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Ques.

NASA is developing two top-secret space shuttles. One has two engines, the other has four. All the engines are identical, and have the same probability of failure. Each is designed to fly if at least half of its engines work. A visiting scientist says, "The four-engine shuttle is more reliable, isn't it?" The NASA technician replies that the probability of failure is top secret, but that in fact both shuttles have the same probability of flying. The visitor then says, "Aha! Never mind, now I know both the probability an engine will fail and the probability that the shuttle will fly." How did he figure this out, and what are the two probabilities?

Attempt:

Let $x$ be the probability that an engine will work.

Then the probability that an engine won't work is $1-x$.

Space shuttle $1$ will fly when at least one engine will work= probability that one engine will work + probability that both engine will work

Probability that space shuttle $1$ (with two engines) will fly $=x(1-x)+x^2$

Probability that space shuttle $2$ (with four engines) will fly $=x^2(1-x)^2+x^3(1-x)+x^4$

Now, we are given that both shuttles have same probability of flying $\Rightarrow x(1-x)+x^2=x^2(1-x)^2+x^3(1-x)+x^4$

On solving we get $(x^2-x)(x^2+1)=0$

As $x$ should be real no. we get $x=0$ or $1$.

Am I right ?

User
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  • If $x = {1 \over 2}$ then probability of first shuttle flying is obviously $3/4$. Your estimate gives $1/2$ ergo your estimate is wrong. – Abstraction Aug 17 '16 at 15:11
  • So where did I go wrong ? – User Aug 17 '16 at 15:14
  • In your formulae of probabilities of flight. For the first one, it is $1-(1-x)^2$ ("not both engines failing"). For the second - $1-(1-x)^4-4x(1-x)^2$. For self-check, find both probabilities of flying and not flying and verify that their sum equals $1$. – Abstraction Aug 17 '16 at 15:17
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    @Abstraction Wouldn't your second probability have $(1-x)$ raised to $3$ since you've got $3$ failed engines? In the term with $4x$, I mean. – Jam Aug 17 '16 at 15:40
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    @Jam ...yes. It really should, my typo. Too late to edit - in the comment above second expression should be $1 - (1-x)^4 - 4x(1-x)^3$. – Abstraction Aug 17 '16 at 15:43

2 Answers2

1

Let $p$ denote the probability that an engine will be good.

The probability that the $\color\red2$-engine shuttle will be good is:

$$\sum\limits_{n=\color\red2/2}^{\color\red2}\binom{\color\red2}{n}\cdot(p)^{n}\cdot(1-p)^{\color\red2-n}$$

The probability that the $\color\green4$-engine shuttle will be good is:

$$\sum\limits_{n=\color\green4/2}^{\color\green4}\binom{\color\green4}{n}\cdot(p)^{n}\cdot(1-p)^{\color\green4-n}$$

We know that:

$$\sum\limits_{n=\color\red2/2}^{\color\red2}\binom{\color\red2}{n}\cdot(p)^{n}\cdot(1-p)^{\color\red2-n}=\sum\limits_{n=\color\green4/2}^{\color\green4}\binom{\color\green4}{n}\cdot(p)^{n}\cdot(1-p)^{\color\green4-n}$$

Therefore:

$$\tiny\binom21\cdot(p)^{1}\cdot(1-p)^{2-1}+\binom22\cdot(p)^{2}\cdot(1-p)^{2-2}=\binom42\cdot(p)^{2}\cdot(1-p)^{4-2}+\binom43\cdot(p)^{3}\cdot(1-p)^{4-3}+\binom44\cdot(p)^{4}\cdot(1-p)^{4-4}$$

Therefore:

$$2p(1-p)+p^2=6p^2(1-p)^2+4p^3(1-p)+p^4$$

Therefore:

$$-3p^4+8p^3-7p^2+2p=0$$

I'm not sure how to solve this equation, but WolframAlpha gives the following solutions, all within the legal range (under the definition of probability):

  • $p=0$
  • $p=2/3$
  • $p=1$

From here you can calculate both the probability that an engine will fail and the probability that the shuttle (either one of them) will fly...

barak manos
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    I get the same solution. For the record, I think it's easier if you approach it from the perspective of "the probability each shuttle won't fly" and then equate those probabilities since you get $(1-x)^2=(1-x)^4-4x(1-x)^3$, which simplifies nicely to $x(1-x)^2(2-3x)=0$ and the solutions just fall straight out. – Jam Aug 17 '16 at 15:47
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    @Jam: You are possibly correct. I found it a bit confusing to relate to the probability that an engine would be bad (fail) and that a shuttle would be good (fly), so I "aligned" both of them to "good". – barak manos Aug 17 '16 at 15:50
  • One way to solve the equation by hand would be to note that it is obvious from the context that $p=0$ and $p=1$ are solutions. Thus you can factor out $p^2-p$ and solve the quadratic that is left behind after the polynomial division. – Ian Aug 17 '16 at 15:59
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The probability that the two-engine shuttle will fly is the probability that just one engine works plus the probability that both engines work. You used that fact, but you neglected the fact that when one engine fails, the engine that works could be the first engine or the second engine. So the probability of flying is not $x(1-x)+x^2$, as you wrote, but rather $2x(1-x)+x^2$.

Similarly, in the four-engine probability you neglected to count the number of ways it could happen that three engines work or that just two engines work.

It seems easier to write the probability that the shuttle will not fly, writing $q$ for the probability that an engine fails. Then you only have to account for one possibility for two engines (that is, you need the probability that both engines fail, which is $q^2$), and for four engines you need only to consider three or four engines failing (which comes to $4q^3(1-q)+q^4$).

We therefore have $$ q^2 = 4q^3(1-q)+q^4 $$ Clearly this polynomial in $q$ will have zeros at $q=0$ and $q=1$, because if the engines work with probability $1$ then both shuttles fly with probability $1$, whereas if the engines fail with probability $1$ then both shuttles fly with probability $0$. But we can factor out $q^2$ on each side of the equation above, so if $q\neq1$ then $$ 1 = 4q(1-q)+q^2 = 4q-3q^2, $$ from which we find that $$ 3q^2 - 4q +1 = 0. $$ Applying the quadratic formula, we find two roots. One root is the one we already know must exist, $q=1$, and the other is $q=\frac13$. Now recall that the probability an engine works is $1-q$, and that the probability that the first shuttle flies is $1-q^2$, and you can find the probabilities the visiting scientist found.

(I think we are supposed to assume that the probability of each engine working is neither $0$ nor $1$ because such results would not be "realistic".)

David K
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