Problem 19 - 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 landing uppermost on the first die is a 3 , and let \(F\) denote the event that the sum of the numbers landing uppermost is 7. Determine whether \(E\) and \(F\) are independent events.

Step-by-Step Solution

Answer

Events E and F are independent because P(E∩F) = P(E)*P(F) = \(\frac{1}{36}\).

Step 1: Find the probabilities of E, F, and E∩F

First, let's find the probability of each event: - Event E: Number landing uppermost on the first die is 3. As there are 6 faces on a die, the probability of this event is \(\frac{1}{6}\). - Event F: The sum of the numbers landing uppermost is 7. There are 6 possible ways to get a sum of 7 with two dice (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1), out of a total of 36 possible outcomes for rolling two dice. So the probability of event F is \(\frac{6}{36}\) or, simplified, \(\frac{1}{6}\). - Event E∩F: Both events E and F occur, which means the first die shows a 3, and the sum of the numbers is 7. There is only one way for this to happen (3+4). So the probability of event E∩F is \(\frac{1}{36}\).

Step 2: Check for independence

Now let's check if events E and F are independent by verifying if P(E∩F) = P(E)*P(F). Based on the probabilities we found above: P(E) = \(\frac{1}{6}\) P(F) = \(\frac{1}{6}\) P(E∩F) = \(\frac{1}{36}\) P(E)*P(F) = \(\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}\) Since P(E∩F) = P(E)*P(F), events E and F are independent.

Key Concepts

Independent Events

In probability theory, two events are considered independent if the occurrence of one does not affect the probability of the other occurring. This means that the outcome of one event does not influence or change the outcome probability of another event. In mathematical terms, two events, say, \(E\) and \(F\), are independent if:

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

In the dice rolling exercise above, the independence test involved calculating whether the probability of both events \(E\) and \(F\) occurring simultaneously was equal to the product of their individual probabilities. Since the calculated value \(P(E \cap F) = \frac{1}{36}\) was the same as \(P(E) \times P(F)\), which also equals \(\frac{1}{36}\), the events \(E\) and \(F\) were shown to be independent.

Probability of Events

The probability of an event is a measure of the likelihood that the event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes in a sample space. Probability values range from 0 to 1, where 0 indicates impossibility, and 1 indicates certainty.

• If an event is certain, its probability is 1.
• If an event is impossible, its probability is 0.

For example, in rolling a single six-sided die, the probability of rolling a 3 (event \(E\)) is \(\frac{1}{6}\), since only one out of the six possible dice faces can land as a 3. Similarly, the probability of rolling two dice and getting a sum of 7 (event \(F\)) is \(\frac{6}{36}\) or simplified, \(\frac{1}{6}\). Understanding probability helps in predicting which outcomes are more likely than others.

Dice Probability

Dice probability deals with determining the likelihood of outcomes when rolling one or more dice. Each die is a cube with six faces, numbered 1 through 6. When determining probability with dice, consider the possible outcomes:

• Single Die: Each of the six faces is equally likely, so the probability of any specific face showing is \(\frac{1}{6}\).
• Two Dice: There are 36 possible combinations of the results because each die has 6 faces, resulting in 6 \(\times\) 6 = 36 outcomes.

In the exercise, the probability of getting a sum of 7 when rolling two dice is determined by counting the successful combinations: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). This yields 6 successful outcomes. The probability is therefore \(\frac{6}{36}\), or \(\frac{1}{6}\).

Sample Space in Probability

The sample space in probability is the set of all possible outcomes of a random experiment. When rolling two dice, each die can land on any number between 1 and 6, making the total combinations of the two dice a part of the sample space. The sample space for rolling two dice can be represented by a set \(S\), which includes 36 ordered pairs such as (1,1), (1,2), ..., (6,6).
An important aspect of the sample space is that it must include every possible outcome of an experiment, ensuring the total probabilities sum to 1. For example, in our exercise, the probability of an event is calculated by dividing the number of successful outcomes (e.g., pairs that sum to 7) by the number of outcomes in the sample space (36 in the case of rolling two dice).
Creating a sample space allows for comprehensive analysis of potential outcomes, which is essential for calculating event probabilities accurately. It forms the foundation for assessing how likely different results are, based on all possible occurrences.