1367-Minta Yuwana-Probability1

8/3/2019 1367-Minta Yuwana-Probability1

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Probabilitas dan Distribusi

Densiti


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PROBABILITAS

SUMBER PROBABILITAS: Model

Nilai ditaksir dari model (sederhana)

Contoh: Coin simetri, diundi menghasilkan probabilitas

Data Hasil experiment

Contoh:

John Kerrich mengundi coin 10000 kali, hasil ~


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PROBABILITAS

SUMBER PROBABILITAS:

Subjective

Perkiraan seseorang berdasar

pengetahuan/pengalaman Contoh:

Fatality rates of nuclear reactors accidents

1 minggu: 1 dalam 300,000,000

20 minggu: 1 dalam 16,000,000

Dasar data tsb. chain mechanism + subjectiveestimate


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PROBABILITAS

MODEL PROBABILITAS SEDERHANA:

Gambaran mekanisme suatuperistiwa/kejadian secara sederhana.

Outcomes

Sample Spaces

Events


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Outcomes

Keluaran/hasil suatu peristiwa atau

experiment Contoh:

outcomes undian coin: Hatau T

outcomes undian dadu: 1,2,3,4,5,6 outcomes WTC ditabrak pesawat: terbakar,

roboh sebagian, runtuh total, .

PROBABILITAS



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PROBABILITAS


Sample spaces (Ruang sample), S

himpunan (set) seluruh keluaran yang

mungkin (all possible outcomes) dariexperimen.

Contoh: 2 coin diundi bersama

1kali: S={HH,TH,HT,TT} 2 kali: S={HHHH, HHHT, HHTH, HTHH, THHH,

HHTT, HTTH, TTHH, HTHT, THHT, THTH,HTTT, THTT, TTHT, TTTH, TTTT}


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PROBABILITAS


Events

kumpulan keluaran (outcomes)

terjadi bila keluaran (outcome) yangmendukung terjadi

Contoh:

Event A=satu Hkeluar dari undian sekali 2coin A={TH, HT}

Event A terjadi bila outcome THatau HTterjadi


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PROBABILITAS


Sample space: event lengkap

Event subset dari sample space.

A

Sample space, S

Event, A


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PROBABILITAS


Sebuah event dapat hanya terdiri dari

satu keluaran (outcome) Contoh:

A={TT} B={HH}


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PROBABILITAS


Komplemen dari event A, dengan notasiA, terjadi bila A tidak terjadi

S

AA


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PROBABILITAS


Gabungan Event

Union,

AB

A B


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PROBABILITAS


Gabungan event

Intersection

A B

AB


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PROBABILITAS


Gabungan event

Mutually exclusive

A B

Mutually exclusive


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HUKUM PROBABILITAS

AXIOMA:

Probability sample space, S

P[S] =1 pasti terjadi

Probability event, A

0 < P[A] < 1 mungkin terjadi

Event kosong

P[] = 0


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Partisi

Event C1, C2, C3, .., Ck membentuk

partisi (pembagian) sample space bilaevent-event tersebut mutually exclusivedan menghabiskan seluruh sample space.

C1C

2C

3...C

k= S

P[C1C2C3...Ck] = P[C1] + P[C2] +

P[C3] + + P[Ck]

HUKUM PROBABILITAS

TEOREMA:


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Partisi

P[A] = P[AC1] + P[AC2] + + P[ACk]

= P[ACi]

HUKUM PROBABILITAS

TEOREMA:

C6

C3C2

C1

C5C4

Ci

AC1

C2 C3

C4C5

C6

A Ci (diarsir gelap)

i=1

k


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HUKUM PROBABILITAS

TEOREMA: Probabilitas Bersyarat (Conditional

Probability)

Event A terjadi bila event B terjadi Probability-nya

P[A|B] =

P[AB]

P[B] syarat P[B

]

0


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Example of Conditional Probability:Proportion of women with the given eyesight

grades

Grade of Left Eye

GradeofRightEye

Highest ThirdSecond Lowest

Highest

Third

Second

Lowest

Total

Total

.237

.058

.017

.024

.027

.010

.009

.112

.066

.328

.301

.265

1

.106

.336

.203

.016

.005

.255

.031

.036

.048

.011

.291

.202

Anggapan: sample cukup banyak dan mewakilipopulasi, proporsi mewakiliprobabilitas


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What is the probability of women with righteye 2nd grade given her left eye highest

grade ?

