1. DABC DBAC DCBA ADBC DACB CABD
Permutations
4 CDs fit the CD wallet you carry to school each day.
You can arrange them in the holder by release date,
alphabetically by artist or title,
or ranked in order of preference.
How many distinct arrangements are possible?
For simplicity label them ABCD and try writing out
an exhaustive listing of all possible arrangements:
ABCD BACD CBAD DBCA ACBD ABDC
BCDA ACDB BADC BCAD CBDA BDCA
CDAB CDBA ADCB CADB BDAC DCAB
DABC DBAC DCBA ADBC DACB CABD
If none have been repeated or missed,
24 different arrangements are possible.

2. There are 4 titles to choose from as 1st in the holder...
4 different ways to start the arrangement.
That leaves 3 from which to select the 2nd...
then two to pick from to hold the next to last slot.
By the time you have made that choice,
there is only one left to be last.
That makes for
(4 possible 1st place holders)
(3 possible remaining 2nd place holders)
(2 ways to fill the next-to-the-last slot)
(1 last-place-holder)
= 4·3·2·1=24.

3. 6! = 6·5·4·3·2·1=720 N! = N (N - 1) (N - 2) …3 · 2 · 1
Each arrangement is called a permutation,
and we employ the special notation of N!
to represent the string of factors counting down from N to 1:
4! (read "four factorial") = 4·3·2·1=24 or
6! = 6·5·4·3·2·1=720
N! = N (N - 1) (N - 2) …3 · 2 · 1

4. Your group of 10 finds 10 seats together
in a front row of the theater!
How many different seating arrangements are possible?
10!
= 10  9  8  7  6  5  4  3  2  1
= 3,628,800
Express using factorials the number of ways
a deck of 52 cards can be shuffled.
52!
= 1067

5. Combinations Four cards are set out on the table.
You are asked to pick any three.
How many different choices are available to you?
In other words:
How many different (sub)sets of 3
can you build out of a pool of 4 objects?
“Picking 3” in this case is the same as “rejecting 1”
(deciding which one NOT to pick)
there are obviously 4 ways to do this.
We say 4C3 = 4
(the number of different combinations of
3 taken from a total of 4 is equal to 4).
Listing them is easy:
ABC ABD ADC BCD.

6. 4C2 = 6 4C1 = 4 4C3 = 4C2= We can count off other combinations:
AB AC AD BC BD CD
4C1 = 4
A B C D
Just check out:
4C3 =
4C2=

7. 10C3 = 52C5 = In a club with 10 members, how many
ways can a committee of 3 be selected?
10C3 =
5 cards are drawn (one at a time)
from a well-shuffled deck. How
many different hands are possible?
52C5 =

8. A “fair” coin is flipped at the start of a football
game to determine which team receives the ball.
The “probability” that the coin comes up HEADs
is expressed as
A. 50/50 “fifty-fifty”
B / “one-half”
C : “one-to-one”
In betting parlance the odds are 1:1;
we say the chances are 50/50, but
the mathematical probability is ½.

9. There are 36 possible outcomes for the toss of
2 six-sided dice. If each is equally likely, the
most probable total score of any single roll is?
A B C. 6
D E F. 9

10. A green and red die are rolled together.
What is the probability of scoring an 11?
A. 1/4 B. 1/6 C. 1/8
D. 1/12 E. 1/18 F. 1/36

11. “Snake eyes” give
the minimum roll.
“Boxcars” give
the maximum roll.
The probability of rolling any even number,
Probability(even) ______ Probability(odd),
the probability of rolling any odd number.
A. > B. = C. <

12. Number of ways to score die totals

13. A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2 B. 1/4
C. 1 D. 0

14. A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2 B. 1/4
C. 1 D. 0
It is equally likely to observe
two heads (HH) as two tails (TT)
T) True. F) False.

15. A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2 B. 1/4
C. 1 D. 0
It is equally likely to observe
two heads (HH) as two tails (TT)
T) True. F) False.
It is equally likely for the two outcomes
to be identical as to be different.

