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
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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.
There are 4 titles to choose from as 1st in the holder...
4 different ways to startthe 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 nexttothelast slot)
(1 lastplaceholder)
= 4·3·2·1=24.
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
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!
= 8.0658175171067
Combinations
Four cards are set out on the table.
You are asked to pick anythree.
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:
ABCABDADCBCD.
We can count off other combinations:
4C2 =
6
AB AC AD BC BD CD
4C1 =
4
A B C D
Just check out:
4C3 =
4C2=
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 wellshuffled deck. How
many different hands are possible?
52C5 =
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“fiftyfifty”
B.1/2 “onehalf”
C. 1:1 “onetoone”
In betting parlance theoddsare1:1;
we say thechancesare50/50, but
the mathematicalprobabilityis½.
There are 36 possible outcomes for the toss of
2 sixsided dice. If each is equally likely, the
most probable total score of any single roll is?
A. 4 B. 5 C. 6
D. 7 E. 8 F.9
A green and red die are rolled together.
What is the probability of scoring an 11?
A. 1/4B. 1/6C. 1/8
D. 1/12E. 1/18F. 1/36
“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. <
Number of ways to score die totals
Die Total
A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A.1/2B.1/4
C. 1D.0
A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2B. 1/4
C. 1D. 0
It is equally likely to observe
two heads (HH) as two tails (TT)
T) True.F) False.
A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2B. 1/4
C. 1D. 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.
T) True.F) False.
A coin is tossed twice in succession.
The probability of observing two heads
(HH) is expressed as
A. 1/2B. 1/4
C. 1D. 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.
T) True.F) False.
The probability of at least one head is
A. 1/2B. 1/4
C. 3/4D. 1/3
6
5
4
3
2
1
2 3 4 5 6 7 8 9 10 11 12
The probability of “throwing a 7”
P(7) = 6/36 = 1/6
P(7±5) = ?
6
5
4
3
2
1
2 3 4 5 6 7 8 9 10 11 12
P(7) = 6/36 = 1/6
P(7±5) = 36/36 =1
P(7±1) =
P(7±2) =
the full
range!
?
?
6
5
4
3
2
1
2 3 4 5 6 7 8 9 10 11 12
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!
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.
P(1)Probability of the first count occurring in
in 1st second
P(10)Probability of the first count occurring in
in 10th second
i.e., it won’t happen until the 10th second
???
P(1) = P(10)
???
= P(100)
???
= P(1000)
???
= P(10000)
???
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) =
Probability of a head on your 1st try,
P(1) =1/2
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) =
What is the total probability of
ALL OCCURRENCES?
P(1) + P(2) + P(3) + P(4) + P(5) + •••
=1/2+1/4 + 1/8 + 1/16 + 1/32 + •••
?
A sixsided die is rolled
repeatedly until it gives a 6.
What is the probability that one roll is enough?
A sixsided 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?
A sixsided 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)=
A sixsided 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?
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
Why?
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?
0 sectime
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).
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 ± ?
The probability of a single COSMIC RAYpassing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
Example: a measured rate of
1200 Hz = 1200/sec
Example: a measured rate of
1200 Hz = 1200/sec
1 millisec = 103 second
Example: a measured rate of
1200 Hz = 1200/sec
1 nanosec = 109 second
The probability of a single COSMIC RAYpassing
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
= 0.0012/msec
=0.0000012/nsec
The probability of a single COSMIC RAYpassing
through a small area of a detector
within a small interval of time Dt
can be very small:
p << 1
The probability of NOcosmic rays passing
through that area during that interval Dtis
A. pB. p2C. 2p
D.( p  1) E. ( 1  p)
The probability of a single COSMIC RAYpassing
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. pB. p2C. 2p
D.( p  1) E. ( 1  p)
The probability of a single COSMIC RAYpassing
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
exactlyn events?
pn
n “hits”
× ( 1 p )???
??? “misses”
× ( 1 p )Nn
Nn“misses”
While waiting N successive intervals
(where the total time is t = NDt )
what is the probability that we observe
exactlyn events?
P(n) = nCNpn( 1 p )Nn
From the properties of logarithms
ln (1p)Nn = ln (1p)
ln (1p)Nn = (Nn)ln (1p)
???
ln x loge x
e=2.718281828
ln (1p)Nn = (Nn)ln (1p)
and since p << 1
ln (1p)
 p
ln (1p)Nn = (Nn) (p)
from the basic definition of a logarithm
this means
e???? = ????
ep(Nn)= (1p)Nn
P(n) = pn( 1 p )Nn
P(n) = pn ep(Nn)
If we have to wait
a large number of intervals, N, for a
relatively small number of counts,n
n<<N
P(n) = pn epN
P(n) = pn epN
And since
N  (n1)
N (N) (N) … (N) = Nn
for n<<N
P(n) = pn epN
P(n) = pn epN
P(n) = eNp
Consider an example…
P(n) = e 4
e4 = 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 recordingn events
in 10 seconds?
P(0) = P(4) =
P(1) = P(5) =
P(2) = P(6) =
P(3) = P(7) =
0.195366816
0.156293453
0.104195635
0.059540363
0.018315639
0.073262556
0.146525112
0.195366816
P(n) = eNp
Hey!
What does Np represent?
Another useful series we can exploit
m, mean =
n=0 term
n / n! = 1/(???)
¥
m
å
(
N
p
)

e
N
p
=
(N
p
)
(
m
)!
=
m
0
m, mean
let m = n1
i.e., n =
what’s this?
m, mean
m = (Np) eNp eNp
m = Np
m = Np
P(n) = em
Poisson distribution
probability of finding exactlyn
events within time t when the events
occur randomly, but at an average
rate of m(events per unit time)
m=1
m=4
m=8