Problem 46 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$ \begin{array}{lccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

Step-by-Step Solution

Answer

a. \(P(A) = \frac{6}{14}\) b. \(P(B) = \frac{3}{14}\) c. \(P(C) = 1\)

Step 1: Recall the rule for finding the probability of an event

To find the probability of an event, sum the probabilities of the individual outcomes that make up the event.

Step 2: Find the probability of event A

Event \(A\) consists of the outcomes \(s_{1}, s_{2},\) and \(s_{4}\). To find the probability of A, add the probabilities of these outcomes: \(P(A) = P(s_1) + P(s_2) + P(s_4) = \frac{1}{14} + \frac{3}{14} + \frac{2}{14}\)

Step 3: Calculate P(A)

Now, add the fractions together: \(P(A) = \frac{1 + 3 + 2}{14} = \frac{6}{14}\)

Step 4: Find the probability of event B

Event \(B\) consists of the outcomes \(s_{1}\) and \(s_{5}\). To find the probability of B, add the probabilities of these outcomes: \(P(B) = P(s_1) + P(s_{5}) = \frac{1}{14} + \frac{2}{14}\)

Step 5: Calculate P(B)

Now, add the fractions together: \(P(B) = \frac{1 + 2}{14} = \frac{3}{14}\)

Step 6: Find the probability of event C

Event \(C\) consists of the entire sample space \(S\), which includes all possible outcomes \(s_{1}, s_{2}, s_{3}, s_{4},\) and \(s_{5}\). To find the probability of C, add the probabilities of all outcomes: \(P(C) = P(s_{1}) + P(s_{2}) + P(s_{3}) + P(s_{4}) + P(s_{5})\)

Step 7: Calculate P(C)

Now, add the fractions together: \(P(C) = \frac{1}{14} + \frac{3}{14} + \frac{6}{14} + \frac{2}{14} + \frac{2}{14} = \frac{1+3+6+2+2}{14} = \frac{14}{14}\)

Step 8: Present the final probabilities

We found the probabilities of events \(A\), \(B\), and \(C\): a. \(P(A) = \frac{6}{14}\) b. \(P(B) = \frac{3}{14}\) c. \(P(C) = \frac{14}{14} = 1\)

Key Concepts

Sample Space

The concept of a sample space is fundamental to probability. We define a sample space as the set of all possible outcomes of an experiment. For any random experiment, knowing all potential outcomes is vital to understanding probability distribution.

In our problem, the sample space is denoted as \(S=\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\}\). Each element \(s_i\) in the set \(S\) represents a possible outcome of the experiment. The sample space encompasses all outcomes we can expect to observe.

• A complete sample space means no outcomes are left out.
• It provides a comprehensive list of all viable scenarios.
• Understanding the sample space helps to compute probabilities effectively.

Knowing the sample space allows us to assign and understand the probability of each event in a structured and organized manner.

Probability of an Event

Once the sample space is established, we can focus on finding the probability of an event. An event is a subset of the sample space, consisting of one or more outcomes. The probability of an event is calculated by summing the probabilities of the individual outcomes that comprise the event.

For example, event \(A=\{s_{1}, s_{2}, s_{4}\}\) includes three outcomes from our sample space. The probability of this event, \(P(A)\), is found by adding the probabilities of \(s_1\), \(s_2\), and \(s_4\). Thus,

\[P(A) = P(s_1) + P(s_2) + P(s_4) = \frac{1}{14} + \frac{3}{14} + \frac{2}{14} = \frac{6}{14}\]
Event \(B\) involves fewer outcomes, \(s_1\) and \(s_5\), leading to:

\[P(B) = P(s_1) + P(s_5) = \frac{1}{14} + \frac{2}{14} = \frac{3}{14}\]

• The process requires identifying all included outcomes within an event.
• Correctly sum all associated probabilities.

Understanding this approach is essential to mastering probability calculations.

Summation of Probabilities

The summation of probabilities is a key principle in probability theory. It states that the sum of probabilities for all outcomes in a sample space must equal 1. This principle ensures that all possible scenarios have been accounted for without any missing or overlapping probabilities.

In our current exercise, the event \(C = S\) represents the entire sample space. Therefore, calculating the probability of \(C\) involves summing the probabilities of all individual outcomes:

\[P(C) = P(s_{1}) + P(s_{2}) + P(s_{3}) + P(s_{4}) + P(s_{5}) = \frac{14}{14} = 1\]

• This confirms that all probabilities in the distribution are properly allocated across the sample space.
• Ensures that no probability is missing or duplicated.

Adhering to this rule validates the integrity of the probability distribution and guarantees that the experiment's total probability is sound, making this a fundamental aspect of probability calculations.