Essential Techniques for Probability Calculation and Counting Methods
1.19k likes | 1.26k Vues
Learn fundamental procedures for probability calculation and counting methods, including permutations and combinations, to solve various mathematical problems effectively. Explore binomial coefficients and useful expansions in combinatorial analysis.
Essential Techniques for Probability Calculation and Counting Methods
E N D
Presentation Transcript
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 ?