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15 Cards in this Set
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KEY WORDS

PERMUTATION
COMBINATION RATIO 


DESC 1

IF YOU CAN DETERMINE THE NUMBER OF OUTCOMES IN AN EVENT AND THE NUMBER OF OUUTCOMES IN THE SAMPLE SPACE, YOU CAN DETERMINE THE PROBABILITY. IT IS JUST A RATION



DESC 2

USE PERMUTATIONS, COMBINATIONS AND RATIONS OF GEOMETRIC AREAS TO SOLVE PROBABILITY PROBLEMS



deffinitions and explictations
SAMPLE SPACE 
TEH SET OF ALL POSSIBLE OUTCOMES EXP: IF YOU ROLL A DIE, THE SET OF ALL POSSIBLE OUTCOMES

SAMPLE SPACE


A SPECIFIC OUTCOME OR TYPE OF OUTCOME. EVENT
POSSIBLE RESULTS OF A PROBABILITY EVENT. OUTCOMES THE RATIO OF THE NUMBER OF WAYS AN EVENT CAN OCCUR TO THE NUMBER IF POSSIBLE OUTCOMES. THEORETICAL PROBABILITY TWO EVENTS IN WHICH EITHER ONE MUST HAPPEN BUT THEY CANNOT HAPPEN AT THE SAME TIME. COMPLEMENTARY EVENTS With selected items: Study Print Export Play Memory Add to Cardfile Remove from Clipboard Copyright © 20052007 Tuolumne Technology Group, Inc.Terms of Service — 
A SPECIFIC OUTCOME OR TYPE OF OUTCOME. EVENT
POSSIBLE RESULTS OF A PROBABILITY EVENT. OUTCOMES THE RATIO OF THE NUMBER OF WAYS AN EVENT CAN OCCUR TO THE NUMBER IF POSSIBLE OUTCOMES. THEORETICAL PROBABILITY TWO EVENTS IN WHICH EITHER ONE MUST HAPPEN BUT THEY CANNOT HAPPEN AT THE SAME TIME. COMPLEMENTARY EVENTS With selected items: Study Print Export Play Memory Add to Cardfile Remove from Clipboard Copyright © 20052007 Tuolumne Technology Group, Inc.Terms of Service — 


NUMBER OF WAYS EVENT A or Event B can occur n(A or B)=n(A) * (b)

A or Event B can occur n(A or B)=n(A) * (b)



Outcomes of an event B)

numbers of way sEvents A AND THEN Event B can occur\
N(a AND THEN b)= N(a) * (b) 


Outcomes of an event c)

number of ways Events A or B, but not A and B could occur n(A or B) = n(A) *(B)n(upside down u)B)



probaility of event

P(E)=# of outcomes in an event/# of outcomes in a sample space =P(E)=n(E)/n(S)



what is the symbol of probabiltiy and event does not occur?

1 P(E)



define perutation:

AN ARRANGEMENT OF THE ELEMENTS FROM A GIVEN SET IN A DEFINTE ORDR



PERMUTAION EXAMPLE: gIVEN{A,b,c] there are 6 arrrangemtnes. even though each arrangemetn has the same letters, the order is different in each

pe ABC, ACB, BAC, BCA, CAB, CBA



Combination: uniques subsets of the elements of set, regardless how they are arranged show example

EX: Given [A,B, C], THERE IS 1 ARRANGEMENT BECAUSE EACH RRANGEMENT HAS TEH SAME LETTERS,
ABC, ACB, BAC, BCA, CAB, CBA 


DO EXAMPLE 1 PAGE 93

D) 26*26*26*10*10*10REPLACEMENT ALLOWED

(09) STANDS FOR 10 FIGURES


DO EXAMPLE 2 PAGE 93

SOLUTIONS=GO OVER WITH AMY

