Math 464 Homework 2 Spring 2013 The Following Assignment Is

Math 464homework 2spring 2013the Following Assignment Is To Be Turned

Write down a sample space and give the probability of each outcome for selecting a letter at random from the word MISSISSIPPI. Determine the percentage of students who either smoke or drink, given the percentages that smoke, drink, and do both. Calculate the probability P(B) given P(A), P(A^c∩B^c), and P(A∩B). Find the maximum and minimum possible values of P(A∩B) based on given P(A) and P(P(B)). Prove the inclusion-exclusion formula for three events. Find the probability that Alice, flipping a fair coin first, wins a game where the first to heads wins, with Bob allowing Alice to flip first. For an unfair coin with specific head/tail probabilities, compute the probability that it takes exactly 8 flips given it took at least 6 flips. Prove that P(A^c|B) = 1 - P(A|B) using the definition of conditional probability. Show that if A and B are independent events, then their complements are also independent. When rolling a six-sided die twice, define events A (first roll odd), B (second roll even), and C (both rolls same parity), and demonstrate that A, B, and C are pairwise independent but not independent jointly.

Paper For Above instruction

The following paper explores fundamental probability concepts through a series of problems reflective of real-world and theoretical scenarios, emphasizing understanding of sample spaces, probability calculations, independence, and combinatorial methods essential for students engaged in advanced probability coursework.

Introduction

Probability theory is a mathematical framework that models uncertainty and randomness. It provides tools for quantifying the likelihood of events and understanding their relationships. This paper addresses several core notions, including sample space construction, probability calculation, inclusion-exclusion principle, independence, and conditional probability, through practical problems designed for students studying advanced probability such as those in a Math 464 course.

Sample Space and Probability Distributions

The first problem involves selecting a letter at random from the word "MISSISSIPPI." The sample space consists of the individual letters: {M, I, S, S, I, S, S, I, P, P, I}. Since the selection is random with equal likelihood for each letter, the sample space can be represented as a set of outcomes, with respective probabilities based on their frequencies. The letter S appears four times, I four times, P twice, and M once. Therefore, the probability of selecting a particular letter equals its frequency divided by the total number of letters, which is 11.

Specifically, P(M) = 1/11, P(I) = 4/11, P(S) = 4/11, P(P) = 2/11. This distribution exemplifies the construction of a finite sample space with assigned probabilities, a fundamental aspect of probability theory.

Probability of Students Smoking or Drinking

In a hypothetical student population, 25% smoke, 60% drink, and 15% do both. To compute the percentage of students who either smoke or drink, we employ the inclusion-exclusion principle: P(smoke ∪ drink) = P(smoke) + P(drink) - P(both). Plugging in the values gives 0.25 + 0.60 - 0.15 = 0.70, indicating that 70% of students either smoke or drink. This illustrates how overlapping probabilities impact the calculation of union probabilities.

Conditional Probability and Event Relationships

Given that P(A) = 1/3, P(A^c∩B^c) = 1/2, and P(A∩B) = 1/4, the task is to find P(B). Recognizing that P(A^c∩B^c) = 1 - P(A ∪ B), and that P(A ∪ B) = P(A) + P(B) - P(A∩B), we set up the equation:

1 - [P(A) + P(B) - P(A∩B)] = 1/2, which leads to P(A) + P(B) - P(A∩B) = 1/2. Substituting the known values yields (1/3) + P(B) - 1/4 = 1/2. Solving this for P(B) gives P(B) = 1/4 + 1/2 - 1/3 = (3/12 + 6/12 - 4/12) = 5/12. This calculation demonstrates the use of basic probability identities to find unknown probabilities.

Maximum and Minimum of P(A∩B)

Given P(A) = 0.4 and P(B) = 0.7, the possible values of P(A ∩ B) are bounded by the rules: P(A ∩ B) ≥ max(0, P(A) + P(B) - 1) and P(A ∩ B) ≤ min(P(A), P(B)). Therefore, the minimum is max(0, 0.4 + 0.7 - 1) = 0.1, and the maximum is min(0.4, 0.7) = 0.4. These bounds illustrate the range of joint probabilities consistent with specified marginals, crucial for understanding dependence.

