The lifetime (in hours) of an electronic component is a random variable with density function given by

$f(y)=\left\{\begin{array}{ll}

\frac{1}{100} e^{-y / 100}, & y>0, \

0, & \text { elsewhere. }

\end{array}\right.

$

Three of these components operate independently in a piece of equipment. The equipment fails if at least two of the components fail. Find the probability that the equipment will operate for at least 200 hours without failure.

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Anshuyadav Anshuyadav · Answer 1 · 2022-11-17T11:12:04+01:00

The probability that the equipment will operate for at least 200 hours without failure is [tex](1-e^{-2} )^2[2e^-2+1][/tex].

Given:

mean = 1/100

Let X = the number of components that fail before 200 hours . The X has a binomial distribution n = 3 and p = 1 – e^-2.

probability p = 3/2[tex]p^{2} (1-p)[/tex] + 3/3 [tex]p^{3}[/tex].

= 3([tex]1-e^-2)^2[/tex][tex]e^-2[/tex] + ([tex]1-e^-2)^3[/tex]

= [tex](1-e^-^2)^2[3e^-^2 - e^-^2+1][/tex]

= [tex](1-e^{-2} )^2[2e^-2+1][/tex]

Therefore the probability that the equipment will operate for at least 200 hours without failure is [tex](1-e^{-2} )^2[2e^-2+1][/tex].

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