Question

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

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

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.

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: \(P(X_1 < X_2 - t) = \int_{0}^{\infty} \int_{x_1+t}^{\infty} f(x_1, x_2) \,dx_2 \,dx_1\)

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: \(P(X_1 < X_2 - t) = \frac{\lambda_1}{\lambda_1+\lambda_2} e^{-\lambda_2 t}\)

Final answer

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