P[right eye 2nd grade|left eye highest grade] =

P[right eye 2nd grade left eye highest grade]

P[left eye highest grade]

Example of Conditional Probability:Proportion of women with the given eyesight

grades

0.0310.225

= 0.1216


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Consider 3 special cards below

Example of Conditional Probability:3 Cards Trick

Randomized& draw oneblue

red

blue

blue

red

red

blue

?

Probability of blue bottom = ?

Probability of red bottom = ?


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Probability of blue top faces:

P[blue top] = 3/4

Probability of blue bottom:P[blue bottom] = 1/2

Probability of a card having blue bottom face

given blue top face:P[blue bottom|blue top] = (1/2)/(3/4) = 2/3



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Top

Bottom

Blue Red

Blue

Red 1/6 2/6

1/62/6 1/2

1/2

1/2 1/2 1total

total


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blue top

red top

3/6

3/6

blue bottom

blue bottom

red bottom

red bottom

2/3

1/3

2/3

1/3

3/6 x 2/3 = 2/6

3/6 x 1/3 = 1/6

3/6 x 2/3 = 2/6

3/6 x 1/3 = 1/6

P[A] P[B|A] P[AB]


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Independence and Multiplication

Rule Independent Events

Two events are independent if one may

occur irrespective of the other

Event A and B are independent if and

only if P[AB] = P[A] P[B]


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Example:

A1 = sample contains Pb P[A1] = 0.32

A2 = sample contains Hg P[A2] = 0.16 sample contains both P[A1A2] = 0.10

What is the probability of a samplecontains Pb will also contains Hg?

Independence and Multiplication Rule


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Answer

P[A2|A1] = P[A1A2] / P[A1]

= 0.10/0.32 = 0.31

P[A2] = 0.16

A1 and A2 are not independent

Independent if

P[A2|A1] = P[A2]

Independence and Multiplication Rule


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Bayes Theorem

Let A1, A2, A3, , Anbe a collection ofevents which partition S.

Let B be event such that P[B] 0 For any events Aj, j= 1,2,3,,n then

ni

ii

jjj

APABP

APABPBAP

1

][]|[

][]|[]|[


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Example Blood type distribution

in a country is

type A = 41%

type B = 9%

type AB = 4%

type O = 46%

Bayes Theorem

Estimated that during handlingemergency there are

probability of wrong blood typeidentification

type O identified A = 4%

type A identified A =

88% type B identified A = 4%

type AB identified A =10%


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Someone having a traffic accident his/herblood identified as type A. What is the

probability that this is a correct type? Let

A : someone has type A blood

B : someone has type B blood

AB : someone has type AB bloodO : someone has type O blood

TA : someone identified as type A blood

Bayes Theorem


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Bayes Theorem

TA

OABB

A OTAABTA

BTAATA

TA = (ATA)(BTA)(ABTA )(OTA )


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Lets try to use conditional probability

Where are P[ATA] and P[TA] ??P[ATA] = P[TA|A] P[A]

= (0.88)(0.41) = 0.36P[TA] = P[ATA] + P[BTA] + P[ABTA]

+ P[OTA]

Bayes Theorem

][

][]|[

TAP

TAAPTAAP


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DISTRIBUSI PROBABILITAS DISKRIT(Discrete Probability Distribution)

DISKRIT

VARIABLE & VALUE

RANDOM VARIABLE & RANDOMVALUE


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DISKRIT

terpisah jelas satu dengan lainnya dapat dihitung/cacah (countable)

masing-masing bernilai sendiri dan

terpisah tidak dapat dipecah atau dibagi


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VARIABLE & NILAI (VALUE)

Variable: suatu kuantitas (event) yang

mempunyai satu nilai tertentu daribanyak kemungkinan nilai

Nilai (Value): ukuran yang dikaitkandengan variable


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RANDOM VARIABLE & RANDOMVALUE

