Problem 16 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A pair of fair dice is rolled. Let \(E\) denote the event that the number falling uppermost on the first die is 4 , and let \(F\) denote the event that the sum of the numbers falling uppermost is 6 . a. Compute \(P(F)\). b. Compute \(P(E \cap F)\). c. Compute \(P(F \mid E)\). d. Compute \(P(E)\). e. Are \(E\) and \(F\) independent events?

Step-by-Step Solution

Answer

a. \(P(F) = \frac{5}{36}\) b. \(P(E \cap F) = \frac{1}{36}\) c. \(P(F \mid E) = \frac{1}{6}\) d. \(P(E) = \frac{1}{6}\) e. Events E and F are not independent events.

Step 1: Calculate P(F)

To calculate the probability of event F (sum of the numbers on the uppermost faces is 6), we need to find all the possible combinations of two dice that result in a sum of 6. Those combinations are: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) There are 5 possible outcomes for event F to happen. Since there are a total of 6x6 = 36 possible outcomes when rolling two dice, the probability of event F is: P(F) = # of successful outcomes / total # of outcomes = 5 / 36

Step 2: Calculate P(E ∩ F)

To calculate the probability of both events E and F occurring, we need to find the combinations that satisfy both event E (the first die rolls a 4) and event F (sum of the numbers is 6). There is only one combination that satisfies both conditions: (4, 2). There are still a total of 36 possible outcomes when rolling two dice, so the probability of both events E and F happening is: P(E ∩ F)= # of successful outcomes / total # of outcomes = 1 / 36

Step 3: Calculate P(F|E)

To calculate the conditional probability of event F occurring given event E has occurred, we can use the formula: P(F|E) = P(E ∩ F) / P(E) First, we need to calculate the probability of event E (the first die rolls a 4). Since each of the 6 faces of the first die is equally likely, the probability of event E is: P(E) = 1 / 6 Now we have all the probabilities needed to calculate P(F|E): P(F|E) = (1 / 36) / (1 / 6) = 1 / 6

Step 4: Verify if events E and F are independent

To check if events E and F are independent, we need to verify if the following condition is true: P(E ∩ F) = P(E) * P(F) We already calculated P(E), P(F), and P(E ∩ F) from the previous steps, so now we just need to check if their product is equal: (1 / 36) = (1 / 6) * (5 / 36) Multiplying the probabilities of P(E) and P(F) we get (5 / 216), which is not equal to (1 / 36), therefore events E and F are NOT independent. The answers to the exercise are: a. P(F) = 5 / 36 b. P(E ∩ F) = 1 / 36 c. P(F | E) = 1 / 6 d. P(E) = 1 / 6 e. Events E and F are not independent events.

Key Concepts

Conditional Probability

In probability, we often want to find the likelihood of an event occurring given that another event has already happened. This is called conditional probability. It is denoted by \( P(F \mid E) \), representing the probability of event \( F \) occurring given that \( E \) has occurred. To calculate this, we use the formula:

• \( P(F \mid E) = \frac{P(E \cap F)}{P(E)} \)

This formula says that the conditional probability is the probability of both events happening together divided by the probability of the given event.
In our dice example, event \( E \) is rolling a 4 on the first die, and \( F \) is getting a total sum of 6. We calculated that \( P(E \cap F) = \frac{1}{36} \) and \( P(E) = \frac{1}{6} \). Therefore, the conditional probability \( P(F \mid E) \) becomes \( \frac{1}{6} \).
Teaching yourself how to interpret and calculate conditional probabilities can help when dealing with real-world situations where one event affects another.

Independent Events

Two events are considered independent if the occurrence of one does not influence the probability of the other. Mathematically, events \( E \) and \( F \) are independent if:

• \( P(E \cap F) = P(E) \times P(F) \)

If this equation holds true, it means the events have no effect on each other's outcome. In our dice scenario, \( P(E \cap F) = \frac{1}{36} \) and \( P(E) \times P(F) = \frac{1}{6} \times \frac{5}{36} = \frac{5}{216} \).
Clearly, \( \frac{1}{36} eq \frac{5}{216} \), so events \( E \) and \( F \) are not independent. Knowing whether events are independent can help when predicting probabilities in complex scenarios.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations and permutations of sets. It's very useful in probability to calculate the number of possible outcomes. When rolling two dice, combinatorics helps us determine the total number of potential outcomes and combinations that meet specific criteria.

• Total Outcomes: For two six-sided dice, each die has 6 faces, leading to \( 6 \times 6 = 36 \) total outcomes.
• Specific Combinations: To find combinations that result in a sum of 6, we identify pairs: \( (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) \), totaling 5 combinations.

Combinatorics provides a systematic method to count and organize these possibilities, making it efficient to solve probability problems.