1. 1200 Hz = 1200/sec would mean in 5 minutes we
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 5 minutes we
should expect to count about
6,000 events B. 12,000 events
C. 72,000 events D. 360,000 events
E. 480,000 events F. 720,000 events

2. 1200 Hz = 1200/sec would mean in 3 millisec we
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 3 millisec we
should expect to count about
0 events B. 1 or 2 events
C. 3 or 4 events D. about 10 events
E. 100s of events F. 1,000s of events
1 millisec = 10-3 second

3. 1200 Hz = 1200/sec would mean in 100 nanosec we
Example: a measured rate of
1200 Hz = 1200/sec
would mean in 100 nanosec we
should expect to count about
0 events B. 1 or 2 events
C. 3 or 4 events D. about 10 events
E. 100s of events F. 1,000s of events
1 nanosec = 10-9 second

4. pn × ( 1 - p )N-n × ( 1 - p )??? p << 1 ( 1 - p )  1
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt is
p << 1
the probability
that none pass in
that period is
( 1 - p )  1
While waiting N successive intervals
(where the total time is t = NDt )
what is the probability that we observe
exactly n events? pn
n “hits”
× ( 1 - p )N-n
N-n“misses”
× ( 1 - p )???
??? “misses”

5. P(n) = nCN pn ( 1 - p )N-n ln (1-p)N-n = ln (1-p)
While waiting N successive intervals
(where the total time is t = NDt )
what is the probability that we observe
exactly n events?
P(n) = nCN pn ( 1 - p )N-n
From the properties of logarithms
you just reviewed
ln (1-p)N-n = ln (1-p)
ln (1-p)N-n = (N-n) ln (1-p)
???
ln x  loge x e=

6. e???? = ???? e-p(N-n) = (1-p)N-n ln (1-p)N-n = (N-n) ln (1-p)
and since p << 1
ln (1-p) 
- p
ln (1-p)N-n = (N-n) (-p)
from the basic definition of a logarithm
this means
e???? = ????
e-p(N-n) = (1-p)N-n

7. P(n) = pn ( 1 - p )N-n P(n) = pn e-p(N-n) n<<N P(n) = pn e-pN
If we have to wait
a large number of intervals, N, for a
relatively small number of counts,n
n<<N
P(n) = pn e-pN

8. P(n) = pn e-pN  N (N) (N) … (N) = Nn And since N - (n-1)
for n<<N

9. P(n) = pn e-pN
P(n) = pn e-pN
P(n) = e-Np

10. P(n) = e-Np P(0) = P(4) = P(1) = P(5) = P(2) = P(6) = P(3) = P(7) =
If the average rate of some random event is
p = 24/min = 24/60 sec = 0.4/sec
what is the probability of recording n events
in 10 seconds?
P(0) = P(4) =
P(1) = P(5) =
P(2) = P(6) =
P(3) = P(7) =

11. P(n) = e- 4
e-4 =
If the average rate of some random event is
p = 24/min = 24/60 sec = 0.4/sec
what is the probability of recording n events
in 10 seconds?
P(0) = P(4) =
P(1) = P(5) =
P(2) = P(6) =
P(3) = P(7) =

12. P(n) = e-Np
Hey!
What does Np represent?

13. m, mean = Another useful series we can exploit n=0 term
n / n! = 1/(???)

14. m, mean å = - N )! ( ) (N m p e
let m = n-1
i.e., n =
what’s this?

15. m, mean
m = (Np) e-Np eNp
m = Np

16. P(n) = e-m m = Np Poisson distribution
probability of finding exactly n
events within time t when the events
occur randomly, but at an average
rate of m (events per unit time)

17. Recall: The standard deviation s is
a measure of the mean (or average)
spread of data away from its own
mean. It should provide an estimate
of the error on such counts.

18. The standard deviation s should provide
an estimate of the error in such counts

19. What is n2 for a Poisson distribution?
first term in the series is zero
factor out e-m which is independent of n

20. What is n2 for a Poisson distribution?
Factor out a
m like before
Let j = n-1  n = j+1

21. What is n2 for a Poisson distribution?

22. What is n2 for a Poisson distribution?
This is just
em again!

23. s 2 = m s = m The standard deviation s should provide
an estimate of the error in such counts
In other words
s 2 = m
s = m