Problem 25 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(E\) and \(F\) be two events that are mutually exclusive, and suppose \(P(E)=.2\) and \(P(F)=.5\). Compute: a. \(P(E \cap F)\) b. \(P(E \cup F)\) c. \(P\left(E^{c}\right)\) d. \(P\left(E^{c} \cap F^{c}\right)\)

Step-by-Step Solution

Answer

a. \(P(E \cap F)=0\) b. \(P(E \cup F)=0.7\) c. \(P\left(E^{c}\right)=0.8\) d. \(P\left(E^{c} \cap F^{c}\right)=0.3\)

Step 1: Understanding the Given Information

The problem states that events E and F are mutually exclusive, meaning that they can't occur at the same time. Additionally, we have been given the probabilities of E occurring \(P(E)=0.2\) and F occurring \(P(F)=0.5\).

Step 2: Calculating \(P(E \cap F)\)

Remember, mutually exclusive events cannot occur at the same time; hence, the intersection of E and F, denoted as \(E \cap F\), is an impossible event. Throughout probability theory, the probability of an impossible event is 0. So, \(P(E \cap F)=0\).

Step 3: Calculating \(P(E \cup F)\)

The union of E and F, denoted as \(E \cup F\), represents either event E occurring, event F occurring or both occurring. But since these events are mutually exclusive, we don't have to worry about them occurring simultaneously. The formula for computing the union of two mutually exclusive events is: \(P(E \cup F) = P(E) + P(F)\). Plugging in our values, our calculations will be: \(P(E \cup F) = 0.2 + 0.5 = 0.7\).

Step 4: Calculating \(P(E^{c})\)

The notation \(E^{c}\) denotes the complement of E, meaning the event of E not occurring. In any probability space, the probability of an event plus the probability of its complement equals 1. Hence, \(P(E) + P(E^{c}) = 1\). Solving for \(P(E^{c})\), we get: \(P(E^{c}) = 1 - P(E) = 1 - 0.2 = 0.8\).

Step 5: Calculating \(P(E^{c} \cap F^{c})\)

The expression \(E^{c} \cap F^{c}\) represents the probability of both E and F not occurring, which is the intersection of their individual complementary events. As in Step 4, we can use the complements to calculate this. However, these two complementary events aren't mutually exclusive so we cannot simply add their probabilities. Fortunately, the complement of \(E \cup F\) is exactly \(E^{c} \cap F^{c}\) (an intersection distributes over a union). So, we calculate the probability of \(E^{c} \cap F^{c}\) as follows: \(P(E^{c} \cap F^{c}) = P((E \cup F)^{c}) = 1 - P(E \cup F) = 1 - 0.7 = 0.3\). So, the answers to the given prompts are: a. \(P(E \cap F)=0\) b. \(P(E \cup F)=0.7\) c. \(P(E^{c})=0.8\) d. \(P(E^{c} \cap F^{c})=0.3\)

Key Concepts

Mutually Exclusive Events

Mutually exclusive events are common concepts in probability theory and are critical for understanding how different outcomes in a probability space interact with each other.
In simple terms, two events are considered mutually exclusive if they cannot both happen at the same time.

• Example: Flipping a coin can result in either heads or tails, but not both. Thus, the event of getting heads and the event of getting tails are mutually exclusive.

For mutually exclusive events, the probability of both events happening together is zero.
In mathematical terms, if events E and F are mutually exclusive, then:
\[ P(E \cap F) = 0 \]This property is fundamental when calculating the union of mutually exclusive events, simplifying the process considerably.

Complementary Events

Complementary events are pairs of outcomes that together cover all possible outcomes of a given probability experiment. If one event occurs, the complementary event does not occur, and vice-versa.

• Example: Consider the event of rolling a 3 on a six-sided die. The complement of this event is not rolling a 3, which includes rolling a 1, 2, 4, 5, or 6.

The probability of a complement is calculated as:
\[ P(E^c) = 1 - P(E) \]This formula simply means that the total probability of all possible outcomes is always 1.
Thus, the probability of an event not occurring is precisely the total probability minus the probability of the event happening.
This knowledge can help solve various probability exercises, such as finding the probability of not seeing a certain outcome.

Intersection and Union in Probability

These terms are crucial for understanding how multiple events interact within a probability framework.
The intersection of events, denoted as \( E \cap F \), refers to the likelihood that both events occur simultaneously. In case of mutually exclusive events, this probability is zero.

• For example, if two dice are rolled, the intersection might represent the probability that both die show a 3 simultaneously.

The union of events, denoted as \( E \cup F \), is the probability of either event occurring or both. For mutually exclusive events:
\[ P(E \cup F) = P(E) + P(F) \]The union accounts for all outcomes that lead to either event taking place, ensuring no double-counting in mutually exclusive situations.

Probability Calculations

Understanding basic probability calculations allows us to determine the likelihood of various scenarios.
Key calculations include finding the probability of an event, its complement, intersections, and unions among others.

• Probability of an event: \( P(E) = \text{Number of favorable outcomes} / \text{Total outcomes} \)
• Complement: \( P(E^c) = 1 - P(E) \)
• For mutually exclusive events, \( P(E \cap F) = 0 \) and \( P(E \cup F) = P(E) + P(F) \)
• Non-mutually exclusive events require adjustments. Use: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)

These mathematical rules support a comprehensive approach to solving probability problems, guiding calculations through practical and theoretical scenarios.