Question

The time (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ = 1. What is a) the probability that a repair time exceeds 2 hours? b) the conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours?

Answer 1

Explanation:

Given

`\lambda =1`

Let X denote the time to repair a machine

(a)Probability that a repair time exceeds 2 hours

`P(X>2)=1-P(x\leq 2)`

`P(X>2)=1-\int_(0)^(2)(1)/(2)e^{-(x)/(2)}dx`

`=1+\left [ e^{-(x)/(2)}\right ]_0^2`

`P(X>2)=e^(-1)`

(b)Probability that a repair takes at least 10 hr, given that its duration exceeds 9 hours

`P\left ( X\geq 10|X>9\right )=(P\left ( X\geq 10\right ))/(P\left ( X>9\right ))`

`=\frac{\int_(10)^(\infty )(1)/(2)e^{-(x)/(2)}dx}{\int_(9)^(\infty )(1)/(2)e^{-(x)/(2)}dx}`

`=\frac{2e^{-(1)/(2)}|_(10)^(\infty )}{2e^{-(1)/(2)}|_(9)^(\infty )}`

`=e^{-(1)/(2)}`