Please help me solve the following question with two parts.

$T$ is the time required to repair a machine. We have that $T$ is exponentially distributed with a mean of $\frac{1}{2}$ hours.

For the first part of the question, I am asked to find the probability that repair time exceeds $\frac{1}{2}$ hours. I find $$P(T> \frac{1}{2})= \frac{1}{e}$$

I am a bit stuck on the second part to the question: "What is the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours"?

I was thinking the answer is

$$P(T\geq 12.5 \, | \, T>12)=P(T\geq .5) = \frac{1}{e} \, \text{ ,}$$

since exponential distribution is memoryless. Is this correct?

  • 1
    $\begingroup$ Sounds right to me. $\endgroup$ – Dilip Sarwate Apr 8 '13 at 2:46
  • $\begingroup$ @DilipSarwate Would this just equal $\frac{1}{e}$, like the first part? $\endgroup$ – Alti Apr 8 '13 at 2:54
  • $\begingroup$ $P(T\ge{12.5}|T\gt{12})=\frac{P(T\ge{12.5})}{P(T\gt{12})}$. And yes, since it is memoryless, = $P(T\ge{.5})$. $\endgroup$ – Eleven-Eleven Apr 8 '13 at 3:01
  • $\begingroup$ Yes, the answer to the second part is also $\frac{1}{e}$. As a practical, non-mathematical, issue, I would worry that the conditioning event $\{T > 12\}$ is such a rare event that the exponential model might not be a good fit to the real-life problem any more. $\endgroup$ – Dilip Sarwate Apr 8 '13 at 12:19
  • $\begingroup$ Duplicate (but which is a duplicate of which?) $\endgroup$ – Jacob Akkerboom Mar 10 '14 at 15:23

Yes, this is correct, this is indeed what is meant by memorylessness.

In general, for all $s\geq 0$, $t \geq 0$, we have

$$P(T>s+t\mid T>s) = P(T>t) = e^{-\lambda t}.$$

We can prove the memorylessness by transforming the expression on the LHS using the definition of conditional probability, as follows

$$ \begin{align} P(X>s+t\mid X>s) &= \frac{P\big(X>s+t\cap X>s\big)}{P(X>s)} = \frac{P(X>s+t)}{P(X>s)} \\ &= \frac{\int_{s+t}^{\infty}\lambda e^{-\lambda x}dx}{\int_{s}^{\infty}\lambda e^{-\lambda x}dx}= \frac{0- -e^{-\lambda (s+t)}}{0- -e^{-\lambda s}} = e^{-\lambda t} = P(X>t) \end{align} $$

Accreditation: This is a slight modification of this answer by Tianyu Zheng

  • $\begingroup$ Because I have made this answer a community wiki post, I feel I can ask for an upvote, so please upvote this to 1. This will enable me to suggest to close the linked question in the comments as a duplicate of this one. $\endgroup$ – Jacob Akkerboom Mar 10 '14 at 16:14
  • $\begingroup$ But this question would seem to be the duplicate, being asked 3 months after the other. $\endgroup$ – robjohn Mar 12 '14 at 7:17
  • $\begingroup$ @robjohn thanks for your reply. I realise time order is regarded as important when closing as duplicate, but I've never quite understood why, though maybe it is to prevent this kind of situation. In the other question, the OP is unclear about more than one thing and I would say the mistake about the events there will usually only distract users. Jeff seems to approve of my strategy here. $\endgroup$ – Jacob Akkerboom Mar 12 '14 at 9:06

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