Suppose a radioactive source is metered for two hours, during which time the total number of alpha particles counted is 482. What is the probability that exactly three particles will be counted in the next two minutes?
Answer the question by defining X for poisson distribution to be the number of particles counted in one minute.
I know how to do it When I take X : number of particles in next two minutes,
answer comes out to be 0.028
but not when x= no. of particles in one minute
Best Answer
It certainly seems more natural to take a $2$ minute unit of time.
But if you are required to take a $1$ minute unit, then let $X$ be the number of particles counted in the first minute; and let $Y$ be the number of particles counted in the second minute. These should be (independent) Poisson random variables, each with the (empirically determined) parameter of $\lambda=\frac{482}{120}\approx 4.0167$.
We obtain $$P(X+Y=3)=P(X=3)P(Y=0)+P(X=2)P(Y=1)+P(X=1)P(Y=2)+P(X=0)P(Y=3)$$ This seems to work out to the same answer suggested in the original post.