Random variable: nilainya random

Random value (nilai random):

besar pastinya tidak diketahui kisaran nilainya diketahui dari sample

spacenya

terkait probabilitas


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Cara penulisan

Nilai: x

Variabel: X Probabilitas variabel Xbernilai x:

P[X=x]

Probabilitas bila xbermacam-macam:f(x)= P[X=x] x= macam-macam


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Distribusi Probabilitas (ProbabilityDistribution)

gambaran/fungsi sebaran probabilitas adalah f(x)

dinyatakan dalam bentuk

distribusi densitas distribusi kumulatif


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Distribusi Densitas Probabilitas

menyatakan nilai probabilitas pada setiap

nilai xf(x) =P[X=x]

x

f(x)

Total probabilitas =

f(x)= 1 utk seluruh xBila xkontinyuf(x)dx=1


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Distribusi Kumulatif Probabilitas

menyatakan nilai kumulatif probabilitas

sampai dengan nilai x

F(x) =P[X < x]

limit F(x) = 1

x

F(x)



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PARAMETER DISTRIBUSI:

NILAI HARAPAN (Expected Value)

Definisi Nilai Harapan (Expected Value),

H(x) AndaikanX : variabel random diskrit

dengan distribusi densiti f(x)

AndaikanH(X) : variabel random (yanglain)


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Nilai harapandari H(X), yi. E[H(X)]:

xsemua

H(x)f(x)E[H(X)]

PARAMETER DISTRIBUSI:NILAI HARAPAN (Expected Value)

semua xSyaratH(X)f(x) bernilai tertentu

Perjumlahan untuk semua nilai x dengan P[X=x]0

Statistik:nilai harapan = mean = matau

mx


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PARAMETER DISTRIBUSI:NILAI HARAPAN (Expected Value)

TEOREMA AndaikanXdanYadalah variabel

random

Andaikancadalah sebarang bilanganreal E[X + Y] = E[X] + E[Y]

E[cX ] = cE[X]

E[c] = c


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Definisi Variance

Andaikan X adalah random variable

denganMeanm Variance X

Var X =s2 = E[(X-m)2] hitungmdulu

ataus2 = E[X2] - (E[X])2 lebih praktis


VARIANCE dan STANDARD DEVIASI


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Defini Standard Deviasi, s Penyimpangan standar dari mean

Akar dari variances=Var X =s2

Derajad Kebebasan (Degrees of

Freedom), nn= n - 1 n = banyaknya X




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AndaikanXdanYadalah variabelrandom dancadalah bilangan real

Var c = 0Var cX = c2Var X

Var(X + Y) = Var X + Var Yuntuk X dan Y tak saling terikat

(independent)



TEOREMA


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CIRI

Outcomes: sukses dan gagal Undian identik dan independent

Probabilitas sukses,p, tetappada setiap undian

Undian Bernoulli

(Bernoulli Trial)


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Undian Bernoulli

(Bernoulli Trial)

Trial

n kali trial hinggasukses pertama

kali

jml suksesdalam n kali

trial

DistribusiGeometrik

DistribusiBinomial


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Distribusi Geometrik

Variabel random Xberdistribusigeometrik dengan parameterpapabila

densitas probabilitasnya

f(x) = (1- p)x-1puntuk 0


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Berlaku Bernoulli trial

Tiap trial identik dan independent Probability sukses ptetap

Xmenyatakan jumlah trial sampai

sukses yang pertama kali

Distribusi Geometrik

Ciri


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Trial Geometrik:

Coba terus sampai berhasil!!

s

p = 3/4 q = 1 - p = 1/4

f

f(1)=3/4

s

s

s

f

f

f(3)=1/4 x 1/4 x 3/4 = 3/64

f(4)=1/4 x 1/4 x 1/4 x 3/4 = 3/256

f(2)=1/4 x 3/4 = 3/16

1x

2x

3x

4x

Probabilitas kumulatif s/d 4x =255/256 = 0.9961


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Trial Geometrik:Coba terus sampai berhasil!!

