# What is the probability emission of next 2 alpha particles will take at least 2 seconds?

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.

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 $9\exp(-8)\approx 3\times 10^{-3}$.
• @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. Commented Nov 23, 2016 at 19:11