Problem 43 - Algebra and Trigonometry

Question

\(43-44\) . A die is rolled, and the number showing is observed. Determine whether the events \(E\) and \(F\) are mutually exclusive. Then find the probability of the event \(E \cup F .\) (a) \(E :\) The number is even. F: The number is odd. (b) \(E :\) The number is even. F: The number is greater than \(4 .\)

Step-by-Step Solution

Answer

(a) Mutually exclusive, \(P(E \cup F) = 1\). (b) Not mutually exclusive, \(P(E \cup F) = \frac{2}{3}\).

Step 1: Understanding Mutually Exclusive Events for Part (a)

Event \(E\) is the die showing an even number: \(2, 4, 6\) and Event \(F\) is the die showing an odd number: \(1, 3, 5\). Mutually exclusive events are those that cannot happen at the same time. Since a number on a die cannot be both even and odd at the same time, Events \(E\) and \(F\) are mutually exclusive.

Step 2: Calculating the Probability of \(E \cup F\) for Part (a)

To find the probability of \(E \cup F\), since \(E\) and \(F\) are mutually exclusive, you add their probabilities. \(P(E) = \frac{3}{6} = \frac{1}{2}\) and \(P(F) = \frac{3}{6} = \frac{1}{2}\). So, \(P(E \cup F) = P(E) + P(F) = \frac{1}{2} + \frac{1}{2} = 1\).

Step 3: Understanding Mutually Exclusive Events for Part (b)

Event \(E\) is the die showing an even number: \(2, 4, 6\) and Event \(F\) is the die showing a number greater than \(4\): \(5, 6\). Events \(E\) and \(F\) are not mutually exclusive because the outcome \(6\) is common to both events.

Step 4: Calculating the Probability of \(E \cup F\) for Part (b)

To find the probability of \(E \cup F\), where they are not mutually exclusive, use the formula: \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). \(P(E) = \frac{3}{6} = \frac{1}{2}\), \(P(F) = \frac{2}{6} = \frac{1}{3}\), and \(P(E \cap F) = \frac{1}{6}\) because \(6\) is common. So, \(P(E \cup F) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}\).

Key Concepts

Mutually Exclusive Events

Mutually exclusive events are events that cannot occur at the same time. They are like opposite sides of the same coin. When one event happens, the other cannot. An example of this is when you roll a fair six-sided die: getting an even number (2, 4, or 6) and getting an odd number (1, 3, or 5) are mutually exclusive events.

• In a single roll of a die, it's impossible for the outcome to be both even and odd at the same time.
• The events "even number" and "odd number" clearly do not overlap.

Understanding mutually exclusive events helps in computing probabilities because when events are mutually exclusive, the probability of either event occurring is simply the sum of their individual probabilities. If we denote two mutually exclusive events as Event A and Event B, then \( P(A \cup B) = P(A) + P(B) \).
This principle is crucial for quickly finding the likelihood of one event or the other happening.

Event Outcomes

Event outcomes are the possible results that arise from an event. In probability, each outcome is a way things can happen. For example, when you roll a die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. In our exercise:

• Event \(E\) corresponds to obtaining an even number: 2, 4, or 6.
• Event \(F\) in part (a) corresponds to obtaining an odd number: 1, 3, or 5.
• Event \(F\) in part (b) corresponds to obtaining a number greater than 4: 5 or 6.

By examining the outcomes, you can better understand the characteristics of each event.

• Event \(E\) has three outcomes, and thus the probability of \(E\) is \( P(E) = \frac{3}{6} = \frac{1}{2} \).
• For \(F\) in part (a), probability is similarly \( P(F) = \frac{3}{6} = \frac{1}{2} \).
• For \(F\) in part (b), this changes to \( P(F) = \frac{2}{6} = \frac{1}{3} \).

Understanding event outcomes and how each contributes to the overall probability helps determine whether events might overlap or be exclusive.

Addition Rule for Probability

The addition rule for probability is a fundamental rule that allows you to find the probability of either of two events occurring. This is crucial for solving problems where we are interested in the probability of at least one event happening.
There are two scenarios with this rule:

• If the events are mutually exclusive, use: \[ P(A \cup B) = P(A) + P(B) \]
• If the events are not mutually exclusive, modify it to: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

This subtraction occurs because for overlapping events (not mutually exclusive), the probability of the common outcome would be counted twice without this adjustment.
For our die rolling example:

• In part (a), where \(E\) and \(F\) are mutually exclusive, the probability is:\[ P(E \cup F) = \frac{1}{2} + \frac{1}{2} = 1 \]
• In part (b), since events overlap (as both include 6), use:\[ P(E \cup F) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3} \]

Understanding how to apply the correct probability rule ensures accurate results in probability problems.