Stat 301.001 Fall 2016 Homework 5 Due September

Stat 301.001 Fall 2016 Homework 5 Due: September

Find the values (in the form of integers) of the following expressions:

• (a) 4!
• (b) (5)4
• (c) (100)2
• (d) ( 5 3 )
• (e) ( 100 98 )

A sequence consisting of only 0’s and 1’s is called a binary sequence, e.g. .

• (a) Find the number of different binary sequences of length 6.
• (b) Find the number of different binary sequences of length 6 in which exactly three of the digits are 0.
• (c) Find the number of different binary sequences of length 6 in which at least one of the digits is 0.

In this problem, an example of using product principle incorrectly will be shown.

Please read it carefully and keep it in mind when you use the product principle in the future. A hand of 5 cards is drawn from 52 cards. Our goal in this problem is to find the probability of the event A = {the five-card hand contains at least two aces and two 7’s}. The sample space Ω is the set of all possible five-card hand, as shown in Page 7 in Lecture 13 slides. The following is a WRONG way to use the product principle to count the number of outcomes in this event: The task can be divided into 3 steps: Step 1. Choose 2 cards of rank ace. Step 2. Choose 2 cards of rank 7. Step 3. Choose 1 card from the rest cards. Then by the product principle, the number of outcomes is the product of the number of different choices in the three steps. The above procedure is wrong, because it may count the same outcome multiple times. Choosing differently in the steps may lead to the same result. For example, the following two different combinations of choices:

• • Step 1: Choose â™ A and ♥A; Step 2: Choose â™ 7 and ♥7; Step 3: Choose ♦A.
• • Step 1: Choose â™ A and ♦A; Step 2: Choose â™ 7 and ♥7; Step 3: Choose ♥A.

result in the same hand â™ A♥A♦Aâ™ 7♥7. As a result, the answer from the wrong procedure may be larger than the correct number. The main reason that different ways may lead to the same outcome is that the card chosen in Step 3 may be an ace or 7. Therefore, for the correct answer, we may first partition the event A into three events, depending on what the “5th card†is: A1 = {the five-card hand contains three aces and two 7’s} A2 = {the five-card hand contains two aces and three 7’s} A3 = {the five-card hand contains exactly two aces and exactly two 7’s} By addition principle, #A = #A1 + #A2 + #A3. By symmetry, #A1 and #A2 should be the same. You are asked to complete the rest of the calculation: (a) Find #Ω

• (b) Find #A1.
• (c) Find #A3.
• (d) Find #A and P (A).

Suppose that tomorrow will be either a rainy day or sunny day, and you may go hiking tomorrow. The probability that tomorrow will be a rainy day is 0.4. The probability that you will go hiking tomorrow is 0.8. The probability that you will go hiking tomorrow on a rainy day is 0.3. (a) What is the probability that you will go hiking tomorrow given that it is a rainy day tomorrow? (b) What is the probability that you will go hiking tomorrow given that it is a sunny day tomorrow? (c) What is the probability that tomorrow will be a rainy day given that you will not go hiking tomorrow? (d) Is the two events “You will go hiking tomorrow†and “It is a rainy day tomorrow†independent?

A hand of 5 cards is drawn from 52 cards. Suppose that after you draw the first two cards, you see that they are ♠A and ♠7. (a) The sample space Ω is the same as in Problem 3. Let A = {The hand consists of ♠A and ♠7} What is the number of different hands in A? (b) What is the probability that the 5-card hand is a flush (all 5 cards are of the same suit) given that two cards are ♠A and ♠7 ? Hint: Define B = {The hand is a flush}. Then A∩B = {The hand is a flush consists of ♠A and ♠7} The counting task is only to determine the other three ♠cards. (c) What is the probability that the 5-card hand is a full house (3 cards are of the same rank and the other two are of another rank) given that two cards are ♠A and ♠7 ? Hint: For the intersection event, you may need to partition it into two disjoint events. Review Problem 3 if necessary. For the event of drawing a full house, the number of possible outcomes in this event is already calculated in class, so you can use the result directly in this problem.

