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Example_class_1_slides

Example_class_1_slides - Example class 1 STAT1301...

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Example class 1 STAT1301 Probability and Statistics I Chan Chi Ho, Chan Tsz Hin & Shi Yun September 14, 2011
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Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. I Deadline of Assignment 1: Sep. 26th
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Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. I Deadline of Assignment 1: Sep. 26th
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Information about tutors Tutors: Chan Chi Ho (MW515), Chan Tsz Hin (MW518) and Shi Yun (MW518) Emails: [email protected], [email protected], [email protected], respectively. Assignment: Put assignments in assignment box outside MW515. I Deadline of Assignment 1: Sep. 26th
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Review: Combinatorial Analysis Multiplication Principle: If an operation consists of a sequence of k separate steps of which the first can be performed in n 1 ways, followed by the second n 2 ways, and so on until the k th can be performed in n k ways, then the operation consisting of k steps can be performed by n 1 × n 2 × ... × n k = k i = 1 n i ways.
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Review: Combinatorial Analysis (cont’d) I Selection of r from n distinct objects: 1 Ordered selection with replacement: Consider drawing r balls from n distinct balls. The total number of ways is n × n × ... × n | {z } r times = n r . 2 Ordered selection without replacement: Consider drawing r balls from n distinct balls without replacement. The total number of ways is n × ( n - 1 ) × ( n - 2 ) × ... × ( n - r + 1 ) = n ! ( n - r )! .
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Review: Combinatorial Analysis (cont’d) II 3 Unordered selection with replacement: Consider drawing r balls from n distinct balls with replacement and without regard to order. This is the same as placing r identical objects into n distinct containers (why?). The problem is the same as arranging the “0” and “|” in, say, 00 | 000 || 00 | 0 ||| 00 ..... | 0 || 0 | {z } r "0"s and n - 1 "|"’s The total number of ways is r + n - 1 r = ( r + n - 1 )! r !( n - 1 )! . 4 Unordered selection without replacement: Consider drawing r balls from n distinct balls without replacement and without regard to order. The total number of ways is n r = n ! r !( n - r )! .
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Review: Combinatorial Analysis (cont’d) III Arrangement of n objects, with r distinct types. In other words, there are n 1 objects of type 1, n 2 objects of type 2, ..., n r objects of type r . (e.g. number of different letter arrangements can be formed using the letters STATISTICS.) Total number of ways is n ! n 1 ! × n 2 ! ×···× n r ! . Partition of n distinct objects into r distinct groups with specified size n 1 , n 2 , ··· , n r . (e.g. number of ways in dividing a class of 40 into groups of 10, 10 and 20.) Toal number of ways is n n 1 , n 2 , ··· , n r = n ! n 1 ! × n 2 ! ×···× n r ! .
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Review: Combinatorial Analysis (cont’d) IV Partition of n indistinguishable objects into r distinct groups (i.e.
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