Question

An experiment consists of tossing a single die and observing the number of dots that show on the upper face. Events \(A, B\), and \(C\) are defined as follows: A: Observe a number less than 4 \(B\) : Observe a number less than or equal to 2 \(C\) : Observe a number greater than 3 Find the probabilities associated with the events below using either the simple event approach or the rules and definitions from this section. a. \(S\) b. \(A \mid B\) c. \(B\) d. \(A \cap B \cap C\) e. \(A \cap B\) f. \(A \cap C\) g. \(B \cap C\) h. \(A \cup C\) i. \(B \cup C\)

Short answer

Answer: The probabilities of the events are as follows: - P(A) = 1/2 - P(B) = 1/3 - P(C) = 1/2 - P(A|B) = 1 - P(A∩B) = 1/3 - P(A∩C) = 0 - P(B∩C) = 0 - P(A∪C) = 1 - P(B∪C) = 5/6

Step-by-step solution

1. Event Interpretation

First, let's interpret the events given: \(A\): Observe a number less than 4, i.e., observe a 1, 2, or 3. \(B\): Observe a number less than or equal to 2, i.e., observe a 1 or 2. \(C\): Observe a number greater than 3, i.e., observe a 4, 5, or 6. Now, let's find the probabilities of each requested event.

a. Event S

Event \(S\) is the sample space, which means all possible outcomes. In this case, there are 6 possible outcomes: {1, 2, 3, 4, 5, 6}. The probability of each outcome is equal, which means \(P(S) = 1\).

b. Event A | B

Event \(A \mid B\) means the probability of event \(A\) occurring given that event \(B\) has occurred. Since event \(B\) is a subset of event \(A\), then observing a number less than or equal to 2 will always result in a number less than 4. Therefore, \(P(A \mid B) = 1\).

c. Event B

Event \(B\) has two possible outcomes: {1, 2}. There are a total of 6 possible outcomes for the die roll. Thus, the probability \(P(B) = \frac{2}{6} = \frac{1}{3}\).

d. Event A ∩ B ∩ C

In order for events \(A\), \(B\), and \(C\) to occur simultaneously, an outcome must be less than 4, less than or equal to 2, and greater than 3. However, no such outcome exists. Hence, \(P(A \cap B \cap C) = 0\).

e. Event A ∩ B

The intersection of events \(A\) and \(B\) are the outcomes that are common to both events. The outcomes are {1, 2}, so the probability \(P(A \cap B) = \frac{2}{6} = \frac{1}{3}\).

f. Event A ∩ C

The intersection of events \(A\) and \(C\) implies an outcome that is both less than 4 and greater than 3. However, no such outcome exists. So, \(P(A \cap C) = 0\).

g. Event B ∩ C

The intersection of events \(B\) and \(C\) requires an outcome that is less than or equal to 2 and greater than 3. Since no such outcome exists, \(P(B \cap C) = 0\).

h. Event A ∪ C

The union of events \(A\) and \(C\) includes outcomes that belong to either event. In this case, outcomes are {1, 2, 3, 4, 5, 6}, which are all possible outcomes for a die roll. So, \(P(A \cup C) = 1\).

i. Event B ∪ C

The union of events \(B\) and \(C\) contains outcomes from both events. Here, the outcomes are {1, 2, 4, 5, 6}. Consequently, \(P(B \cup C) = \frac{5}{6}\).

Key concepts

Probability of Events

Probability theory is all about measuring how likely an event will occur. When dealing with probability of events, we focus on calculating the likelihood of specific events happening.
For example, when we roll a single die, each face, or outcome, has an equal chance of occurring. The probability of any particular face landing up is \(rac{1}{6}\).
If we define events based on certain conditions, like having numbers less than 4 (Event \( A \)) or numbers less than or equal to 2 (Event \( B \)), we can calculate the probability by counting the favorable outcomes and dividing by the total number of possible outcomes.
This is why the probability of \( B \) is \( \frac{2}{6} \).To summarize, the probability of an event \( E \) happening is given by \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in the sample space}}. \]

Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring, given that another event has already occurred.
In simple terms, it's about narrowing down the possible outcomes based on additional information. It is denoted as \( P(A \mid B) \), which reads as "the probability of \( A \) given \( B \)."Using our die example, if we know that event \( B \) has occurred (observing a number less than or equal to 2), and we want to find the probability of event \( A \) (observing a number less than 4), we consider only those outcomes that are in \( B \).
Since \( B \) is a subset of \( A \), the occurrence of \( B \) ensures \( A \), making \( P(A \mid B) = 1 \).The formula for conditional probability is: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \]provided \( P(B) eq 0 \). This formula shows how to compute the likelihood of event \( A \) when we know \( B \) has happened.

Sample Space

The sample space is a fundamental concept in probability theory. It represents the set of all possible outcomes of a random experiment.
In terms of tossing a die, the sample space \( S \) is \({1, 2, 3, 4, 5, 6}\), capturing every possible result when the die is rolled.The sample space is essential because it serves as the reference set for calculating probabilities of events.
For instance, with a die, each number has a probability of \( \frac{1}{6} \), derived from the sample space.
The completeness of the sample space allows for the precise computation of probabilities for any defined event.Remember, without defining the sample space, gauging probabilities becomes ambiguous as we lack the full picture of possible outcomes. This is why understanding the sample space is the first step in approaching any probability problem.

Intersection and Union of Events

Events can interact with each other in ways that can be measured by intersection and union, two key concepts in probability theory.The intersection of two or more events (denoted by \( A \cap B \)) refers to outcomes that are common to all the events involved.
For example, the intersection of events \( A \) and \( B \) is the set \({1, 2}\), thus \( P(A \cap B) = \frac{1}{3} \) because these are shared outcomes.Union, on the other hand, combines the outcomes from either or both events. It is denoted by \( A \cup B \).
The union of events \( A \) and \( C \) (outcomes: \( 1, 2, 3, 4, 5, 6 \)) covers all possible outcomes, so \( P(A \cup C) = 1 \).These concepts help in understanding how events overlap or spread, which is crucial for broader probability assessments.