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LennoxLewis
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Homework Statement
A detector is made up of 64 by 64 detection strips - perpendicular, i.e. 64 strips along the x-axis and 64 along the y-axis. Particles impact along the z-axis, evenly distributed among the detector surface. When a particle hits the detector, one x-strip and one y-strip will fire, and the intersection of the two gives the impact position. The time window for the strips to reset is 2 us.
Now, if two particles impact within 2 us, a false positive (or better put: a wrong location) can be gotten, if the wrong strips are combined. Exactly how this happens is not very relevant, what matters here is the simple probability of getting two hits within 2 us.
The particle influx is 500 Hz, randomly distributed in both time and space.
Homework Equations
The binomial distribution being defined by three parameters:
n: number of trials
p: chance of succes
v: number of successes with n trials
The Attempt at a Solution
Since the time resolution is 2us, you can divide one second up into 500.000 segments of 2us each. distributing 500 particles randomly among these segments, what are the odds of two ending up in the same segment?
From what i gather, it's equivalent to throwing dice and getting the same number twice, only the dice has 500.000 numbers, and you throw 500 times.
Of course, having 3 hits in 2us would also yield a false result, but i think this probability will be another factor of 1000 or so smaller, and i suspect the probability of two hits is pretty low in the first place, so let's forget about that for the moment...
But does anyone know how to calculate this probability?