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Chapter 2 Combinatorial Analysis. 主講人 : 虞台文. Content. Basic Procedure for Probability Calculation Counting Ordered Samples with Replacement Ordered Samples without Replacement Permutations Unordered Samples without Replacement Combinations Binomial Coefficients
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Chapter 2Combinatorial Analysis 主講人:虞台文
Content • Basic Procedure for Probability Calculation • Counting • Ordered Samples with Replacement • Ordered Samples without Replacement Permutations • Unordered Samples without Replacement Combinations • Binomial Coefficients • Some Useful Mathematic Expansions • Unordered Samples with Replacement • Derangement • Calculus
Chapter 2Combinatorial Analysis Basic Procedure for Probability Calculation
Basic Procedure for Probability Calculation • Identify the sample space . • Assign probabilities to certain events in A, e.g., sample point event P(). • Identify the events of interest. • Compute the desired probabilities.
Chapter 2Combinatorial Analysis Counting
Goal of Counting Counting the number of elements in a particular set, e.g., a sample space, an event, etc.
Cases • Ordered Samples w/ Replacement • Ordered Samples w/o Replacement • Permutations • Unordered Samples w/o Replacement • Combinations • Unordered Samples w/ Replacement
Chapter 2Combinatorial Analysis Ordered Samples with Replacement
Ordered Samples eat elements in samples appearing in different orders are considered different. ate tea
Ordered Samples w/ Replacement meet 1. Elements in samples appearing in different orders are considered different. 2. In each sample, elements are allowed repeatedly selected. teem mete
k Ordered Samples w/ Replacement n distinct objects • Drawing k objects, their order is noted, among n distinct objects withreplacement. • The number of possible outcomes is
Example 1 How many possible 16-bit binary words we may have? 16 2 distinct objects
Example 2 Randomly Choosing k digits from decimal number, Find the probability that thenumber is a valid octal number. For any , P()=1/10k.
Chapter 2Combinatorial Analysis Ordered Samples without Replacement Permutations
清 以 心 也 可 Permutations 可以清心也 以清心也可 清心也可以 心也可以清 也可以清心
k Ordered Samples w/o Replacement Permutations n distinct objects • Drawing k objects, their order is noted, among n distinct objects withoutreplacement. • The number of possible outcomes is
Example 3 Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events? 1. Begin with an s. 2. Contains novowel. 3. Begins and ends with a consonant. 4. Contains onlyvowels.
E1: word begins with an s. E2: word contains no vowel. E3: word begins and ends with a consonant. E4: word contains only vowels. Define Example 3 P(E1)=? P(E2)=? P(E3)=? P(E4)=? Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events? 1. Begin with an s. 2. Contains novowel. 3. Begins and ends with a consonant. 4. Contains onlyvowels.
E1: word begins with an s. E2: word contains no vowel. E3: word begins and ends with a consonant. E4: word contains only vowels. Define Example 3 P(E1)=? P(E2)=? P(E3)=? P(E4)=? For any , P()=1/||.
Chapter 2Combinatorial Analysis Unordered Samples without Replacement Combinations
Combinations ndistinct objects Choose k objects How many choices?
This notation is preferred Combinations • Drawing kobjects, their order is unnoted, among ndistinct objects w/o replacement, the number of possible outcomes is
Example 4 • x • Denoting the all-assistant event as E, The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.
Example 5 A poker hand hasfive cards drawn from an ordinary deck of 52 cards. Find theprobability that the poker hand has exactly2 kings. • x • Denoting the 2-king event as E,
m n 1 2 3 r 1 2 3 r Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers. Example 6 E P(“k matches”) = ? ||=? |E|=?
m n 1 2 3 r # possible outcomes from the 1st box. # possible k-matches. 1 2 3 r # possible outcomes from the 2nd box for each k-match. Example 6
m n 1 2 3 r 1 2 3 r Example 6
m n 1 2 3 r 1 2 3 r Example 6 樂透和本例有何關係?
m n 1 2 3 r 1 2 3 r Example 6 本式觀念上係由第一口箱子出發所推得
m n 1 2 3 r 1 2 3 r Example 6 觀念上,若改由第二口箱子出發結果將如何?
m n 1 2 3 r 1 2 3 r Example 6
m n 1 2 3 r 1 2 3 r Exercise
Chapter 2Combinatorial Analysis Binomial Coefficients
n terms Binomial Coefficients
n boxes Binomial Coefficients
Facts: Binomial Coefficients
Properties of Binomial Coefficients Exercise
Properties of Binomial Coefficients n不同物件任取k個 第一類取法: 第二類取法:
Properties of Binomial Coefficients Pascal Triangular
Properties of Binomial Coefficients Pascal Triangular 1 1 1 2 1 1 3 1 3 1 6 4 1 4 1
Fact: Example 7-2 kx+1 ?