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A factory contains 8 machines. Any one of the machines can breakdown independently of the other machines following an exponential distribution with a mean value of 10 days. What is the probability that the time until the first machine break-down is at least 5 days?

obviously $\mu = 10$ and hence, $\lambda = 1/10$

I think we have that $P(X>5) = 1-P(X<5) = 1-(1-e^-5\times0.1)) = 0.60653$ but this is just for one machine. I'm not sure how to deal with the fact that there are 8 machines.

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1 Answers1

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Hint. Let $X_1, \dots, X_n$ be independent exponentially distributed random variables with rate parameters $\lambda_1,\dots, \lambda_n$. Then

$$\min \left \{X_1,\dotsc,X_n \right \}$$

is also exponentially distributed, with parameter

$$\lambda = \lambda_1+\dotsb+\lambda_n.$$ (Source: Wikipedia.)

So you need to calculate $\mathbb P(X\geq 5)$ where $X$ has an exponential distribution with parameter $\frac8{10}$.