# Solved: The time (in hours) required to repair a machine is an exponential distributed random variable with parameter

#### ByDr. Raju Chaudhari

Feb 24, 2021

The time (in hours) required to repair a machine is an exponential distributed random variable
with parameter $\lambda =1/2$. 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?

## Solution

Let $T$ denote the time (in hours) required to repair a machine. Given that $T$ is exponentially distributed with $\lambda = 1/2$.

The pdf of $T$ is

 \begin{aligned} f(t) &= \frac{1}{2}e^{-t/2},\; t>0. \end{aligned}

The distribution function of $T$ is

 \begin{aligned} F(t) &= P(T\leq t) = 1- e^{-t/2}. \end{aligned}

(a) The probability that a repair time exceeds 2 hours is

 \begin{aligned} P(t> 2) &= 1- P(t\leq 2)\\ & = 1- F(2)\\ & = 1- \big[1- e^{-2/2}\big]\\ &= e^{-1}\\ & = 0.3679 \end{aligned}

(b) The conditional probability that a repair takes at least 10 hours, given that its duration exceeds 9 hours is

 \begin{aligned} P(t \geq 10|t>9) &= \frac{P(t\geq 10)}{P(t>9)}\\ & = \frac{1- P(t<10)}{1-P(t<9)}\\ & = \frac{1- F(10)}{1-F(9)}\\ &= \frac{1-(1-e^{-10/2})}{1-(1-e^{-9/2})}\\ & = \frac{e^{-10/2}}{e^{-9/2}}\\ &=0.6065 \end{aligned}