would mean in 5 minutes we

1
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
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
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
The probability of a single COSMIC RAY
passing through a small area of a detector
within a small interval of time Dt is p ltlt 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?
( 1 - p )??? ??? misses
pn n hits
( 1 - p )N-n N-nmisses
5
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 e2.718281828
6
ln (1-p)N-n (N-n) ln (1-p) and since p ltlt
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)
If we have to wait a large number of intervals,
N, for a relatively small number of counts,n nltltN
P(n) pn e-pN
8
P(n) pn e-pN
And since
N - (n-1)
? N (N) (N) (N) Nn for nltltN
9
P(n) pn e-pN
P(n) pn e-pN
P(n) e-Np
10
P(n) e-Np
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 0.018315639
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) 0.018315639
P(4) 0.195366816 P(1) 0.073262556 P(5)
0.156293453 P(2) 0.146525112 P(6)
0.104195635 P(3) 0.195366816 P(7) 0.059540363
12
P(n) e-Np
Hey!
What does Np represent?
13
Another useful series we can exploit
m, mean
n0 term
n / n! 1/(???)
14

m, mean
let m n-1 i.e., n
whats this?
15
m, mean
m (Np) e-Np eNp
m Np
16
m Np
P(n) e-m
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 j1
21
What is n2 for a Poisson distribution?
22
What is n2 for a Poisson distribution?
This is just em again!
23
The standard deviation s should provide an
estimate of the error in such counts
In other words
s 2 m
s ?m