sum of the probabilities of exactly $3$ failures and exactly $4$ failures. Again, we assumed independence.

Since the probability of failure of the system in the two cases is said to be the same, we have $$p^2=4p^3(1-p)+p^4=4p^3-3p^4.$$ Presumably $p\ne 0$, so $3p^2-4p+1=0$. This has roots $p=1/3$ and $p=1$. Presumably $p\ne 1$!

• so a = 1/3 and b = 1? The rocket will always fail? – user3777772 Mar 7 '16 at 1:36
• On the assumption that $p$ is different from $0$ and $1$, we have $p=1/3$. So the answer to a) is $1/3$. The answer to b) is $p^2$, in our case $1/9$. – André Nicolas Mar 7 '16 at 1:38
• You are welcome. I hope that everything is clear. – André Nicolas Mar 7 '16 at 1:40

Hint: The engine failure probability satisfies the equation: $$p^2=4p^3(1-p)+p^4$$ which has 2 trivial solutions $p=0,1$ and a non-trivial solution...