Problem 45 - 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 following probability distribution: $$ \begin{array}{lcccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} & s_{6} \\ \hline \text { Probability } & \frac{1}{12} & \frac{1}{4} & \frac{1}{12} & \frac{1}{6} & \frac{1}{3} & \frac{1}{12} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{3}\right\\}\) b. \(B=\left\\{s_{2}, s_{4}, s_{5}, s_{6}\right\\}\) c. \(C=S\)

Step-by-Step Solution

Answer

The probabilities of events A, B, and C are: a. \(P(A) = \frac{1}{6}\) b. \(P(B) = \frac{5}{6}\) c. \(P(C) = 1\)

Step 1: Identify outcomes in event A

Event A includes outcomes \(s_1\) and \(s_3\).

Step 2: Add probabilities of the respective outcomes

The probability of event A = probability of \(s_1\) + probability of \(s_3\). \[ P(A) = \frac{1}{12} + \frac{1}{12} \]

Step 3: Simplify the expression

Sum the fractions to find the probability of event A. \[ P(A) = \frac{2}{12} = \frac{1}{6} \] #b. Find the probability of event B#

Step 1: Identify outcomes in event B

Event B includes outcomes \(s_2\), \(s_4\), \(s_5\), and \(s_6\).

Step 2: Add probabilities of the respective outcomes

The probability of event B = probability of \(s_2\) + probability of \(s_4\) + probability of \(s_5\) + probability of \(s_6\). \[ P(B) = \frac{1}{4} + \frac{1}{6} + \frac{1}{3} + \frac{1}{12} \]

Step 3: Simplify the expression

Sum the fractions to find the probability of event B. \[ P(B) = \frac{3}{12} + \frac{2}{12} + \frac{4}{12} + \frac{1}{12} = \frac{10}{12} = \frac{5}{6} \] #c. Find the probability of event C#

Step 1: Identify outcomes in event C

Event C includes all outcomes in the sample space S, which are \(s_1\), \(s_2\), \(s_3\), \(s_4\), \(s_5\), and \(s_6\).

Step 2: Add probabilities of the respective outcomes

The probability of event C = probability of \(s_1\) + probability of \(s_2\) + probability of \(s_3\) + probability of \(s_4\) + probability of \(s_5\) + probability of \(s_6\). \[ P(C) = \frac{1}{12} + \frac{1}{4} + \frac{1}{12} + \frac{1}{6} + \frac{1}{3} + \frac{1}{12} \]

Step 3: Simplify the expression

As event C includes the entire sample space S, its probability is always 1. \[ P(C) = 1 \] The probabilities of events A, B, and C are: a. \(P(A) = \frac{1}{6}\) b. \(P(B) = \frac{5}{6}\) c. \(P(C) = 1\)

Key Concepts

Sample Space

In probability, a sample space is a comprehensive list of all possible outcomes of a particular experiment. Think of it as the universe of possibilities that could happen when you conduct your experiment. If you imagine rolling a six-sided die, the sample space is the set of all numbers that could appear on the face: \(S = \{1, 2, 3, 4, 5, 6\}\). An important aspect of a sample space is its completeness. It needs to encompass every conceivable outcome.For example, in the exercise provided, the sample space is \(S = \{s_1, s_2, s_3, s_4, s_5, s_6\}\). This indicates there are six distinct outcomes, each labeled uniquely. This uniqueness allows us to easily identify individual outcomes and group them into events. Determining the sample space correctly is crucial for calculating probabilities since it forms the foundation of all further probability analysis.

Probability Distribution

A probability distribution assigns a probability to each outcome in a sample space. It tells us how likely each outcome is to occur. The probabilities in a probability distribution must sum up to 1, as this covers the certainty of one of the outcomes happening.In our exercise, the distribution is displayed in a table with outcomes and their associated probabilities:

• ext{Outcome } s_1: \( \frac{1}{12} \)
• ext{Outcome } s_2: \( \frac{1}{4} \)
• ext{Outcome } s_3: \( \frac{1}{12} \)
• ext{Outcome } s_4: \( \frac{1}{6} \)
• ext{Outcome } s_5: \( \frac{1}{3} \)
• ext{Outcome } s_6: \( \frac{1}{12} \)

The sum of all these probabilities is:\[ \frac{1}{12} + \frac{1}{4} + \frac{1}{12} + \frac{1}{6} + \frac{1}{3} + \frac{1}{12} = 1 \]Confirming that it correctly represents a full probability distribution. When working with probability distributions, it’s crucial that all probabilities are non-negative and their total equals 1 to maintain the validity of the distribution.

Events in Probability

In probability theory, an event is a specific outcome or a set of outcomes from a sample space. Events can consist of a single outcome or multiple outcomes, and the probability of an event is obtained by summing the probabilities of the individual outcomes it comprises.For instance, let's consider the events in the exercise:**Event A** comprises outcomes \(s_1\) and \(s_3\). Thus, the probability of event A is computed as:\[ P(A) = P(s_1) + P(s_3) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6} \]**Event B** consists of outcomes \(s_2, s_4, s_5, \text{and } s_6\). We find the probability of event B by summing the probabilities of these outcomes:\[ P(B) = \frac{1}{4} + \frac{1}{6} + \frac{1}{3} + \frac{1}{12} = \frac{5}{6} \]**Event C** is perhaps the simplest, as it encompasses all possible outcomes, meaning \(C = S\) which results in a total probability of 1:\[ P(C) = 1 \]Understanding events in probability enables us to evaluate how likely specific occurrences are, based on the outcomes they consist of in the context of the complete sample space.