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Assume $T_1,T_2...T_{300}$ are the time one talking in the lesson before he was stopped by his teacher.$T$ are $iid$ and follow an exponential distribution with mean $exp({\theta})$ what is the probability that he will not be punished.Noted that teacher will aware that he is talking if $T_{min}\ge \frac{ln2}{300}$

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Hence one is NOT punished if teacher catches one chatting 299 times during the lesson? – Did Jan 3 '12 at 11:39

He will not be punished if teacher is not aware he is talking. Since the problem doesn't talk about whether the student is punished or not when noticed by teacher for his talking, I assume that by "not punished", you actually mean "not noticed by teacher".

The teacher notices when $T_{min} \ge \frac{ln 2}{300}$.

Hence, probability of teacher not noticing is $Pr(T_{min} < \frac{ln 2}{300})$.

Thus, required probability= $1-Pr(T_{min} \ge \frac{ln 2}{300}) \cdots (1)$.

Now, I assume $T_{min}$ is the minimum amongst $T_1,T_2,\cdots T_{300}$. All are $iid$ and follow $exp(\theta)$ distribution. Thus, we have:

$Pr(T_{min}\ge k)=\prod_{i=1}^{300} Pr(T_i\ge k)=(e^{-\theta k})^{300}.$

Thus, $Pr(T_{min}\ge k)=e^{-300 \theta k}$.

and Thus, $Pr(T_{min}\ge \frac{ln 2}{300})=e^{-300 \theta {\frac{ln 2}{300}}}=e^{-\theta ln2}=2^{-\theta}$.

Thus, from $(1)$, Required probability= $1- 2^{-\theta}$ where $\theta$ is the value of exponential parameter. Once you specify the value of $\theta$, the required probability can be calculated from the derived formula.

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