slide 03 elements of probability

7/27/2019 Slide 03 Elements of Probability

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Topic 3Elements of Probability

Fasilkom UI, 2013


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` Introduction to Probability and Statistics for Engineers &

` Sheldon M. Ross, Elsevier, 2004, Chapter 3

` Modern Introduction to Probability and Statistics,

Understanding Why and How,` re er c e e ng et a ., pr nger, . apter -

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` Sample space & events

` Axioms of Probability

amp e space av ng equa y e y ou comes

`

Conditional Probability` ayes ormu a

` Independent Events

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erminolo ies Sample space (S): the set of all possible outcomes of an experiment

Examples:

tossin a coin: S={Head, Tail}; E={H}

Throwing a die: S={1,2,3,4,5,6}; E={3}, {3,5,2}

Questions

Suppose we are interested in determining the amount of dosage that must begiven to a patient until the patient reacts positively. S = ?

Occurrence of an outcome is assumed to be random

Random = having no specific pattern

Since S & E are set, operations on set are applied.

Recall also: relation on set.4


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Venn Diagram: graphical presentation of set, illustrate

Algebra and basic properties of U &

Commutative, associative and distributive laws

De-Morgans law of complement

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Examples

Experiment: Fair coin tossing:

, .

A defect coin has P(H)P(T)

= = , , , .

E=at least one Head; P(E)=3/4.

One roll: S={1,2,3,4,5,6};

.

Experiment : Month of birth of a mystery guest

an, e , , ec; ong mon ; =

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` Experiment : Tossing a dart on a plate of 1 m2

x,y x;y assum ng e ar mus e p a e

somehow

` = = < = =

` What is P(E), P(A) and P(B)

` Need to model the robabilit function

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1. ( ) 0c

P =

= .

3. ( ) ( ) ( ) ( )P A B P A P B P A B = +

. en =

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Sample space with equally likely

`

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` A committee of 5 is to be selected from 6 men and 9 women.

,

of having 2 men and 3 women in the committee?

` There are C(15, 5) possible committee to form. Selecting

2 men and 3 women are independent, and thus productrule is applied

` So the probability is C(6, 2) * C(9, 3) / C(15, 5) = ?

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LetA and B are two events. The intersection of the two

=. ., .

E.g:A = born in month of June; B = Female; C = Female born in

The probability of eventA given that the event B has

Notation: P(A|B) probability of A given B

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A Formal Definition of Conditional robabilit

` Example:

New (Reduced) Sample Space

` A = born in a month with the letter r [Jan, Feb, Mar, Apr, Sep, Oct, Nov, Dec]

` C= born in long month [Jan, Mar, Mei, Jul, Aug, Oct, Dec]

` Suppose we met someone on the street and he told us that he was born in

long month. What is the probability that he is born in month with the letter r?

` P(A)=8/12; P(C)=7/12; P(AC)=4/12; so P(A|C)=4/7

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Multi lication Rule For any eventsA and B

( ) ( | ) ( )P AB P A B P B=

s g ves t e ormu a or comput ng t e pro a ty o a o nt

event

Example:

Ali believes that 70% he will graduate this year. If he does, then he will have 80%.

and being recruited by the MNC?

G=Ali graduates this year; R|G = recruited by MNC if graduated

RG = graduated and recruited: P(RG)=0.7x0.8 = 0.56

Note:SinceAB=BA,thenP(AB)=P(B|A)P(A); makesureyouusetherightinfo

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cE EF EF=

w ere s any event n . ote t at an are

mutually disjoint.This leads us toc

( | ) ( ) ( | )[1 ( )]cP E F P F P E F P F

=

= +

i.e., the probability of an event E is the weightedaverage of its conditional probability that an event F

as occurre or not occurre

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Example : mad cow disease (BSE) test

` The test-kit has the following effectiveness

` Probability of positive result for an infected cow is 70% P(T|B)

` Probability of positive result for an healthy cow is 10% P(T|Bc)

` Suppose the BSE risk is P(B)=0.01, then P(T)=P(T|B)P(B)+ P(T|Bc)[1 P(B)] = 0.7x0.01+0.1 x0.99 = 0.106

` If m cow is tested ositive then what is the robabilit that it reall is BSE?

( ) ( | ) ( ) 0.7 0.01( | ) 0.066

( ) ( ) 0.106

P TB P T B P BP B T

P T P T

= = = =

` What is P(B|Tc)? What do these tell you?

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An eventA is said to be independent ofB is P(A|B) = P(A)

It tells us that the occurrence of B has nothing to do with the

occurrence of A

It is easily seen that ifA and B are independent events,

then P(AB)=P(A)P(B) Example: tossing a coin and rolling a die; what is the

probability of having even outcome if the coin show Tail

P(E|T)

P T =0.5 P E =0.5 P ET =0.25 Thus P E T =0.5=P E

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Inde endent events cont.`

P A A A P A P A P A=L L

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The End of Topic 3

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slide 03 elements of probability

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