Question

A fair die is rolled, and a fair coin is tossed. The sample space is taken to be \(\Omega=\) \(\Omega_{1} \times \Omega_{2}\) where \(\Omega_{1}\) is the six- clement sample space for the die and \(\Omega_{2}\) is the twoelement sample space for the coin. Let \(A \subseteq \Omega_{1}\) be the event "a 5 is rolled." Let \(B \subseteq \Omega_{2}\) be the event "heads." Let \(C \subseteq \Omega\) be the event "at most two spots on the top face of the die (with heads or tails on the coin) or at least five spots on the top face of the die together with heads on the coin." Let \(D\) be the event "at least a 5 on the die (with heads or tails on the coin)." Which of the following sets of events are independent sets? Explain your answer. (a) \(\\{A, B\\}\) (b) \(\\{A, B, C\\}\) (c) \(\\{B, C]\) (d) \(\\{B, C, D\\}\)

Short answer

Set (a), \(\{A, B\}\), and set (b), \(\{A, B, C\}\), are independent.

Step-by-step solution

Understanding Event Independence

Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. For events \(A\) and \(B\) to be independent, \(P(A \cap B) = P(A)P(B)\). We'll apply this for each set of events.

Identifying Probabilities for Events

We begin by identifying the probabilities of each event. Since the die is fair, the probability of rolling a 5, \(P(A) = \frac{1}{6}\). The coin is fair, so the probability of heads, \(P(B) = \frac{1}{2}\). For event C, we must consider two scenarios: either at most two spots on the die with any coin side or at least five spots with heads. \(P(C) = \left( \frac{2}{6}\right)\left(1\right) + \left(\frac{1}{6}\right)\left(\frac{1}{2}\right) = \frac{1}{3} + \frac{1}{12} = \frac{5}{12}\). Event D requires at least a 5 on the die, with either heads or tails: \(P(D) = \frac{2}{6} \times 1 = \frac{1}{3}\).

Checking Independence for Set (a): {A, B}

Calculate \(P(A \cap B)\), which requires rolling a 5 and getting heads. Since these are independent actions, \(P(A \cap B) = P(A) * P(B) = \frac{1}{6} * \frac{1}{2} = \frac{1}{12}\). Checking independence: \(\frac{1}{12} = \frac{1}{6} * \frac{1}{2} = \frac{1}{12}\). Thus, events \(A\) and \(B\) are independent.

Checking Independence for Set (b): {A, B, C}

Recalculate \(P(A \cap B \cap C)\) which requires rolling a 5 with heads and fitting conditions of C: This can only occur if we have 5 with heads, contributing \(\frac{1}{12}\). We already calculated \(P(A \cap B)\) and hence find \(P(A \cap B \cap C) = \frac{1}{12} = P(A)P(B)P(C) = \frac{1}{6} \times \frac{1}{2} \times \frac{5}{12}\). Since these are equal, events \(A, B, C\) are independent.

Checking Independence for Set (c): {B, C}

Calculate \(P(B \cap C)\), which includes scenarios defined by C and heads: \(\frac{1}{12} + \frac{1}{4} = \frac{1}{3}\). Check \(P(B)P(C) = \frac{1}{2}\times \frac{5}{12} = \frac{5}{24}\). Since \(\frac{1}{3} eq \frac{5}{24}\), events \(B\) and \(C\) are not independent.

Checking Independence for Set (d): {B, C, D}

Calculate \(P(B \cap C \cap D)\), \(P(C \cap D)\), and \(P(B \cap D)\), and verify independence with \(P(B)P(C)P(D)\). For \(P(B \cap C \cap D)\) it requires both a head, a qualified condition of \(C\) and \(D\). We find \(P(B \cap C \cap D) = \frac{1}{12}\) and \(P(B)P(C)P(D) = \frac{1}{2}\times\frac{5}{12}\times\frac{1}{3}= \frac{5}{72}\). These are not equal, so events \(B\), \(C\), \(D\) are not independent.

Key concepts

Sample Space

In probability theory, the concept of a sample space is foundational. A sample space, often denoted by the symbol \( \Omega \), represents all the possible outcomes of a probability experiment. For example, when tossing a fair coin, the sample space is \( \{ \text{heads}, \text{tails} \} \), because these are the only two outcomes that can occur. Similarly, when rolling a fair six-sided die, the sample space is \( \{ 1, 2, 3, 4, 5, 6 \} \). Each number represents a side of the die that could face up after a roll.
The exercise provides a more complex sample space by combining two independent actions: rolling a die and tossing a coin. The combined sample space (\( \Omega \)) includes every possible combination of die rolls and coin flips. For instance, the outcome (3, heads) is in the sample space. When determining probabilities and analyzing events, we use the sample space to ensure all outcomes are considered.

Event Independence

Events are said to be independent if the occurrence of one event does not affect the probability of the other. Mathematically, two events \( A \) and \( B \) are independent if the probability of both events occurring is equal to the product of their individual probabilities: \( P(A \cap B) = P(A) \times P(B) \).
Imagine rolling a fair die and flipping a fair coin. If rolling the die gives you a 5, it doesn't influence whether the coin lands on heads or tails—these events are independent. Understanding event independence helps in calculating probabilities by simplifying the complexities of multiple events.
In the exercise, several sets of events were analyzed for independence. By verifying through calculations that \( P(A \cap B) \) equaled \( P(A) \times P(B) \), students could determine if those events were indeed independent.

Discrete Mathematics

Discrete mathematics serves as a fundamental branch of mathematics dealing with distinct and usually finite sets of objects. It underpins many probability problems involving discrete outcomes such as coin tosses or die rolls. In our exercise, we explore concepts like permutations and combinations within discrete sample spaces, which are examples of discrete math in action.
This form of mathematics is essential in defining clear and countable outcomes, allowing probability predictions to be systematically calculated. For events such as those in the exercise, discrete mathematics helps students understand the underlying structure of the problem, by illustrating how different individual outcomes contribute to the overall sample space and subsequent probabilities.

Fair Die and Coin Experiments

Fair die and coin experiments are classic examples used in probability to illustrate randomness and equality of outcomes. A fair die means each number from 1 to 6 has an equal chance of appearing face up, corresponding to a probability of \( \frac{1}{6} \) for each outcome. Similarly, a fair coin implies that both heads and tails have an equal chance of landing face up, with each having a probability of \( \frac{1}{2} \).
In the given exercise, these concepts explain the process of creating a sample space using both a fair die and a coin. By leveraging these fair conditions, students can apply probability laws accurately. These experiments form the backbone of many introductory lessons in probability, serving as a relatable means to engage with abstract statistical concepts. Understanding the fairness ensures accurate calculations when assessing the independence of events, as shown in the original exercise.