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Permutations 4 CDs fit the CD wallet you carry to school each day.

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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 next-to-the-last slot)

(1 last-place-holder)

= 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

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:

ABC ABD ADC BCD.

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 =

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.1/2 “one-half”

C. 1:1 “one-to-one”

In betting parlance theoddsare1:1;

we say thechancesare50/50, but

the mathematicalprobabilityis½.

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. 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. 1/4 B. 1/6 C. 1/8

D. 1/12 E. 1/18 F. 1/36

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.1/2 B.1/4

C. 1D.0

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.

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.

T) True.F) False.

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.

T) True.F) False.

The probability of at least one head is

A. 1/2 B. 1/4

C. 3/4 D. 1/3

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 + •••

?

repeatedly until it gives a 6.

What is the probability that one roll is enough?

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?

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)=

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?

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)

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?

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?

Random events arrive independently

- unaffected by previous occurrences
- unpredictably

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

- cosmic rays arrive at a fairly stable, regular rate
- when averaged over long periods

- the rate is not constant nanosec by nanosec
- oreven second by second

- this average, though, expresses the
- probabilityper unit time
- of a cosmic ray’s passage

1200 Hz = 1200/sec

- would mean in 5 minutes we
- should expect to count about
- 6,000 eventsB. 12,000 events
- C. 72,000 events D. 360,000 events
- E. 480,000 eventsF. 720,000 events

1200 Hz = 1200/sec

- would mean in 3 millisec we
- should expect to count about
- 0 eventsB. 1 or 2 events
- C. 3 or 4 events D. about 10 events
- E. 100s of eventsF. 1,000s of events

1 millisec = 10-3 second

1200 Hz = 1200/sec

- would mean in 100 nanosec we
- should expect to count about
- 0 eventsB. 1 or 2 events
- C. 3 or 4 events D. about 10 events
- E. 100s of eventsF. 1,000s of events

1 nanosec = 10-9 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 )N-n

N-n“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 )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=2.718281828

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

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

n<<N

P(n) = pn e-pN

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 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) = e-m

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)

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