Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}\right\\}\) be the sample space associated with an experiment having the probability distribution shown in the accompanying table. If \(A=\left\\{s_{1}, s_{2}\right\\}\) and \(B=\left\\{s_{1}, s_{5}, s_{6}\right\\}\), find a. \(P(A), P(B)\) b. \(P\left(A^{c}\right), P\left(B^{c}\right)\) c. \(P(A \cap B)\) d. \(P(A \cup B)\) e. \(P\left(A^{c} \cap B^{c}\right)\) f. \(P\left(A^{c} \cup B^{c}\right)\) $$ \begin{array}{cc} \hline \text { Outcome } & \text { Probability } \\ \hline s_{1} & \frac{1}{3} \\ \hline s_{2} & \frac{1}{8} \\ \hline s_{3} & \frac{1}{6} \\ \hline s_{4} & \frac{1}{6} \\ \hline s_{5} & \frac{1}{12} \\ \hline s_{6} & \frac{1}{8} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

a. \(P(A) = \dfrac{5}{8}\) and \(P(B) = \dfrac{7}{12}\) b. \(P\left(A^{c}\right) = \dfrac{3}{8}\) and \(P\left(B^{c}\right) = \dfrac{5}{12}\) c. \(P(A \cap B) = \dfrac{1}{3}\) d. \(P(A \cup B) = \dfrac{11}{12}\) e. \(P\left(A^{c} \cap B^{c}\right) = \dfrac{1}{3}\) f. \(P\left(A^{c} \cup B^{c}\right) = \dfrac{1}{12}\)

Step 1: a. Finding \(P(A)\) and \(P(B)\)

To find the probability of events A and B, we simply need to add the probabilities of their elements. Event A = {s1, s2}: \(P(A) = P(s_{1}) + P(s_{2}) = \dfrac{1}{3} + \dfrac{1}{8} = \dfrac{5}{8}\) Event B = {s1, s5, s6}: \(P(B) = P(s_{1}) + P(s_{5}) + P(s_{6}) = \dfrac{1}{3} + \dfrac{1}{12} + \dfrac{1}{8} = \dfrac{7}{12}\)

Step 2: b. Finding \(P(A^{c})\) and \(P(B^{c})\)

Now we will find the probabilities for the complements of events A and B by subtracting the probability of each event from 1: \(P(A^{c}) = 1 - P(A) = 1 - \dfrac{5}{8} = \dfrac{3}{8}\) \(P(B^{c}) = 1 - P(B) = 1 - \dfrac{7}{12} = \dfrac{5}{12}\)

Step 3: c. Finding \(P(A \cap B)\)

Next, we find the probability of the intersection of events A and B, which is the common outcome(s) between both events: A ∩ B = {s1}: \(P(A \cap B) = P(s_{1}) = \dfrac{1}{3}\)

Step 4: d. Finding \(P(A \cup B)\)

Now, let's find the probability of the union of events A and B: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \dfrac{5}{8} + \dfrac{7}{12} - \dfrac{1}{3} = \dfrac{11}{12}\)

Step 5: e. Finding \(P(A^{c} \cap B^{c})\)

To find the probability of the intersection of the complements of events A and B, we can first identify the elements in both complements: \(A^{c} = \{s_{3}, s_{4}, s_{5}, s_{6}\}\) \(B^{c} = \{s_{2}, s_{3}, s_{4}\}\) \(A^{c} \cap B^{c} = \{s_{3}, s_{4}\}\) \(P(A^{c} \cap B^{c}) = P(s_{3}) + P(s_{4}) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}\)

Step 6: f. Finding \(P(A^{c} \cup B^{c})\)

Finally, let's find the probability of the union of the complements of events A and B: \(P(A^{c} \cup B^{c}) = P(A^{c}) + P(B^{c}) - P(A^{c} \cap B^{c}) = \dfrac{3}{8} + \dfrac{5}{12} - \dfrac{1}{3} = \dfrac{1}{12}\)

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics dealing with counting, arrangement, and combination of objects. In probability, combinatorics helps us determine the number of ways an event can occur, which is crucial for calculating probabilistic outcomes.
When dealing with sets and probability, combinatorics tells us how many subsets or arrangements can be formed. For instance, when you have a sample space, calculating the probabilities of different events often relies on understanding combinations and permutations.
In this specific exercise, combinatorics helps us identify the elements within sets A and B, and their respective intersections and unions. This approach aids in laying the groundwork for subsequent probability calculations, showing the importance of effectively counting the arrangements or combinations involved.

Event Probability

Event probability is the measure of how likely an event is to occur, expressed as a value between 0 (impossible event) and 1 (certain event). To find this, divide the number of favorable outcomes by the total number of possible outcomes in the sample space.
In the exercise, we calculated the event probabilities for A and B by summing the probabilities of individual outcomes in each event. This approach gives us the likelihood of event A or B occurring by adding the associated probabilities:

• Event A: {s1, s2}, with probabilities for each element added together.
• Event B: {s1, s5, s6}, similarly computed by summing the probabilities of these particular outcomes.

Understanding event probability allows you to predict how often an event will occur in repeated trials of an experiment.

Sample Space

The sample space is the complete set of all possible outcomes of an experiment. In probability theory, it's crucial to precisely define this space to effectively analyze any event.
For the exercise, the sample space is defined as \(S=\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}\}\). Each outcome has an associated probability, summing to 1, as it covers all possible results. By considering different subsets of outcomes, we can calculate the probabilities of more complex events, like unions or intersections of sets.
The sample space acts as the foundational structure upon which we build further probability concepts, ensuring that all potential results are considered when calculating probabilities.

Probability Distribution

A probability distribution assigns probabilities to each outcome in the sample space. These probabilities indicate how likely each outcome is to occur within the context of an experiment.
The table from the exercise presents a probability distribution over the sample space \(S\), with each outcome \(s_{i}\) having a specific probability:

• \(s_{1}: \frac{1}{3}\)
• \(s_{2}: \frac{1}{8}\)
• \(s_{3}: \frac{1}{6}\)
• \(s_{4}: \frac{1}{6}\)
• \(s_{5}: \frac{1}{12}\)
• \(s_{6}: \frac{1}{8}\)

It's important to note that the probabilities in the distribution must sum to 1, confirming the completeness of the sample space. Understanding the probability distribution allows us to calculate event probabilities more efficiently, as it provides the likelihood of each basic outcome in a structured format.