Probabilities

PROBABILITIES

Probabilities• Basic concepts:

• Random experiment: • All possible outcomes have to be known in advance• The result of a particular experimen tcannot be predicted (randomness).• The experiment can be repeated under identical conditions

• Sample space Ω:• Set of all possible outcomes

• Events• A sub-set of the sample space, i.e. a set of possible otucomes

• Singular event:• A sub-set of the sample space with one element

Probability model

1. Set S of possible outcomesa) Throw the coin, S = (pitch, toss)b) Throw the dice, S = (1, 2, 3, 4, 5, 6)

2. Probabilites Pia) Throw the coin, S = (pitch, toss)

P(pitch) = P(toss) = 1/2b) Throw the dice, S = (1, 2, 3, 4, 5, 6)

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

Probabilities• Dice 1, 2, 3, 4, 5, 6• P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

• Two dice are thrown

1 2 3 4 5 6

1

2

3

4

5

6

Model

W2W1

1/6

1/6

1/6

1/6

1/6

1/6

1/6 1/6 1/6 1/6 1/6 1/6

1/36

Probabilities1. P(both dice show 1) = P(1) P(1) = 1/6 1/6 = 1/36

1 2 3 4 5 6

1

2

3

4

5

6

Model

W2W1

1/6 1/6 1/6 1/6 1/6 1/6

1/6

1/6

1/6

1/6

1/6

1/6

Probabilities2. Event E: get a „2“ and a „3“: P(E) = 2 P(2) P(3) = 2 1/6

1/6 = 1/18 Event F: first dice a „2“, second dice a „3“, P(F) = 1/6 1/6 = 1/36

1 2 3 4 5 6

1

2

3

4

5

6

Model

W2W1

1/6 1/6 1/6 1/6 1/6 1/6

1/6

1/6

1/6

1/6

1/6

1/6

Probabilities3. P(double) = P(double 1) + ... + P(double 6) = 6 1/36 = 1/6

1 2 3 4 5 6

1

2

3

4

5

6

Model

W2W1

1/6 1/6 1/6 1/6 1/6 1/6

1/6

1/6

1/6

1/6

1/6

1/6

Probabilities4. P(sum of both dice is 7) = 6 1/36 = 1/6

1 2 3 4 5 6

1

2

3

4

5

6

Model

W2W1

1/6 1/6 1/6 1/6 1/6 1/6

1/6

1/6

1/6

1/6

1/6

1/6

Probabilities• Two hunters try to shoot a fox.• Each hunter strikes with probability 1/3.• What is the probability, that the fox survives?

ÞEach hunter does not strike with probability 2/3ÞP(fox survives) = P(both hunters do no strike) = 2/3 ∙ 2/3 = 4/9

Further probability exercises

PROBABILITIESFactorial, binomial coefficient and binomial distribution

ProbabilitiesFactorial:

• A number of students is called to the black board. How many orderings are possible?

1A 1 (A)

n ∙(n-1) ∙∙2 ∙1

3 ∙2 = 6CAB; CBA;ACB; BCA;ABC; BAC

2AB; BA

3 (A, B, C)

Number of possible orderings

OrderingsNumber of students

2 (A, B)

n

FactorialIn R: factorial() 0! := 1

1! = 12! = 1∙2 = 2

3! = 1∙2 ∙3 = 6

n! = 1∙2 ∙3 ∙ ∙n

Probabilities• Binomial coefficient

Choose 6 numbers out of 49. How many outcomes are there if the ordering does NOT matter and if numbers can be selected only once?

49 · 48 · 47 · 46 · 45 · 44 choices

49 · 48 choicesFirst and 2nd number:

48 choicesSecond number:49 choicesFirst number:

6 numbers:

Is this the solution? No! – Why?

1. selection: {7; 18; 5; 43; 1; 22}2. selection: {18; 7; 5; 43; 1; 22}

Both selections are equal,since they differ only with respect to the order

How many orderings are there? 6!

Final solution of question above:123456

444546474849

Probabilities

!649!6!49

123456444546474849

Claim:

124243!6124243444546474849

Definition:

649:!649!6

!49

!kn!k!n:k

n

is the number of ways k things can be chosen from n things. Here, the ordering of the k things is not of interest and each element can be chosen only once.

Binomial coefficient (In R: choose(n,k) )

1nn

0n

n1nn

1n

nkfür0kn

kn

nkn

Probabilities• Binomial coefficient

S:

wm

wm

mm

Phenotype: affectedunaffected

PKU is autosomal reccesive hereditary disease.

