Suppose that the space shuttle has three separate computer control systems: the main system and two backup duplicates of it. The first backup would monitor the main system and kick in if the main system failed. Similarly, the second backup would monitor the first. We can assume that a failure of one system is independent of a failure of another system, since the systems are separate. The probability of failure for any one system on any one mission is known to be 0.01.
(a) Find the probability that the shuttle is left with no computer control system on a mission. (Enter your answer to six decimal places.)
(b) How many backup systems does the space shuttle need if the probability that the shuttle is left with no computer control system on a mission must be
1
one hundred thousand
(Enter your answer as a whole number.)
_____ backup systems.

Answer:

a) 0.000001 = 0.0001% probability that the shuttle is left with no computer control system on a mission.

b) 3 backup systems are needed.

Step-by-step explanation:

For each system, there are only two possible outcomes. Either it works, or is does not. Systems are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

Three separate computer control systems

This means that [tex]n = 3[/tex]

The probability of failure for any one system on any one mission is known to be 0.01.

So the probability of working is:

[tex]p = 1 - 0.01 = 0.99[/tex]

(a) Find the probability that the shuttle is left with no computer control system on a mission. (Enter your answer to six decimal places.)

This is [tex]P(X = 0)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{3,0}.(0.99)^{0}.(0.01)^{3} = 0.000001[/tex]

0.000001 = 0.0001% probability that the shuttle is left with no computer control system on a mission.

(b) How many backup systems does the space shuttle need if the probability that the shuttle is left with no computer control system on a mission must be 1 over one hundred thousand

This is n for which:

[tex]P(X = 0) < \frac{1}{100000} = 0.00001[/tex]

With 2 backup systems, the probability is lower than this, so lets see if with only two is possible.

We find [tex]P(X = 0)[/tex] when [tex]n = 2[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{2,0}.(0.99)^{0}.(0.01)^{2} = 0.0001 > 0.00001[/tex]

So 3 backup systems are needed.

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