Emission of alpha-particles occurs according to a Poisson process. Suppose that, on average, the number of alpha-particles emitted from a radioactive substance is 4 every second. What is the probability that emission of the next two alpha-particles will take at least 2 seconds?

My solution:

Defining X to be the time in seconds for one alpha particle to be emitted.

Since it is a Poisson process, for one particle he have P(X>=2)=1-P(x<2)=1-Integral from 0 to 2 of ((e^(-4)*4^x)/(x!)). Since we have 2 alpha particles to be emitted then we square the answer.

I don't know if I got the integration and squaring the answer correctly.


1 Answer 1


For the next two alpha-particles to take at least 2 seconds, means that within that 2 second period you can have zero or 1 emission. Two emissions within that period would mean that the two particles have been emitted in some time less than two seconds, which is not consistent with the requirement.

From the average emission rate, it follows that on average there are 8 emissions in a two second period. The probability that there are zero or 1 emissions, is thus $9\exp(-8)\approx 3\times 10^{-3}$.

  • $\begingroup$ The question says "emission of the next two alpha-particles will take at least 2 seconds" so the two emissions must happen at second 2 or later. So I decided to do the 1-{"a single emission happening from 0 to 2"} it will give me what i need. All I need now is to make is for two emissions. So can I square the answer and get the result? $\endgroup$
    – anon
    Commented Nov 22, 2016 at 19:38
  • $\begingroup$ @ErinAvllazagaj The entire process of emitting the two will have to take more than 2 seconds, that process can thus be partially completed within that time frame, as long as it is not fully competed. Therefore you need to sum over the probability of zero or 1 emission within the two second time frame. $\endgroup$ Commented Nov 23, 2016 at 3:10
  • $\begingroup$ Ye you are right I got the idea. But how did you calculate probability of 0 or 1 emissions? $\endgroup$
    – anon
    Commented Nov 23, 2016 at 8:47
  • 1
    $\begingroup$ @anon You have to add up the probability of 0 and 1 event. The probability of $k$ events is $$p(k) = \frac{\lambda^k}{k!}\exp(-\lambda)$$ and we have that $\lambda = 8$, because with 4 events per second, the average number of events in 2 seconds is 8. $\endgroup$ Commented Nov 23, 2016 at 19:11

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