Problem 15 - 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 5 , and let \(F\) denote the event that the sum of the numbers falling uppermost is 10 . a. Compute \(P(F)\). b. Compute \(P(E \cap F)\). c. Compute \(P(F\lfloor E)\). d. Compute \(P(E)\). e. Are \(E\) and \(F\) independent events?

Step-by-Step Solution

Answer

a. \(P(F) = \frac{1}{12}\) b. \(P(E \cap F) = \frac{1}{36}\) c. \(P(F|E) = \frac{1}{6}\) d. \(P(E) = \frac{1}{6}\) e. Events E and F are independent.

Step 1: Find the sample space

When rolling a pair of fair dice, the sample space consists of all the possible outcomes, which would be the product of the outcomes of each die. Since each die has 6 numbers, the sample space has a total of 6 x 6 = 36 possible outcomes.

Step 2: Compute the probability of Event E

Event E is the event that the first die shows a 5. There are 6 possible outcomes for the second die, so there are 6 favorable outcomes for event E. Therefore, the probability of E is: \(P(E) = \frac{6}{36} = \frac{1}{6}\)

Step 3: Compute the probability of Event F

Event F is the event that the sum of the dice is 10. To find this, we need to count the combinations of rolls that result in a sum of 10. Let's list them: (4, 6), (5, 5), (6, 4) Three combinations out of 36 possible outcomes give us a sum of 10. Therefore, the probability of F is: \(P(F) = \frac{3}{36} = \frac{1}{12}\)

Step 4: Compute the probability of the intersection of E and F

The intersection of E and F (denoted as \(E \cap F\)) is the event that both E and F occur simultaneously. In this case, we need to find the outcomes where the first die shows a 5 and the sum of the dice is 10. This only happens in one case: (5, 5). So, the probability of \(E \cap F\) is: \(P(E \cap F) = \frac{1}{36}\)

Step 5: Compute the conditional probability P(F | E)

To compute the conditional probability of F given E, we need to find the probability of F occurring when E has already occurred. We use the formula: \(P(F|E) = \frac{P(E \cap F)}{P(E)}\) Plugging in the probabilities we computed earlier: \(P(F|E) = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6}\)

Step 6: Determine if Events E and F are independent

Events E and F are considered independent if the following condition holds true: \(P(E \cap F) = P(E) \times P(F)\) Using the probabilities we computed in the previous steps: \(\frac{1}{36} = \frac{1}{6} \times \frac{1}{12}\) After calculating, we see that this condition holds true as: \(\frac{1}{36} = \frac{1}{36}\) Therefore, events E and F are independent. In summary, a. \(P(F) = \frac{1}{12}\) b. \(P(E \cap F) = \frac{1}{36}\) c. \(P(F|E) = \frac{1}{6}\) d. \(P(E) = \frac{1}{6}\) e. Events E and F are independent.

Key Concepts

Sample Space

A critical aspect of probability theory is understanding the concept of a sample space. The sample space is the set of all possible outcomes of a random experiment. In the exercise, a pair of fair dice are rolled, creating a sample space of their outcomes. Each die has 6 sides, leading to a total sample space of \(6 \times 6 = 36\) outcomes.

• Each outcome can be represented as a pair of numbers, like (1, 1), (1, 2), ..., (6, 6).
• Rolling dice is a classic example of creating a sample space, where every combination represents a unique outcome.
• This comprehensive set allows us to calculate probabilities of specific events, like getting a sum of 10 or a first die showing a 5.

By enumerating all possible results, we have a foundation to delve into more complex ideas, such as the probability of individual or joint events.

Independent Events

In probability, understanding whether events are independent is crucial. Two events are independent if the occurrence of one does not affect the occurrence of the other. Mathematically, events \(E\) and \(F\) are independent if:\[ P(E \cap F) = P(E) \times P(F) \]Let's consider our exercise:

• \(E\) is the event where the first die shows a 5.
• \(F\) is the event where the sum of the dice is 10.

Upon calculation, we find:

• \(P(E) = \frac{1}{6}\)
• \(P(F) = \frac{1}{12}\)
• \(P(E \cap F) = \frac{1}{36}\)

Using these values, when we check for independence, it verifies:\[ \frac{1}{36} = \frac{1}{6} \times \frac{1}{12} \]So, in this instance, events \(E\) and \(F\) are indeed independent. Understanding this relationship helps in calculating compound probabilities more effectively.

Conditional Probability

Conditional probability helps us determine the likelihood of an event occurring, given that another event has already taken place. This is specifically helpful when sample spaces or outcomes are restricted due to the occurrence of a specific event. The formula for conditional probability is:\[ P(F|E) = \frac{P(E \cap F)}{P(E)} \]In our exercise, we are finding the probability of event \(F\) (the sum is 10) given that event \(E\) (the first die is 5) has occurred. From the earlier calculations:

• \(P(E \cap F) = \frac{1}{36}\)
• \(P(E) = \frac{1}{6}\)

Plug these into the formula:\[ P(F|E) = \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{6} \]This calculation shows how likely event \(F\) is under the condition that \(E\) has occurred, refining our understanding of probabilities under certain conditions.

Intersection of Events

The intersection of events in probability refers to the occurrence of two events simultaneously. Notationally, the intersection of events \(E\) and \(F\) is expressed as \(E \cap F\). For intersections, the probability calculates how often both events happen at the same time. In our dice example:

• Event \(E\): First die shows a 5.
• Event \(F\): The sum of dice is 10.

We check which outcomes make both conditions true. Here, only the outcome (5, 5) satisfies both events. Thus, we have:\[ P(E \cap F) = \frac{1}{36} \]Calculating intersections helps determine scenarios where specific conditions coincide. It narrows down possibilities to meet all criteria, assisting in multifaceted probability analysis.