0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

0 2 4 6

trial

probability

density

cuml


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Distribusi Binomial

Definisi Variabel random Xberdistribusi

geometrik dengan parameterndanp

apabila densitas probabilitasnyaf(x) = (1- p)n-xpx

untuk 0


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Kombinasi

Distribusi Binomial

Definisi

nx = C(n,x) = n!x! (n-x)!

C(7,3) = 7!3! (7-3)!

= 7.6.53.2.1

= 35


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Andaikan Xvariabel random denganparameter ndan p, maka

Expected Value, E(X)=m= np

-Varian, s2(X)

= npq q= 1 -p

Distribusi Binomial

Teorema


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Berlaku Bernoulli trial

Jumlah ntetap pada setiap trial

Tiap trial identik dan independent

Probability sukses ptetap

Xmenyatakan jumlah sukses dalamsetiap trial n

Distribusi Binomial

Ciri


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Anggap probabilitas hujan setiap hari, p

Anggap ptetap = 0.15

Berapa probabilitas 3 hari hujan dalamseminggu?

n = 7 p = 0.15 q = 0.85

x = 3

P[3] = C(7,3)(0.85)4(0.15)3=0.062

Distribusi Binomial

Contoh


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Sampling

Sampling

withreplacement

withoutreplacement

BinomialDistribution

HypergeometricDistribution


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draws or trials are not independent

probability of success, p, is not constant

variable X, the number of success in ntrials follows hypergeometric distribution

Sampling without Replacement


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A random variable Xhas a hypergeometricdistribution with parameters N, n, and rif itsdensity is given by

Hypergeometric Distribution

rx

N

n

N - rn- xf(x) =

max[0, n- (N - r)] < x< min(n, r)N, r, n positive integers


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Example

There are 20 small balls inside a bowl.

12 of them are black and 8 are white All balls are well mixed inside the bowl

5 balls are sampled from the bowl withoutreplacement

Xrepresents the number of black balls in thesample



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What is the probability of X?

N = 20, r = 12, n = 5

P[X=0] = C(12,0)C(8,5)/C(20,5) = 0.0036P[X=1] = C(12,1)C(8,4)/C(20,5) = 0.0542

P[X=2] = C(12,2)C(8,3)/C(20,5) = 0.2384

P[X=3] = C(12,3)C(8,2)/C(20,5) = 0.3973P[X=4] = C(12,4)C(8,1)/C(20,5) = 0.2554

P[X=5] = C(12,5)C(8,0)/C(20,5) = 0.0511



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Limiting case

if n


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Example of limiting case

100 transistors are sampled from 5000

the number of defective transistors are5%

What is the probability of X, the number

of defective transistors in the sample?



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Hypergeometric solution N = 5000 r = 250 n = 100

P[X=0] = C(250,0)C(4750,100)/C(5000,100) = 0.0056P[X=1] =C(250,1)C(4750,99)/C(5000,100) = 0.0302

P[X=2] = C(250,2)C(4750,98)/C(5000,100) = 0.0800

P[X=3] =C(250,3)C(4750,97)/C(5000,100) = 0.1392

P[X=4] = C(250,4)C(4750,96)/C(5000,100) = 0.1792

P[X=5] =C(250,5)C(4750,95)/C(5000,100) = 0.1818



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Binomial solution

n = 100, p = 0.05

P[X=0] = C(100,0)(0.05)0(0.95)100= 0.0059P[X=1] =C(100,1)(0.05)1(0.95)99= 0.0312

P[X=2] = C(100,2)(0.05)2(0.95)98= 0.0812

P[X=3] =C(100,3)(0.05)3(0.95)97= 0.1396

P[X=4] = C(100,4)(0.05)4(0.95)96= 0.1781

P[X=5] =C(100,5)(0.05)5(0.95)95= 0.1800



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Hypergeometric

P[X=0] = 0.0056

P[X=1] = 0.0302

P[X=2] = 0.0800

P[X=3] = 0.1392

P[X=4] = 0.1792P[X=5] = 0.1818


Limiting Case Comparison

Binomial

P[X=0] = 0.0059

P[X=1] = 0.0312P[X=2] = 0.0812

P[X=3] = 0.1396

P[X=4] = 0.1781P[X=5] = 0.1800


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Poisson Distribution

Definition French mathematician: Simeon Denis

Poisson (1781-1840)

