A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has six identical components, each with a probability of 0.45 of failing in less than 1,000 hours. The sub system will operate if any three of the six components are operating. Assume that the components operate independently. Find the probability that:________.
a. exactly two of the four components last longer than 1000 hoursb. the subsystem operates longer than 1000 hours

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Answer:

a) [tex]P( X=2) = 0.368[/tex]

b) P(the subsystem operates longer than 1000 hours) = 0.758

Step-by-step explanation:

Note: I believe that the  correct statement is that one subsystem has four identical components because that is what is clearly stated in the question.

The number of components, n = 4

Probability that a component fails in less than 1000 hours, q = 0.45

Probability that a component lasts longer than 1000 hours, p = 1 - q = 1 - 0.45

p = 0.55

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

Using binomial distribution formula:

[tex]P( X = r) = nCr p^{r} q^{n-r}\\P( X = 2) = 4C2 * 0.55^{2} * 0.45^{4-2}\\P(X=2) = 6 * 0.2025* 0.3025\\P( X=2) = 0.368[/tex]

b) Probability that the subsystem operates longer than 1000 hours

Since there are four components, the system will operate longer than 1000 hours if more than two of its components operate more than 1000 hours.

[tex]P(X\geq 2) = 1 - P(X<2)\\P(X\geq 2) = 1 - [P(X-0) + P(X=1)]\\P(X\geq 2) = 1 - [(4C0 * 0.55^0 * 0.45^4) + (4C1 * 0.55^1 * 0.45^3)]\\P(X\geq 2) = 1 -(0.041 + 0.201)\\P(X\geq 2) = 0.758[/tex]

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