# Finding probability a particle will appear after t seconds (exponential r.v)

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

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

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