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Suppose you are watching a radioactive source that emits particles at a rate described by the exponential density with $\lambda=1$

The probability $P(0,T)$ that a particle will appear in the next T seconds is $P ([0, T ])$ = $\int_0^T\lambda e ^{-\lambda t}$

Find the probability that a particle (not necessarily the first) will appear after 4 seconds from now.


How can I setup my equation? My hunch is that this has something to do with the memoryless property of the exponential r.v.

Would appreciate any guidance. thanks

NOTE: The answer given is 1

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  • $\begingroup$ it is probalbty zero $\endgroup$
    – terrace
    Apr 14, 2016 at 19:56
  • $\begingroup$ @terrace the answer given is 1. But why? $\endgroup$
    – misheekoh
    Apr 14, 2016 at 20:04
  • $\begingroup$ is it poisson distribution"? $\endgroup$
    – terrace
    Apr 14, 2016 at 20:05
  • $\begingroup$ @ No. It's the exponential $\endgroup$
    – misheekoh
    Apr 14, 2016 at 20:09
  • $\begingroup$ look at my answer - I think that is correct $\endgroup$
    – terrace
    Apr 14, 2016 at 20:09

1 Answer 1

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If you wanted to find the probability that the first particle would appear after $4$ seconds from now, I think it would be this: $$1-\int_0^4 \lambda e^{-\lambda t} \,\mathrm{d}t$$

But you are asked only the probability that some particle will appear after $4$ seconds from now is just a certainty, or $1$. This is because whenever a particle spawns the probability resets (is this the memorylessness you were talking about?).

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  • $\begingroup$ Hmm, that seems right. But why the probability that some particle will spawn after 4 seconds from now is 1? Because $\lambda = 1$? $\endgroup$
    – misheekoh
    Apr 14, 2016 at 20:39

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