Paper For Above instruction

The assignment encompasses a set of combinatorial and probability problems aimed at exploring factorial calculations, binary sequences, card game probabilities, conditional probabilities, and independence of events within a probabilistic framework. These problems are designed to reinforce the understanding of fundamental principles such as permutations, combinations, the product principle, and partitioning of sample spaces to avoid double counting, which are essential for advanced statistical reasoning and analysis. The next sections provide a detailed exploration and solutions to each problem, drawing on core concepts and methodologies in probability theory and combinatorics.

Introduction

Understanding basic combinatorial calculations and probability principles is vital for analyzing real-world scenarios, especially in card games and binary data sequences. Through this set of exercises, we aim to develop proficiency in factorial computation, counting binary sequences under specific constraints, calculating probabilities of complex events, and understanding the importance of correct application of the product principle. The context spans from simple factorial expressions to more complex conditional probability assessments involving drawing cards and analyzing event independence, which are foundational topics in statistical education and applications.

Problem 1: Calculations of Factorials and Combinations

Problems involving factorials and combinations such as 4!, (5)4, (100)2, ( 5 3 ), and ( 100 98 ), serve as fundamental tools for enumerating arrangements and selections. These calculations form the bedrock of probability computations, particularly in determining total sample spaces and favorable outcomes in scenarios involving permutations and combinations. For instance, 4! calculates the total permutations of four distinct elements, while combinations like ( 5 3 ) determine the number of ways to choose 3 items from 5 without regard to order, crucial in card selection probabilities.

Problem 2: Binary sequences enumeration

The enumeration of binary sequences examines the total possible arrangements of 0s and 1s of a given length, highlighting the exponential growth of possibilities. Specifically, for length 6, the total number is 2^6. When restrictions are applied, such as having exactly three zeros, combinations are used to select the positions of zeros. These principles underpin coding theory and digital data analysis, illustrating how combinatorial reasoning scales in binary data contexts.

Problem 3: Probabilities in Card Draws and Proper Application of Product Principle

This problem emphasizes the correct methodology for counting outcomes in card games, illustrating why naively applying the product principle can lead to overcounting. The key is understanding how to partition the problem space based on specific attributes—here, the composition of the hand regarding aces and 7s—and applying the addition principle for mutually exclusive events. The partitioning into three events (A1, A2, A3) ensures accurate counts of outcomes aligned with the sample space.

Calculations involve determining the total number of possible hands (#Ω), the counts corresponding to each partitioned event (#A1, #A3), and ultimately the probability P(A). Accurate combinatorial counting—using the appropriate binomial coefficients—is essential for valid probability estimation and avoiding double counting.

Problem 4: Conditional Probability and Event Independence

This problem explores probability laws, including conditional probability and the concept of independence. By applying Bayes' theorem and the law of total probability, it demonstrates how to find the probability of hiking given weather conditions, and vice versa. The question of independence hinges on whether the joint probability factors into the product of individual probabilities, a fundamental property in probability theory used in assessing relationships between events.

Problem 5: Card Hands, Conditional Probabilities, and Advanced Card Counting

This section extends earlier card-related problems, focusing on conditional probabilities for specific types of hands given partial information. It involves counting specific hands (e.g., flushes, full houses) conditioned on known cards, which requires understanding of specific combinations, suits, and ranks. The problem underscores the importance of partitioning and proper enumeration for accurate probability calculations in complex combinatorial contexts. The hints provided guide the correct approach to count intersections and to apply prior results directly where applicable.

Conclusion

This collection of problems illustrates critical concepts in probability and combinatorics, emphasizing accurate enumeration, avoidance of overcounting, the application of the addition and multiplication principles, and understanding of conditional probability and independence. Mastery of these topics lays the groundwork for more advanced statistical analysis and supports decision-making in uncertain scenarios encountered in everyday life and research.

References

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