Question

The time (in hours) required to repair a machine is an exponentially distributed random variable with parameters$\lambda =\frac{1}{2}$. What is

$\left(a\right)$the probability that a repair time exceeds$2$hours?

$\left(b\right)$the conditional probability that a repair takes at least$10$hours, given that its duration exceeds$9$hours?

Short answer

$\left(a\right)$the probability that a repair time exceeds$2$hours is $0.3679$

$\left(b\right)$the conditional probability that a repair takes at least$10$hours, given that its duration exceeds$9$hours is$0.6065$

Step-by-step solution

Given information.

$\lambda =\frac{1}{2}$

Explanations.

Let$x$be the time required to fix a machine.

$x~Exp(\lambda =\frac{1}{2})$

$$\begin{gathered} pdf:f\left(x\right)=\lambda e^{\lambda x},0<x \\ cdf:p(X<x)=1-e^{-\lambda x} \end{gathered}$$

$$\begin{gathered} p(X\le x)=1-e^{(-0.5x)}forx>0 \\ \Rightarrow p(X>x)=e^{-\lambda x} \end{gathered}$$

Part a Step 3 Explanation.

The probability that the repair time exceeds $2$hours:

$$\begin{gathered} p(X>2)=e^{\frac{-1}{2}} \\ =0.367879 \end{gathered}$$

Part b Explanation.

$$\begin{gathered} p(X\ge 10|x\ge 9) \\ =\frac{p(x>10)}{p(x>9)} \\ =\frac{0.006738}{0.011109} \\ \approx 0.6065 \end{gathered}$$