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Re: Geiger, not binomial ?



I am not comfortable with the p of a binomial distribution.
Since the mean=p*n (using n for the number of attempts
and r for the number of successes) then p depends on n.
This conflict with a view that p is an intrinsic property of
a setup. Is it not true that p-->r/n when n-->oo? Defined
in that way it should not depend on n.

Example: trying to hit a target with an arrow. My skill
for being successful can be described by the probability
of a hit in a single trial. If p=0.2 then I am expected to
hit the target 2 out of 10 attempts. In reality the outcome
will fluctuate around 2. In this case n is given and all is
fine.

But in the case of a Geiger counter (where the mean was
2.20) both n=170 and n=340 give a satisfactory fit to
experimental data. This implies that both p=0.0129 and
p=0.00647 are OK. How can it be? Isn't the objective
probability of recording one event per unit time equal
to 0.24 (7278/30058)? Is the word "probability" used
for two different things in this "paradox"?

John Denker wrote:

At 09:47 PM 3/23/00 -0600, Jack Uretsky wrote:
Something is wrong here. Let N_n be the number of occurrences
of n counts in the interval T, and let r be the counting rate. Then,
for a Poisson distribution the total number of counts N is:
N=Sum of n= 0 to infinity of N_n,
where N_n = N{(rT)^n/n!}exp(-rT)
and both the mean and variance of the distribution = rT.

OK.

In other words,
nothing but the total no. of counts depends on N.

I'm not sure what that means. "Nothing" seems like a too-sweeping term.

In this case, r is the probability per unit time that one nucleus
will decay - that in fact is one way to derive the Poisson distribution.

If r is the counting rate (which I take to mean average counting rate) then
don't you mean r/N (not plain r) is the probability per unit time that one
nucleus will decay?