16. A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2 B. 1/4
C. 1 D. 0
It is equally likely to observe
two heads (HH) as two tails (TT)
T) True. F) False.
It is equally likely for the two outcomes
to be identical as to be different.
The probability of at least one head is
C. 3/4 D. 1/3

17. The probability of “throwing a 7”6 5 4 3 2 1
The probability of “throwing a 7”
P (7) = 6/36 = 1/6
P (7±5) = ?

18. P (7) = 6/36 = 1/6 P(7±5) = 36/36 =1 P(7±1) = P(7±2) = ? ? 6 5 4 3 2 1
P (7) = 6/36 = 1/6
P(7±5) = 36/36 =1
P(7±1) =
P(7±2) =
the full
range! ? ?

19. P (7) = 6/36 = 1/6 P(7±5) = 36/36 =1 P(7±1) = 16/36 = 4/92 1
P (7) = 6/36 = 1/6
P(7±5) = 36/36 =1
P(7±1) = 16/36 = 4/9
P(7±2) = 24/36 = 2/3
the full
range!

20. Height in inches of sample of 100 male adults
Frequency table of the distribution of heights
56 1
57 0
58 1
59 1
60 2
61 2
62 5
63 3
64 3
65 7
66 7
67 8
68 12
69 8
70 10
71 7
72 8
73 5
74 3
75 3
76 2
77 1
78 0
79 1

21. Range = xmax-xmin = 23
Mean = =67.20
Mode = 68
Median = 68.52

22. The range can be misleading if the sample
averages alone don’t
show how tightly
clumped together
the data is
The range can be misleading if the sample
includes rare extreme data points.

23. 3 different distributions with the same mean and range

24. s = (xi – m)2 N-1 mean, m describe the spread in data
by a calculation of
the average distance
each individual data point
is from the overall mean N
(xi – m)2
N-1
s =
i=1
mean, m

25. Range = xmax-xmin = 23
Mean = =67.20
s = 4.357

26. Frequency table of the distribution of heights
Number of classes K = log10N
= log10100
= ×2 = 7.6  8
Frequency table of the distribution of heights
56 1
58 1
59 1
61 2
62 5
64 3
65 7
67 8
68 12
70 10
71 7
73 5
74 3
76 2
77 1
79 1

28. If events (the emission of an  particle
from a uranium sample, or the passage
of a cosmic ray through a scintillator)
occur randomly in time,
repeated measurements of the time
between successive events should
follow a “normal” (Gaussian or
“bell-shaped”) curve
T) True F) False.

29. If events occur randomly in time, the probability that the next event
occurs in the very next second
is as likely as it not occurring
until 10 seconds from now.
T) True F) False.

30. P(1)Probability of the first count occurring in in 1st second
in 10th second
i.e., it won’t happen until the 10th second
???
P(1) = P(10)
= P(100)
= P(1000)
= P(10000)

31. Imagine flipping a coin until you get a head.
Is the probability of needing to flip just once
the same as the probability of needing to flip
10 times?
Probability of a head on your 1st try,
P(1) =
Probability of 1st head on your 2nd try,
P(2) =
Probability of 1st head on your 3rd try,
P(3) =

32. Probability of a head on your 1st try,
Probability of 1st head on your 2nd try,
P(2) =1/4
Probability of 1st head on your 3rd try,
P(3) =1/8
Probability of 1st head on your 10th try,
P(10) =

33. P(1) + P(2) + P(3) + P(4) + P(5) + •••
What is the total probability of
ALL OCCURRENCES?
P(1) + P(2) + P(3) + P(4) + P(5) + •••
=1/2+ 1/ /8 + 1/16 + 1/32 + ••• ?

34. A six-sided die is rolled repeatedly until it gives a 6.
What is the probability that one roll is enough?

35. A six-sided die is rolled repeatedly until it gives a 6.
What is the probability that one roll is enough?
1/6
What is the probability that it will take exactly
2 rolls?