Inclusion-Exclusion Principle for Three Events

The inclusion-exclusion formula for three events A, B, C states:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

Proving this involves considering the union as the sum of individual probabilities minus overlaps, correcting for double counting, and adding back the triple overlap. This principle underpins complex probability calculations involving multiple events and is vital for accurately computing combined event probabilities.

Game of Flipping Coins

In a game where Alice and Bob flip a fair coin, and the first to get heads wins with Bob allowing Alice to flip first, the probability that Alice wins can be derived by considering the sequence of flips. On each turn, Alice has a 1/2 chance of winning immediately by flipping heads. If she flips tails, the game proceeds in a similar fashion, shifted by Bob's turn. The probability that Alice wins starting first is the sum of an infinite geometric series: P(Alice wins) = 1/2 + (1/2)(1/2)(1/2) + ... = 1/3.

Probability in Unfair Coin Flips

When flipping an unfair coin with P(H) = 1/3 and P(T) = 2/3, the probability that the number of flips needed to get heads is exactly 8, given it took at least 6 flips, can be calculated by conditional probability: P(Takes exactly 8 flips | Takes at least 6 flips) = P(Takes exactly 8 flips) / P(Takes at least 6 flips). The numerator is (P(T))⁷ · P(H) = (2/3)⁷(1/3), because the first 7 flips are tails, and the 8th is heads. The denominator is 1 minus the probability it takes fewer than 6 flips, which is 1 − Σ_{k=1}^5 P(taking exactly k flips). This calculation exemplifies usage of conditional probabilities in sequences.

Conditional Probability and Independence

For events A and B with P(B) > 0, the conditional probability P(A^c|B) = 1 - P(A|B) can be shown directly. By the definition of conditional probability:

P(A|B) = P(A ∩ B) / P(B).

Similarly, P(A^c ∩ B) = P(B) - P(A ∩ B), so:

P(A^c|B) = P(A^c ∩ B) / P(B) = [P(B) - P(A ∩ B)] / P(B) = 1 - P(A ∩ B) / P(B) = 1 - P(A|B).

This confirms the complement rule within the conditional framework.

Furthermore, if events A and B are independent, then their complements are also independent. This follows directly from the definition of independence: P(A ∩ B) = P(A)P(B). Therefore, P(A^c ∩ B^c) = P(A^c)P(B^c) = (1 - P(A))(1 - P(B)), confirming independence of complements.

Event Independence in Dice Rolls

When rolling a fair six-sided die twice, define:

A: first roll is odd,

B: second roll is even,

C: both rolls have the same parity.

Calculating probabilities:

P(A) = 3/6 = 1/2,

P(B) = 3/6 = 1/2,

P(C): both are odd or both are even, so P(C) = P(odd, odd) + P(even, even) = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2.

Checking pairwise independence:

P(A ∩ B) = P(odd, even) = (1/2)(1/2) = 1/4.

P(A)P(B) = (1/2)(1/2) = 1/4, so A and B are independent.

Similarly, P(A ∩ C) = P(odd, same parity) = P(odd, odd) = 1/4, and P(A)P(C) = (1/2)(1/2) = 1/4, so A and C are independent.

Likewise, B and C also are independent.

However, the three events collectively are not independent because:

P(A ∩ B ∩ C) = P(odd, even, same parity) which is impossible, since "both same parity" includes both odd/odd and even/even, but if B (second even), then third event's conditions are met if A is odd and second even, so the independence does not extend jointly, illustrating that pairwise independence does not imply full independence.

Conclusion

This collection of problems illuminates core probability notions vital for advanced coursework. From constructing sample spaces to exploring independence and conditional probabilities, understanding these concepts establishes a strong foundation for further study and application in fields like statistics, data science, and decision theory.

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