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Pertemuan 6 - 7 PROBABILITAS Hdi Nasbey, M.Si Jurusan Fisika Fakultas Matematika dan Ilmu Pemgetahuan Alam
[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Learning Outcomes ,[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Outline Materi ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
What is Probability? ,[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
What is Probability? ,[object Object],Relative frequency =  f/n Population Probability ,[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | Sample And “How often” = Relative frequency
Basic Concepts ,[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Experiments and Events ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Basic Concepts ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Not Mutually Exclusive Mutually Exclusive B and  C? B and D? 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Basic Concepts ,[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Example ,[object Object],[object Object],E 1 E 2 E 3 E 4 E 5 E 6 S ={E 1 , E 2 , E 3 , E 4 , E 5 , E 6 } 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | 1 2 3 4 5 6 S ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Basic Concepts ,[object Object],[object Object],[object Object],[object Object],A ={E 1 , E 3 , E 5 } B ={E 3 , E 4 , E 5 ,   E 6 } ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | S B A
The Probability  of an Event ,[object Object],[object Object],[object Object],The relative frequency of event A in the population 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
The Probability  of an Event ,[object Object],[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Example ,[object Object],1st Coin  2nd Coin  E i   P(E i )   H T HH HT TH TT 1/4 1/4 1/4 1/4 P(at least 1 head)  = P(E 1 ) + P(E 2 ) + P(E 3 ) = 1/4 + 1/4 + 1/4 = 3/4 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | H T H T
Counting Rules ,[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Permutations ,[object Object],[object Object],Example:  How many 3-digit lock combinations can we make from the numbers 1, 2, 3, and 4? 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | The order of the choice is important!
Combinations ,[object Object],Example:  Three members of a 5-person committee must be chosen to form a subcommittee. How many different subcommittees could be formed? 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | The order of  the choice is  not important!
Event Relations ,[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | S A B
Event Relations ,[object Object],A C 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | S A
Calculating Probabilities for Unions and Complements ,[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | A B
Calculating Probabilities  for Complements ,[object Object],[object Object],[object Object],[object Object],[object Object],P(A C ) = 1 – P(A) 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | A A C
Conditional Probabilities ,[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | “ given”
Example 1 ,[object Object],[object Object],[object Object],HT TH TT 1/4 1/4 1/4 1/4 P(A|B) = ½ P(A|not B) = ½  HH P(A) does not change, whether B happens or not… 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | A and B are independent!
Example 2 ,[object Object],[object Object],[object Object],P(A|B) =P(2 nd  red|1 st  blue)= 2/4 = 1/2 P(A|not B) = P(2 nd  red|1 st  red) = 1/4 P(A) does change, depending on whether B happens or not… 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | m m m m m A and B are dependent!
Defining Independence ,[object Object],Two events A and B are   independent   if and only if P(A  B) = P(A)   or P(B|A) = P(B) Otherwise, they are   dependent . ,[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
The Multiplicative Rule for Intersections ,[object Object],P(A   B) = P(A) P(B given that A occurred)  = P(A)P(B|A) ,[object Object],P(A   B) = P(A) P(B)  01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Example 1 In a certain population, 10% of the people can be  classified as being high risk for a heart attack. Three people are randomly selected from this population. What is the probability that exactly one of the three are high risk? Define H: high risk N: not high risk P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH) = P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H) = (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9) 2  = .243 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Example 2 Suppose we have additional information in the  previous example. We know that only 49% of the population are female. Also, of the female patients, 8% are high risk. A single person is selected at random. What is the probability that it is a high risk female? Define H: high risk F: female From the example,  P(F) = .49  and  P(H|F) = .08.  Use the  Multiplicative Rule: P(high risk female) = P(H  F) = P(F)P(H|F) =.49(.08) = .0392 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
The Law of Total Probability ,[object Object],[object Object],[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
The Law of Total Probability A P(A) = P(A    S 1 ) + P(A    S 2 ) + … + P(A    S k )  = P(S 1 )P(A|S 1 ) + P(S 2 )P(A|S 2 ) + … + P(S k )P(A|S k ) 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  | A   S k A    S 1 S 2…. S 1 S k
Bayes ’  Rule ,[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
Example We know: P(F) =  P(M) =  P(H|F) =  P(H|M) =  From a previous example, we know that 49% of the population are female. Of the female patients, 8% are high risk for heart attack, while 12% of the male patients are high risk. A single person is selected at random and found to be high risk. What is the probability that it is a male? Define H: high risk  F: female  M: male .12 .08 .51 .49 01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |
[object Object],01/02/11 ©  2010 Universitas Negeri Jakarta  |  www.unj.ac.id  |

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Statistika Dasar (6 - 7) probabilitas

  • 1. Pertemuan 6 - 7 PROBABILITAS Hdi Nasbey, M.Si Jurusan Fisika Fakultas Matematika dan Ilmu Pemgetahuan Alam
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  • 28. Example 1 In a certain population, 10% of the people can be classified as being high risk for a heart attack. Three people are randomly selected from this population. What is the probability that exactly one of the three are high risk? Define H: high risk N: not high risk P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH) = P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H) = (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9) 2 = .243 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
  • 29. Example 2 Suppose we have additional information in the previous example. We know that only 49% of the population are female. Also, of the female patients, 8% are high risk. A single person is selected at random. What is the probability that it is a high risk female? Define H: high risk F: female From the example, P(F) = .49 and P(H|F) = .08. Use the Multiplicative Rule: P(high risk female) = P(H  F) = P(F)P(H|F) =.49(.08) = .0392 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
  • 30.
  • 31. The Law of Total Probability A P(A) = P(A  S 1 ) + P(A  S 2 ) + … + P(A  S k ) = P(S 1 )P(A|S 1 ) + P(S 2 )P(A|S 2 ) + … + P(S k )P(A|S k ) 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | A  S k A  S 1 S 2…. S 1 S k
  • 32.
  • 33. Example We know: P(F) = P(M) = P(H|F) = P(H|M) = From a previous example, we know that 49% of the population are female. Of the female patients, 8% are high risk for heart attack, while 12% of the male patients are high risk. A single person is selected at random and found to be high risk. What is the probability that it is a male? Define H: high risk F: female M: male .12 .08 .51 .49 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
  • 34.