36. A six-sided die is rolled repeatedly until it gives a 6.
What is the probability that one roll is enough?
1/6
What is the probability that it will take exactly
2 rolls?
(probability of miss,1st try)(probability of hit)=

37. A six-sided die is rolled repeatedly until it gives a 6.
What is the probability that one roll is enough?
1/6
What is the probability that it will take exactly
2 rolls?
(probability of miss, 1st try)(probability of hit)=
What is the probability that
exactly 3 rolls will be needed?

38. counts for RANDOM EVENTS fluctuate counting cloud chamber tracks
CROP workshop participants have seen
counts for RANDOM EVENTS fluctuate
counting cloud chamber tracks
Geiger-Meuller tubes clicking in
response to a radioactive source
“scaling” the cosmic ray singles rate
of a detector (for a lights on/lights
off response)

39. Why? Cosmic rays form a steady background impinging on the earth
equally from all directions
measured rates
NOT literally CONSTANT
long term averages
are just reliably consistent
These rates ARE measurably affected by
Time of day
Direction of sky
Weather conditions
Why?

40. Can you conclude the phenomena has a ~1/hour rate of occurring?
You set up an experiment to observe some phenomena
…and run that experiment
for some (long) fixed time…
but observe nothing: You count ZERO events.
What does that mean?
If you observe 1 event in 1 hour of running
Can you conclude the phenomena has
a ~1/hour rate of occurring?

41. Random events arrive independently unaffected by previous occurrences
unpredictably
0 sec time
A reading of 1
could result from the lucky capture of an exceeding
rare event better represented by a much lower rate
(~0?).
or the run period could have just missed an event
(starting a moment too late or ending too soon).

42. A count of 1 could represent a real average
as low as 0 or as much as 2
1 ± 1
A count of 2
2 ± 1? ± 2?
A count of 37
37 ± at least a few?
A count of 1000
1000 ± ?

43. p << 1 probability per unit time of a cosmic ray’s passage
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
cosmic rays arrive at a fairly stable, regular rate
when averaged over long periods
the rate is not constant nanosec by nanosec
or even second by second
this average, though, expresses the
probability per unit time
of a cosmic ray’s passage

44. 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

45. 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

46. 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

47. (even for a fairly large surface area) 72000/min=1200/sec
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
for example
(even for a fairly large surface area)
72000/min=1200/sec
=1200/1000 millisec
=1.2/millisec
= /msec
= /nsec

48. p << 1 A. p B. p2 C. 2p D.( p - 1) E. ( 1 - p)
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
The probability of NO cosmic rays passing
through that area during that interval Dt is
A. p B. p C. 2p
D.( p - 1) E. ( 1 - p)

49. p << 1 A. p B. p2 C. 2p D.( p - 1) E. ( 1 - p)
The probability of a single COSMIC RAY passing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
If the probability of one cosmic ray passing
during a particular nanosec is
P(1) = p << 1
the probability of 2 passing within the same
nanosec must be
A. p B. p C. 2p
D.( p - 1) E. ( 1 - p)

50. 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”

51. 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
ln (1-p)N-n = ln (1-p)
ln (1-p)N-n = (N-n) ln (1-p)
???
ln x  loge x e=

52. 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

53. 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

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

55. P(n) = pn e-pN
P(n) = pn e-pN
P(n) = e-Np
Consider an example…

56. P(n) = e- 4 e-4 = 0.018315639 P(0) = P(4) = P(1) = P(5) =
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) =

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

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

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

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

61. 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)

62. m=1
m=4
m=8

63. Another abbreviation (notation):
mean, m = x (the average x value)
i.e.

64. 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. or
for short

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

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

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

68. What is n2 for a Poisson distribution?

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

70. 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

71. m N N ± N Assuming any measurement N usually
gives a result very close to the true m
the best estimate of error
for the reading is N
We express that statistical error
in our measurement as
N ± N

72. 1000
500
Cosmic
Ray
Rate
(Hz)
Time of day

73. How many pages of text are there in the new
Harry Potter and the Order of the Pheonix?
870
What’s the error on that number? 1
 2
 /870  29.5
E.  870/2 =  435

74. A punted football has a hang time of 4.6 seconds.
What is the error on that number?
Scintillator is sanded/polished to a final thickness
of 2.50 cm. What is the error on that number?