Pedigree:m: mutant allelew: wildtype allel (= „normal“ allele)

Inheritance (PKU)

Probabilities• Binomial distribution

wm

wm

?

S:P(child is ill)= 1/2 ∙ 1/2 =1/4P(child is not ill)= 1- P(child is ill)= 3/4


S:

wm

wm

S:

? ?

wm

wm

P(first child ill, 2nd child not ill) = 1/4 ∙ 3/4 =3/16P(one child ill, one child not ill) = 2 ∙ 1/4 ∙ 3/4 =6/16

Probabilities ShealthychildrenillchildrenP 3,2S:

wm

wm

? ?? ? ?

orderings

25

healthy

43

ill

41

healthy

43

healthy

43

32

43

41

25

ill

41

An experiment with two (->“binomial“) possible outcomes („success“ and „failure“) is independently repeated n times

p: probability of event 1 („success“) per experimentn: number of experimentsk: number of successes

knk p1pknp,k,nB

Binomial distribution (In R: dbinom(k,n,p) ):


Density function

S:

wm

wm

? ?? ? ?

p = 1/4 (affected)Þ (1-p) = 3/4 (not affected)

237,0434105

0 50

SildrenaffectedchP

396,0434115

1 41

SildaffectedchP

264,0434125

2 32

SchildrenaffectedP

088,0434135

3 23

SchildrenaffectedP

015,0434145

4 14

SchildrenaffectedP

001,0434155

5 05

SchildrenaffectedP


0 1 2 3 4 50.00

0.10

0.20

0.30

0.40

Number of affected children

f Density function

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

Distribution function

Number of affected children

F

Probabilities• Random variable

x

y

z

Probability modelS

Probability P

Real numbers

1

2

3

4

5

RfunctionX

X: Random variable

?

Probabilities• Random variable: function X that assigns a real number to an event

S

201P

Examples for possible random variables X, Y

X: number of edges

X 34

X 0 X

Y: red 1brown 2

12 Y

Y


S

201P

Question:

P(X = 3) = ?

X: number of edges

(sloppy)

correct:

P( aS | X(a) = 3) = 205

Use the sloppy notation!

ProbabilitiesS

201P

X: Number of edges

P(X = 4) =

P(X = 0) =

P(X = 2) =

P(X = 3) =

P(X = 5) =

52

208

0

41

205

207

0

P(X 4) =

P(X 0) =

P(X 2) =

P(X 3) =

P(X 5) =

52

2013

1

1

P(X < 0) = 0

52

Distribution function: F(a) = P(X a)

0.0

0.1

0.2

0.3

0.4

0 3 4 5 X

f

Density21

0.0

0.2

0.4

0.6

0.8

1.0

0 3 4 5 X

F

Distribution function1 2

Probabilities• Statistical measures

Sample Model

iixhx iixpμ

ExpectationExample

7/2041/432/50pixi

in1 xx

Mean

15.2430 2074152 μ

2

i1n12 xxs

Empircal variance

2ss

Emp. standard deviation

2ii

2 xp μσ

Variance

2σσ

Standard deviation

23.3215.24207

215.2341215.20522

σ

80.1σ


knk p1pknp,k,nB

Expectation: pnμ

Variance: p1pn2 σ

Standard deviation: p1pnσ

Probabilities• Continuous random variable

0° Random variable W: angleP: all angles equally likely

P(W = 180°) = ?

precision: 1° Þ 360 possibilities with P=1/360precision: 0.1° Þ 3600 possibilities with

P=1/3600precision: 0.01° Þ 36000 possibilities with P=1/36000

n possibilities with P=1/n0n

1n Þ

P(W = 180°) = 0


0 90 180 270 360 X0

0.2

0.4

0.6

0.8

1

F

0°

Distribution function: F(a) = P(W a)

F(0) = P(W 0°) =

0F(90) = P(W 90°) =

41

F(180) = P(W 180°) =

21

F(360) = P(W 360°) =

1F(270) = P(W 270°) =

43


0 90 180 270 360 X0

0.20.4

0.6

0.8

1F

Distribution function F

0 90 180 270 360 X

fDensity function f

3601

1

Ff


discrete continuous

ii xpμExpectation

dxxfxμ

Variance 2ii

2 xp μσ

dxxfx 22 μσ


0 90 180 270 360 X

fDensity function f

3601 Expectation?

Conjecture: 180

dxxfxμ 360

0dxxfx

360

0dxx360

1 360

0

2

2x

3601

180236002

3603601 2

Probabilities

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