13.On many multiple-choice tests, 1 point is given for each correct

A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has eight ident

Fudgin [204]

Answer:

a) 0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b) 0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either they last more than 1,000 hours, or they do not. Components operate independently, which means that the binomial probability distribution is used 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

One subsystem has eight identical components, each with a probability of 0.1 of failing in less than 1,000 hours.

So 1 - 0.1 = 0.9 probability of working for more, which means that $p = 0.9$

a. exactly two of the four components last longer than 1000 hours.

This is P(X = 2) when $n = 4$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.9)^{2}.(0.1)^{2} = 0.0486$

0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b. the subsystem operates longer than 1000 hours.

The subsystem has 8 components, which means that $n = 8$

It will operate if at least 4 components are working correctly, so we want:

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{8,0}.(0.9)^{0}.(0.1)^{8} \approx 0$

$P(X = 1) = C_{8,1}.(0.9)^{1}.(0.1)^{7} \approx 0tex][tex]P(X = 2) = C_{8,2}.(0.9)^{2}.(0.1)^{6} \approx 0$

$P(X = 3) = C_{8,3}.(0.9)^{3}.(0.1)^{5} = 0.0004$

Then

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0 + 0.0004 = 0.0004$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0004 = 0.9996$

0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

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