For what it's worth, here's a condensation of one approach I have used for
this topic:
Starting Points (already learned):
(1): The probability that, of N live nuclides, a specified nuclide will
decay during the time interval dt is proportional to dt and is equal to
lamda*(dt). The probability that any one of the N live nuclides decays in
dt is then equal to (N*lamda*dt).
(2): The Poisson distribution giving the probability that any n nuclides
decay in t seconds is given by
P(n,t) = [(rt)^n]exp(-rt)/n! , where r=N*lamda is the rate of decay
(dN/dt)
Aside: (2) is derivable from (1) and the assumptions (n<<N, etc ) which
carry the Binomial into the Poisson distribution.
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(A): Distribution of times between successive counts, f0(t):
From (2), the probability that zero decay in t seconds is P(0,t) =
exp(-rt).
Then the joint probability that zero decay in t seconds and any one (and
only one) of the N decay in the next dt is equal to:
exp(-rt)*(N*lamda*dt) =
r*exp(-rt)dt. The distribution of times between successive counts is then
f0(t) = r*(exp(-rt).
(B:) Distribution of times between 2 successive counts, f1(t):
The joint probability that any one of the N decays during t and any one of
the (N-1) others decays in the next dt is given by P(1,t)*(N-1)*lamda*dt =
rt*(exp(-rt)*(N-1)*lamda*dt ,
which, for N>>1 (already presumed), is equal to:
t*r^2*exp(-rt)*dt. The distribution of times between every two counts is
f1(t) = t*r^2*exp(-rt).
(note that this does have a peak value)
(C:) The general case - Dist of times between m+1 counts, fm(t):
The joint probability that any m decay during t and one of those left
(N-m) decays in the next dt is [(rt)^m]exp(-rt)*(N-m)*lamda*dt/m! . For
N >> m this gives for the distribution of times between (m+1) counts: