/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 125 A student is to select three cla... [FREE SOLUTION] | 91Ó°ÊÓ

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A student is to select three classes for next semester. If this student decides to randomly select one course from each of eight economics classes, six mathematics classes, and five computer classes, how many different outcomes are possible?

Short Answer

Expert verified
The total number of possible outcomes is 240.

Step by step solution

01

Economics Class Selection

The student has to select one class out of eight economics classes. Therefore, there are 8 ways to make this choice.
02

Mathematics Class Selection

Similarly, the student has to select one class out of six mathematics classes. Hence, there are 6 ways to make this choice.
03

Computer Science Class Selection

Lastly, the student has to select one class from five computer science classes. Consequently, there are 5 ways for this selection.
04

Total Possible Outcomes

As per the counting principle, the total number of outcomes can be found by multiplying the number of choices for each class. Thus, total possibilities = number of economics classes x number of mathematics classes x number of computer classes = \(8 \times 6 \times 5 = 240\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Counting Principle
The counting principle is fundamental in combinatorics for determining the total number of possible outcomes. It states that if you have multiple stages or choices and each stage or choice is independent of others, you multiply the number of ways each stage can occur.
This principle simplifies complex counting problems and is foundational for problems involving sequences of decisions.
For example, when selecting one class from each of the economics, mathematics, and computer classes, you calculate the total outcomes as follows:
  • There are 8 choices for the economics class.
  • There are 6 choices for the mathematics class.
  • There are 5 choices for the computer class.
By multiplying these choices (8 x 6 x 5), you find the total number of possible class combinations, equaling 240.
This method enhances the ability to predict possibilities in various decision-making scenarios by simplifying how outcomes multiply across stages.
Exploring Permutations
Permutations are about arranging elements in a specific order or sequence. Unlike combinations, permutations consider the order of arrangement as important. If you have a set of items and want to know how many different ways you can arrange them, you'll use permutations.
For example, if a student had to take three different subjects but had no restriction on which subjects came first across time slots, the permutations of their schedule would matter. Mathematically, if you have 'n' items and you want to arrange 'r' of them, you use the formula for permutations:
\( P(n,r) = \frac{n!}{(n-r)!} \)
This formula captures all possible sequences of 'r' elements from a total of 'n' elements. It's helpful when the arrangement sequence is strictly crucial, like in seating arrangements, race placements, or scheduling.
Understanding Combinations
Combinations differ from permutations because the sequence or order does not matter. In various scenarios, you may only need to know how many ways you can select items from a larger set without considering different orders or sequences.
Imagine you're choosing which classes to attend but don't care about the order you select them in. In this case, you're dealing with combinations. The formula for finding the number of combinations of selecting 'r' items from a total of 'n' is:
\( C(n,r) = \frac{n!}{r!(n-r)!} \)
Here you're dividing out the number of ways to arrange 'r' items since order doesn't matter. Combinations are particularly useful in scenarios like lottery tickets, committee selections, or grouping items where arrangement is irrelevant.
Calculating Probability
Probability is about determining how likely an event is to occur. It's a key part of combinatorics and is often expressed as a fraction or percentage.
To calculate probability, you determine the number of favorable outcomes divided by the total number of possible outcomes. It's a measure of how likely an event is given all the possible outcomes of a scenario.
For example, if you wanted to calculate the probability of a student randomly selecting a specific set of classes, you'd set up the ratio of one favorable outcome to the total combinations, like the one solved in this exercise where there were 240 total class combinations.
  • Probability formula: \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
This approach allows you to rigorously assess and apply likelihoods in various practical applications like games of chance, risk assessments, or making informed decisions based on uncertainty.

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Most popular questions from this chapter

The probability that a student graduating from Suburban State University has student loans to pay off after graduation is .60. The probability that a student graduating from this university has student loans to pay off after graduation and is a male is \(.24\). Find the conditional probability that a randomly selected student from this university is a male given that this student has student loans to pay off after graduation.

Of the 35 students in a class, 22 are taking the class because it is a major requirement, and the other 13 are taking it as an elective. If two students are selected at random from this class, what is the probability that the first student is taking the class as an elective and the second is taking it because it is a major requirement? How does this probability compare to the probability that the first student is taking the class because it is a major requirement and the second is taking it as an elective?

Explain the meaning of the intersection of two events. Give one example.

Refer to Exercise 4.48. A 2010-2011 poll conducted by Gallup (www.gallup.com/poll/148994/ Emotional-Health-Higher-Among-Older- Americans.aspx) examined the emotional health of a large number of Americans. Among other things, Gallup reported on whether people had Emotional Health Index scores of 90 or higher, which would classify them as being emotionally well-off. The report was based on a survey of 65,528 people in the age group \(35-44\) years and 91,802 people in the age group \(65-74\) years. The following table gives the results of the survey, converting percentages to frequencies. $$ \begin{array}{lcc} \hline & \text { Emotionally Well-Off } & \text { Emotionally Not Well-Off } \\\ \hline \text { 35-44 Age group } & 16,016 & 49,512 \\ \text { 65-74 Age group } & 32,583 & 59,219 \\ \hline \end{array} $$ a. Suppose that one person is selected at random from this sample of 157,330 Americans. Find the following probabilities. i. \(P(35-44\) age group and emotionally not well-off \()\) ii. \(P(\) emotionally well-off and \(65-74\) age group \()\) b. Find the joint probability of the events \(35-44\) age group and \(65-74\) age group. Is this probability zero? Explain why or why not.

Given that \(P(A \mid B)=.44\) and \(P(A\) and \(B)=.33\), find \(P(B)\).

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