/*! 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 133 In how many ways can a sample (w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In how many ways can a sample (without replacement) of 5 items be selected from a population of 15 items?

Short Answer

Expert verified
The number of ways a sample of 5 items can be selected from a population of 15 items is \( C(15, 5) = 3,003 \).

Step by step solution

01

Understand the Problem

We are given a population of 15 items, and we need to select a sample of 5 items without replacement (meaning each item can only be selected once). The important thing to remember here is that the order of chosen items is not important, meaning it's a combination, not permutation.
02

Use Combination Formula

In mathematics, the combination formula is used when we need to determine the number of ways to choose r objects from a set of n objects, which is represented as \( C(n, r) \). Here, the 'n' is the total number of objects (in this case 15 items), and 'r' is the number of objects we select (in this case 5 items). The combination formula is \( C(n, r) = \frac{n!}{r!(n-r)!} \), where '!' denotes factorial, which is the product of all positive integers up to that number.
03

Substitute the Values

By substituting n=15, r=5 into the combination formula, we want to find the value of \( C(15, 5)= \frac{15!}{5!(15-5)!} \).
04

Simplify the Expression

To calculate this expression, first calculate the values of 15!, 5! and 10!(15-5), and then perform the division. This will give us our final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Combination formula
The combination formula is a tool used in combinatorics to find out how many ways you can choose a subset of items from a larger set. When we're talking about combinations, remember that the order in which you choose the items doesn't matter.
  • For example, choosing a team of 3 players out of 5 is the same if you pick A, B, C or C, A, B.
To use the combination formula, which is expressed as \( C(n, r) = \frac{n!}{r!(n-r)!} \), where:
  • \(n\) is the total number of items in your set.
  • \(r\) is the number of items you want to select.
  • \(!\) denotes factorial, the product of all positive integers up to that number.
This formula helps us calculate the possible ways to choose \(r\) items from a set of \(n\) items without caring about the order of selection.
Factorial
A factorial, denoted by an exclamation mark \(!\), represents the product of all positive integers up to a given number. Factorials are a key concept in permutations and combinations, as they help find the total number of ways to arrange a set of items.For example, to calculate 5! (5 factorial), you multiply all whole numbers from 1 to 5:\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]The factorial function grows very quickly. For instance, 0! is defined to be 1 because there's exactly one way to arrange zero items: do nothing!Understanding how to compute factorials is essential when using the combination and permutation formulas, as they are embedded in the mathematical equations for calculating choice and arrangement possibilities.
Sample without replacement
"Sample without replacement" is a concept where each item in a population is selected once and cannot be chosen again. This method affects how combinations and probabilities are calculated. In our example, we take a sample of 5 items from a group of 15. Once an item is selected, it can't be chosen again. This fact highlights why combinations, rather than permutations, are used.
  • In permutations, the order matters, and you can select the same item multiple times.
  • Here, once an item is picked, it can't be included in any further selections, reducing the total number of choices with each pick.
This approach ensures that our calculation remains accurate to real-world scenarios where repetition is not allowed.
Permutation vs Combination
Understanding the difference between permutation and combination is crucial in combinatorics.
  • Permutations are arrangements where the order of items matters. For example, the sequences ABC and BAC are different permutations of the letters A, B, and C.
  • Combinations, on the other hand, focus on selection, where the order does not matter. For instance, choosing teams from a pool of candidates is a classic combination problem because it doesn't matter which order team members are picked.
In combination, the goal is to identify how many ways you can select a group of items, not how you can arrange them. In permutation, it's all about how you can organize a set of items once they're chosen. Thus, when the order doesn't matter, combinations are the go-to method of calculation. When order is essential, like in arranging books on a shelf, permutations are needed. Distinguishing between these scenarios is vital for solving combinatorial problems accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following addition rule to find the probability of the union of two events \(A\) and \(B\) : $$ P(A \text { or } B)=P(A)+P(B)-P(A \text { and } B) $$ When and why is the term \(P(A\) and \(B\) ) subtracted from the sum of \(P(A)\) and \(P(B)\) ? Give one example where you might use this formula.

The probability that a randomly selected college student attended at least one major league baseball game last year is .12. What is the complementary event? What is the probability of this complementary event?

A random sample of 250 juniors majoring in psychology or communication at a large university is selected. These students are asked whether or not they are happy with their majors. The following table gives the results of the survey. Assume that none of these 250 students is majoring in both areas. $$ \begin{array}{lcc} \hline & \text { Happy } & \text { Unhappy } \\ \hline \text { Psychology } & 80 & 20 \\ \text { Communication } & 115 & 35 \\ \hline \end{array} $$ a. If one student is selected at random from this group, find the probability that this student is i. happy with the choice of major ii. a psychology major iii. a communication major given that the student is happy with the choice of major iv. unhappy with the choice of major given that the student is a psychology major v. a psychology major and is happy with that major vi. a communication major \(o r\) is unhappy with his or her major b. Are the events "psychology major" and "happy with major" independent? Are they mutually exclusive? Explain why or why not.

The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2004 and 2008 , based on gender and whether or not they graduated. $$ \begin{array}{lcc} \hline & \text { Graduated } & \text { Did Not Graduate } \\ \hline \text { Male } & 126 & 55 \\ \text { Female } & 133 & 32 \\ \hline \end{array} $$ If one of these players is selected at random, find the following probabilities. a. \(P\) (female or did not graduate) b. \(P(\) graduated or male \()\)

There are 142 people participating in a local \(5 \mathrm{~K}\) road race. Sixty-five of these runners are female. Of the female runners, 19 are participating in their first \(5 \mathrm{~K}\) road race. Of the male runners, 28 are participating in their first \(5 \mathrm{~K}\) road race. Are the events female and participating in their first \(5 \mathrm{~K}\) road race independent? Are they mutually exclusive? Explain why or why not.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.