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Machine 1 is currently working. Machine 2 will be put in use at a time \(t\) from now. If the lifetime of machine \(i\) is exponential with rate \(\lambda_{i}, i=1,2\), what is the probability that machine 1 is the first machine to fail?

Short Answer

Expert verified
The probability that Machine 1 fails before Machine 2 is put to use (time \(t\)) is given by: \(P(X_1 < X_2 - t) = \frac{\lambda_1}{\lambda_1+\lambda_2} e^{-\lambda_2 t}\).

Step by step solution

01

Understanding exponential distributions

Since the lifetimes of Machine 1 and Machine 2 follow exponential distributions, we can represent their lifetimes using probability density functions. The probability density function of an exponential distribution with rate \(\lambda\) is given by: \(f(t) = \lambda e^{-\lambda t}\) for \(t \geq 0\) Machine 1 has a rate \(\lambda_1\), and Machine 2 has a rate \(\lambda_2\). Their lifetimes are independent.
02

Finding the probability that Machine 1 fails before Machine 2

Let \(X_1\) and \(X_2\) be the lifetimes of Machine 1 and Machine 2, respectively. We want to find the probability that Machine 1 fails before Machine 2 is put to use, that is, before time \(t\). We can express this as: \(P(X_1 < X_2 - t)\) To find this probability, we need to analyze the joint distribution of \(X_1\) and \(X_2\). Since Machine 1 and Machine 2 are independent, the joint probability density function can be represented as the product of their individual probability density functions: \(f(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2)\) Now, we can find the probability by integrating the joint probability density function over the appropriate range: \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \int_{x_1+t}^{\infty} f(x_1, x_2) \,dx_2 \,dx_1\)
03

Calculating the integral

Now, substitute the exponential probability density functions for Machine 1 and Machine 2: \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \int_{x_1+t}^{\infty} \lambda_1 e^{-\lambda_1 x_1} \lambda_2 e^{-\lambda_2 x_2}\,dx_2 \,dx_1\) Integrate with respect to \(x_2\) first: \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \lambda_1 e^{-\lambda_1 x_1} \left[\int_{x_1 + t}^{\infty} \lambda_2 e^{-\lambda_2 x_2} \,dx_2 \right] \,dx_1\) The inner integral evaluates to \(e^{-\lambda_2 (x_1 + t)}\). Now the integral looks like: \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \lambda_1 e^{-\lambda_1 x_1} e^{-\lambda_2 (x_1 + t)}\, dx_1\) Simplify and integrate with respect to \(x_1\): \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \lambda_1 (\lambda_1+\lambda_2) e^{-(\lambda_1+\lambda_2) x_1} e^{-\lambda_2 t}\, dx_1\) Now, the integral evaluates to: \(P(X_1 < X_2 - t) = \frac{\lambda_1}{\lambda_1+\lambda_2} e^{-\lambda_2 t}\)
04

Final answer

The probability of Machine 1 failing before Machine 2 is put in use (time \(t\)) is: \(P(X_1 < X_2 - t) = \frac{\lambda_1}{\lambda_1+\lambda_2} e^{-\lambda_2 t}\)

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Most popular questions from this chapter

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). Let \(S_{n}\) denote the time of the \(n\) th event. Find (a) \(E\left[S_{4}\right]\), (b) \(E\left[S_{4} \mid N(1)=2\right]\) (c) \(E[N(4)-N(2) \mid N(1)=3]\)

For the conditional Poisson process, let \(m_{1}=E[L], m_{2}=E\left[L^{2}\right] .\) In terms of \(m_{1}\) and \(m_{2}\), find \(\operatorname{Cov}(N(s), N(t))\) for \(s \leqslant t .\)

Let \(X, Y_{1}, \ldots, Y_{n}\) be independent exponential random variables; \(X\) having rate \(\lambda\), and \(Y_{i}\) having rate \(\mu\). Let \(A_{j}\) be the event that the \(j\) th smallest of these \(n+1\) random variables is one of the \(Y_{i} .\) Find \(p=P\left[X>\max _{i} Y_{i}\right\\}\), by using the identity $$ p=P\left(A_{1} \cdots A_{n}\right)=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) \cdots P\left(A_{n} \mid A_{1} \ldots A_{n-1}\right) $$ Verify your answer when \(n=2\) by conditioning on \(X\) to obtain \(p\).

Let \(X\) be an exponential random variable. Without any computations, tell which one of the following is correct. Explain your answer. (a) \(E\left[X^{2} \mid X>1\right]=E\left[(X+1)^{2}\right]\) (b) \(E\left[X^{2} \mid X>1\right]=E\left[X^{2}\right]+1\) (c) \(E\left[X^{2} \mid X>1\right]=(1+E[X])^{2}\)

Let \(S_{n}\) denote the time of the \(n\) th event of the Poisson process \([N(t), t \geqslant 0\\}\) having rate \(\lambda\). Show, for an arbitrary function \(g\), that the random variable \(\sum_{i=1}^{N(t)} g\left(S_{i}\right)\) has the same distribution as the compound Poisson random variable \(\sum_{i=1}^{N(t)} g\left(U_{i}\right)\) where \(U_{1}, U_{2}, \ldots\) is a sequence of independent and identically distributed uniform \((0, t)\) random variables that is independent of \(N\), a Poisson random variable with mean \(\lambda t\). Consequently, conclude that $$ E\left[\sum_{i=1}^{N(t)} g\left(S_{i}\right)\right]=\lambda \int_{0}^{t} g(x) d x \quad \operatorname{Var}\left(\sum_{i=1}^{N(t)} g\left(S_{i}\right)\right)=\lambda \int_{0}^{t} g^{2}(x) d x $$

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