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Let X1,82 be independent random variables representing lifetimes (in hours) of two key components of a device that fails when and only when both components fail. Say each Xi has an exponential distribution with mean 1000. Let Y1 = min(X1,X2) and Y2 = max(X1,X2), so that the space of Y1,Y2 is 0 < y1 < y2 <. (a) Find G(y1, y2) = P(Y1 = 71,Y2 = y2).

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To find the probability that Y1 = 71 and Y2 = y2, we can write G(y1, y2) = P(Y1 = 71, Y2 = y2).

To find the probability that Y1 = 71 and Y2 = y2, we need to consider the possible values of X1 and X2 that could lead to these values of Y1 and Y2. The probability density functions (PDFs) of an exponential distribution with mean 1000 is f(x) = 1/1000 * e^(-x/1000). Therefore, the probability of this event is

  • P(X1 = 71, X2 = y2) = f(71) * f(y2) = 1/1000 * e^(-71/1000) * 1/1000 * e^(-y2/1000).

Another possibility is that X1 = y2 and X2 = 71. The probability of this event is

  • P(X1 = y2, X2 = 71) = f(y2) * f(71) = 1/1000 * e^(-y2/1000) * 1/1000 * e^(-71/1000).

Since these two events are mutually exclusive and exhaust all possible ways that Y1 can equal 71 and Y2 can equal y2, the probability of these events occurring is the sum of these probabilities. Therefore,

  • G(y1, y2) = P(Y1 = 71, Y2 = y2) = P(X1 = 71, X2 = y2) + P(X1 = y2, X2 = 71) = (1/1000 * e^(-71/1000) * 1/1000 * e^(-y2/1000)) + (1/1000 * e^(-y2/1000) * 1/1000 * e^(-71/1000)) = 2/1000^2 * e^(-(71 + y2)/1000).

Note that this expression is only valid when 0 < 71 < y2 < ∞ since these are the conditions under which Y1 can equal 71 and Y2 can equal y2.

Learn More about Exponential Distribution here:

https://brainly.com/question/13339415

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