Question

An experiment can result in one of five equally likely simple events, \(E_{1}, E_{2}, \ldots, E_{5} .\) Events \(A, B,\) and \(C\) are defined as follows: \(A: E_{1}, E_{3}\) $$P(A)=.4$$ \(B: E_{1}, E_{2}, E_{4}, E_{5} \quad P(B)=.8\) \(C: E_{3}, E_{4}\) $$P(C)=.4$$ Find the probabilities associated with the following events by listing the simple events in each. a. \(A^{c}\) b. \(A \cap B\) c. \(B \cap C\) d. \(A \cup B\) e. \(B \mid C\) f. \(A \mid B\) g. \(A \cup B \cup C\) h. \((A \cap B)^{c}\)

Short answer

Answer: The probability of the complement of the intersection of events A and B is \(\frac{3}{5}\).

Step-by-step solution

a. \(A^{c}\)

The complement of event \(A\) includes all simple events that are not in \(A\). Simple events in \(A\) are \(E_1\) and \(E_3\). Hence, the simple events in \(A^c\) are \(E_2, E_4\), and \(E_5\). Since the events are equally likely, the probability of \(A^c = P(A^c) = 1 - P(A) = 1 - 0.4 = 0.6\).

b. \(A \cap B\)

The intersection of \(A\) and \(B\) includes all simple events that are in both \(A\) and \(B\). Here, the only common event is \(E_1\). Therefore, \(A\cap B = E_1\), and since \(P(E_1) = \frac{2}{5}\) (equal probability = \(P(A)\)), \(P(A\cap B) = \frac{2}{5}\).

c. \(B \cap C\)

The intersection of \(B\) and \(C\) includes all simple events that are in both \(B\) and \(C\). Here, the only common event is \(E_4\). Therefore, \(B\cap C = E_4\), and since \(P(B) = 0.8\) and it contains 4 equally likely events, \(P(E_4)= \frac{2}{5}\). Thus, \(P(B\cap C) = \frac{2}{5}\).

d. \(A \cup B\)

The union of \(A\) and \(B\) includes all simple events that are in either \(A\) or \(B\) or both. As \(A\) contains \(E_1,E_3\) and \(B\) contains \(E_1, E_2, E_4, E_5\), the union will have \(E_1, E_2, E_3, E_4, E_5\). Since all the simple events are included, \(P(A \cup B) = 1\).

e. \(B \mid C\)

This represents the probability of event \(B\) given that event \(C\) has occurred. We will use the conditional probability formula: \(P(B \mid C) = \frac{P(B\cap C)}{P(C)}\). We found that \(P(B\cap C) = \frac{2}{5}\) and \(P(C) = 0.4\), so \(P(B \mid C) = \frac{\frac{2}{5}}{0.4} = 1\).

f. \(A \mid B\)

This represents the probability of event \(A\) given that event \(B\) has occurred. We will use the conditional probability formula: \(P(A \mid B) = \frac{P(A\cap B)}{P(B)}\). We found that \(P(A\cap B) = \frac{2}{5}\) and \(P(B) = 0.8\), so \(P(A \mid B) = \frac{\frac{2}{5}}{0.8} = \frac{1}{2}\).

g. \(A \cup B \cup C\)

The union of \(A, B,\) and \(C\) includes all simple events that are in either \(A\) or \(B\) or \(C\) or any combination of them. Since all the simple events are included, \(P(A \cup B \cup C) = 1\).

h. \((A \cap B)^{c}\)

The complement of the intersection of \(A\) and \(B\) includes all simple events that are not in both \(A\) and \(B\). Since \(A \cap B\) contains only event \(E_1\), \((A \cap B)^{c}\) will contain the other simple events \(E_2, E_3, E_4,\) and \(E_5\). Since these events are equally likely, \(P((A \cap B)^c) = 1 - P(A\cap B) = 1 - \frac{2}{5} = \frac{3}{5}\).

Key concepts

Complement Rule

The Complement Rule in probability is a useful concept to determine the likelihood of an event not occurring. When you have an event, like event \(A\) in our exercise, the complement of \(A\), denoted as \(A^{c}\), includes all outcomes that are not part of \(A\). This is especially handy when the event itself has a probability assigned to it, as the probability of the complement can be calculated simply by subtracting the event's probability from 1.
For example, in our exercise, event \(A\) consists of outcomes \(E_1\) and \(E_3\) with a probability \(P(A) = 0.4\). Thus, the complement includes \(E_2, E_4,\) and \(E_5\). Consequently, the probability of \(A^{c}\) is \(P(A^{c}) = 1 - P(A) = 0.6\).

• Benefits: It provides a quick method to find the probability of an event not happening.
• Usage: The complement rule is widely used when calculating probabilities where other computation methods are complex.

Intersection and Union of Events

The Intersection and Union of Events are fundamental operations in probability theory. They help in understanding how different events relate to each other, especially when they overlap. Let's dive into each:

• Intersection (\(A \cap B\)): This represents the events that occur in both \(A\) and \(B\) simultaneously. In our example, only \(E_1\) is common to both \(A\) and \(B\). The probability of this intersecting event \(A \cap B\) is thus \(\frac{2}{5}\), assuming each event has equal probability.
• Union (\(A \cup B\)): The union signifies the occurrence of at least one of the events \(A\) or \(B\). In our given scenario, this includes all possible outcomes since the collection of \(A\) and \(B\) together encompasses all simple events, resulting in \(P(A \cup B) = 1\).

Significance:

• Intersection is crucial when discerning common outcomes, critical in assessments like risks when two conditions must coexist.
• Union is beneficial in scenarios where any one of the possible outcomes is acceptable or favorable for analysis.

Conditional Probability

Conditional Probability is a concept used to find the probability of an event occurring given that another event has already occurred. It's expressed as \(P(A \mid B)\) or \(P(B \mid C)\), representing the probability of \(A\) given \(B\) or \(B\) given \(C\), respectively.
In our exercise, suppose we want \(B\) given \(C\). We use the formula: \(P(B \mid C) = \frac{P(B \cap C)}{P(C)}\). Since \(B \cap C = E_4\) and \(P(C) = 0.4\), hence \(P(B \mid C) = 1\). Similarly, for \(A\) given \(B\), \(P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{2}\).
The power of conditional probability lies in its ability to refine predictions based on subsequent outcomes.

• Tools: Allows better understanding in scenarios like diagnostics tests, reliability assessments, and risk management.
• Implication: Essential in fields requiring sequential probability measures, such as game theory and AI predictions.

Equally Likely Events

Equally Likely Events are scenarios where each outcome of a probability experiment has the same chance of occurring. This concept simplifies finding probabilities since each outcome's probability is consistent. In our exercise, since all simple events \(E_1, E_2, E_3, E_4, \text{and} E_5\) are equally probable, calculations become straightforward.
The probability for any single event among \(n\) equally likely events is \(\frac{1}{n}\). For instance, if \(E_1\) to \(E_5\) are equally likely, each outcome has a probability of \(\frac{1}{5}\).
This concept is particularly beneficial when the uniformity of outcomes is a given, such as rolling a fair die or drawing a card from a well-shuffled deck.

• Application: Effective in analyzing games of chance and ensuring fairness in random processes.
• Advantage: Simplifies complex problems by leveraging uniform probability distribution.