A random variable X has a Poissondistribution with parameter k if itsdensity is

e-kkx

f(x) = ------------x! x = 0,1,2,k > 0


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Let Xbe a Poisson random variablewith parameter k

E[X] = k

Var[X] = k


Theorem


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Connected to Poisson processes

Poisson process:

observing discrete and infrequent eventsin a continuous interval of time, length,or space

Xis the number of occurrences of theevent in the interval of sunits

k= ls, l= rates of occurrence per

units


Features


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Average white blood-cell count of a healthyperson = 6000/mm3 of blood.

To detect white blood-cell deficiency, a 0.001mm3 of blood is taken and the number ofwhite cells Xis counted.

How many white cells are expected in a

healthy person? If at most 2 cells are found, is there evidence

of white blood-cell deficiency?


Example


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Solutionunit = mm3 s= 0.001l= 6000 ls= (6000)(0.001) = 6

Expected value, E[X]= ls= 6 (for a healthyperson)

How rare is it to see at most two?P[X


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Worksheet Functions

BINOMIAL DISTRIBUTION

BINOMDIST(number_s,trials,probability_s,cumulative)

Returns the individual term binomial distribution

probability.

Number_s : the number of successes in trials.

Trials : the number of independent trials.

Probability_s : the probability of success oneach trial.

Cumulative : a logical value. TRUE for thecumulative distribution function, FALSE for theprobability density/mass function.


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Example

The flip of a coin can only result in heads ortails. The probability of the first flip beingheads is 0.5, and the probability of exactly 6 of10 flips being heads is:

BINOMDIST(6,10,0.5,FALSE) equals 0.205078

Worksheet Functions



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Remarks

Number_s and trials are truncated to integers.

If number_s, trials, or probability_s isnonnumeric, BINOMDIST returns the#VALUE! error value.

If number_s < 0 or number_s > trials,BINOMDIST returns the #NUM! error value.

If probability_s < 0 or probability_s > 1,BINOMDIST returns the #NUM! error value.

Worksheet Functions



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COMBIN(number,number_chosen) Returns the number of combinations for a given number

of items.

Number is the number of items. Number_chosen is the number of items in each

combination.

Example

Suppose you want to form a two-person teamfrom eight candidates, and you want to knowhow many possible teams can be formed.COMBIN(8, 2) equals 28 teams.

Worksheet Functions

COMBINATION


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Remarks

Numeric arguments are truncated to integers.

If either argument is nonnumeric, COMBINreturns the #NAME? error value.

If number < 0, number_chosen < 0, or number

number_population, HYPGEOMDIST returnsthe #NUM! error value.

If population_s < 0 or population_s >number_population, HYPGEOMDIST returnsthe #NUM! error value.

Worksheet Functions

HYPERGEOMETRIC DISTRIBUTION


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If number_population < 0, HYPGEOMDISTreturns the #NUM! error value.

HYPGEOMDIST is used in sampling withoutreplacement from a finite population.

Worksheet Functions

HYPERGEOMETRIC DISTRIBUTION


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POISSON(x,mean,cumulative) Returns the Poisson distribution.

X is the number of events.

Mean is the expected numeric value. Cumulative is a logical value. TRUE for

cumulative Poisson probability and FALSE forprobability density/mass function

Examples

POISSON(2,5,FALSE) equals 0.084224

POISSON(2,5,TRUE) equals 0.124652

Worksheet Functions

POISSON DISTRIBUTION


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Remarks

If x is not an integer, it is truncated.

If x or mean is nonnumeric, POISSON returnsthe #VALUE! error value.

If x 0, POISSON returns the #NUM! errorvalue.

If mean 0, POISSON returns the #NUM!error value.

Worksheet Functions

POISSON DISTRIBUTION


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1367-Minta Yuwana-Probability1