/*! 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 60 Solve by the method of your choi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve by the method of your choice. Fifty people purchase raffle tickets. Three winning tickets are selected at random. If first prize is \(\$ 1000,\) second prize is \(\$ 500\) and third prize is \(\$ 100,\) in how many different ways can the prizes be awarded?

Short Answer

Expert verified
The number of different ways the prizes can be awarded is \(C(50, 3) = 19600\).

Step by step solution

01

Identify the number of possible winners

In this case, there are 50 people who have purchased the raffle tickets and are therefore eligible to win. So, \(n\) is 50.
02

Identify the number of prizes to be awarded

There are three prizes to be awarded. Therefore, \(r\) is 3.
03

Apply the combination formula

The number of different ways the prizes can be awarded is calculated by the combination formula. Plug the values of \(n\) and \(r\) into the formula: \[C(n, r) = \frac{n!}{r!(n-r)!} = \frac{50!}{3!(50-3)!}\]

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.

Permutations and Combinations
Understanding permutations and combinations is essential for solving many probability problems, including raffle draws. Permutations refer to arrangements of elements where the order is important. In contrast, combinations are about selecting elements where the order does not matter.

Let's take our raffle example. If we were to determine the sequence in which the three winners are drawn and declared, we would be discussing permutations because the order in which the tickets are drawn matters. However, since we just want to know how many different sets of three winners can be selected, regardless of the order they were chosen in, we are dealing with combinations.

Using combinations simplifies the calculations because it reduces the number of possible outcomes. Imagine placing the names of the winners on a bulletin board; it does not matter in which sequence they are placed, just that they are the winners. This is the core idea behind using combinations in such raffle ticket scenarios.
Probability in Raffles
The probability of certain outcomes in raffles can be intriguing. Specifically, the probability depends on the number of ways the prizes can be distributed among the participants. In our exercise, where fifty people have bought tickets and there are three prizes, we want to know the various outcomes of how the prizes could be awarded.

In probability terms, we look at the 'sample space'—all possible outcomes. Then, to find the probability of a particular event (like winning the first prize), we count how many outcomes correspond to that event and divide by the total number of outcomes. However, in this exercise, we are not calculating the probability for a specific person to win but rather the total number of different ways the prizes can be distributed, which is an initial step in assessing various probability scenarios in raffles.
Factorial Notation
Factorial notation plays a crucial role in mathematics, especially in calculating permutations and combinations. A factorial is symbolized by an exclamation point (!) and represents the product of all positive integers from 1 to the number itself. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

In the combination formula, factorials are used to account for all the various permutations and then divided out to correct for the fact that order does not matter in combinations. This is why we divide by the factorial of the number of items selected (\(r!\)) and the factorial of the difference between the total number and the number selected (\((n-r)!\)).

This ensures we only count each unique selection once, regardless of order, which is exactly what we need for solving our raffle ticket exercise.

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

Company A pays \(\$ 23,000\) yearly with raises of \(\$ 1200\) per year. Company B pays \(\$ 26,000\) yearly with raises of \(\$ 800\) per year. Which company will pay more in year \(10 ?\) How much more?

Use the formula for \(_{n} P_{r}\) to solve Exercises \(41-48\) How many arrangements can be made using four of the letters of the word COMBINE if no letter is to be used more than once?

Find the term in the expansion of \(\left(x^{2}+y^{2}\right)^{5}\) containing \(x^{4}\) as a factor.

The bar graphs show changes in educational attainment for Americans ages 25 and older from 1970 to 2007 . Exercises \(61-62\) involve developing arithmetic sequences that model the data. (GRAPH CANT COPY) In \(1970,55.2 \%\) of Americans ages 25 and older had completed four years of high school or more. On average, this percentage has increased by approximately 0.86 each year. a. Write a formula for the \(n\) th term of the arithmetic sequence that models the percentage of Americans ages 25 and older who had or will have completed four years of high school or more \(n\) years after 1969 . b. Use the model from part (a) to project the percentage of Americans ages 25 and older who will have completed four years of high school or more by 2019 .

Follow the outline below and use mathematical induction to prove the Binomial Theorem: $$\begin{aligned}(a+b)^{n} &-\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\&+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned}$$ a. Verify the formula for \(n-1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$\begin{array}{l}(a+b)^{k+1}-\left(\begin{array}{c}k \\\0\end{array}\right)a^{k+1}+\left[\left(\begin{array}{c}k \\\0\end{array}\right)+\left(\begin{array}{c}k \\\1\end{array}\right)\right] a^{k} b \\\\+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1}\end{array}$$ e. Use the result of Exercise 84 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\ r+1\end{array}\right)\) $$\begin{aligned}&-\left(\begin{array}{l}n+1 \\\r+1\end{array}\right), \text { then }\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)-\left(\begin{array}{c}k+1 \\\1\end{array}\right) \text { and }\\\&\left(\begin{array}{l}k \\\1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)-\left(\begin{array}{c}k+1 \\\2\end{array}\right)\end{aligned}$$ f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)-\left(\begin{array}{c}k+1 \\ 0\end{array}\right)(\text { why? })\) and \(\left(\begin{array}{l}k \\\ k\end{array}\right)-\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

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.