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I am trying to solve the following problem.

The time $T$ required to repair a machine is an exponentially distributed random variable with mean 10 hours.

a) What is the probability that a repair takes at least 15 hours given that its duration exceeds 12 hours? b) What is the probability that the combined time to repair two machines is at least 20 hours?

Solution Attempt

Since mean is given to be 10 hours hence $\lambda = \dfrac {1}{10}$ and the probability distribution of the time is given as $e^{-\lambda t} = e^{-\dfrac {1}{10} t} $

a) $P(T>15 |T>12) = P(0 $ repairs in $ (12, 15]) = e^{-\dfrac {1}{10} 3}$

b) let $T_1$ be the r.v representing time to repair the first machine and $T_2$ be the r.v representing time to repair the second machine. So we seek to evaluate $P(T_1 + T_2 > 20)$ we know both of these time should be independent as the exponential distribution process to memory less but i am not sure how to proceed from here.

Any help would be much appreciated.

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I think you have a problem. You are assuming that if machine 1 is repaired first then the time to repair machine 2 starts when machine 1 is repaired. This is not the case. but the memoryless property does tell you that given machine one is repaired at time T1 the remaining time to repair machine 2 still has the exponential distribution with rate 1/10. –  Michael Chernick Aug 27 '12 at 20:43
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2 Answers 2

up vote 1 down vote accepted

$$\mathrm P(T_1+T_2\gt20)=\mathrm P(T_1\gt20)+\int_0^{20}\mathrm P(T_2\gt20-t)\cdot\lambda\mathrm e^{-\lambda t}\cdot\mathrm dt $$ $$ \mathrm P(T_1+T_2\gt20)=\mathrm e^{-20\lambda}+\int_0^{20}\mathrm e^{-\lambda (20-t)}\cdot\lambda\mathrm e^{-\lambda t}\cdot\mathrm dt=\ ...$$

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While the numerical answer you get for part (a) is correct, I think that your work indicates some misinterpretation of the concepts. $T$ is the time required to complete a repair, and its complementary cumulative distribution function is $\exp(-t/10)$, that is, $$P\{T > t\} = 1 - F_T(t) = e^{-t/10}.$$ The question in part (a) asks for a conditional probability $P\{T > 15 \mid T > 12\}$ which is by definition given by $$P\{T > 15 \mid T > 12\} = \frac{P(\{T > 15 \}\cap\{T > 12\})}{P\{T > 12\}} = \frac{P\{T > 15 \}}{P\{T > 12\}} = \frac{e^{-15/10}}{e^{-12/10}}= e^{-3/10}$$ which is the same answer as you obtained, but it is not the probability of $0$ repairs in $(12,15]$. The repairing began at time $t =0$ and the question asks: if the repair is still ongoing at time $t = 12$, what is the conditional probability that it is still ongoing at $t = 15$, and thus completes at some time $T$ larger than $15$. Your use of the phrase

$$0 ~ \text{repairs in} ~(12,15]$$

almost makes it sound like the repairs are a Poisson arrivals process.

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