Question:
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.
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 $9\exp(-8)\approx 3\times 10^{-3}$.