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Chapter 4: Problem 35

Express all probabilities as fractions. How many different ways can you make change for a quarter? (Different arrangements of the same coins are not counted separately.)

Short Answer

Expert verified
There are 5 different ways.

Step by step solution

01

Identify Coin Denominations

First, list all U.S. coins that can be used to make change for a quarter, which is 25 cents. The possible coins are pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents).
02

Divide into Scenarios

Separate the problem into different scenarios based on the highest denomination coin used, starting with quarters and ending with pennies.
03

Scenario 1: Using a Quarter

There's one way to make 25 cents using a single quarter.
04

Scenario 2: Using Dimes

Combinations of dimes and other coins: 1. Two dimes and a nickel (20 + 5 = 25)2. One dime and three nickels (10 + 3脳5 = 25)3. One dime, one nickel, and ten pennies (10 + 5 + 10 = 25)4. One dime and fifteen pennies (10 + 15 = 25)5. Five nickels (5脳5 = 25)6. One nickel and twenty pennies (5 + 20 = 25)7. Twenty-five pennies (25脳1 = 25)
05

Total Distinct Arrangements

Now, count each unique combination from each scenario. Combine all distinct methods to achieve 25 cents.
06

Final Calculation

Sum all the combinations: 1 (quarter) + 5 (dimes) + 1 (five nickels) + 1 (nickel & pennies) + 1 (all pennies) = 5 distinct ways.

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Key Concepts

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

Probability in Coin Combinations
When discussing probabilities in the context of coin combinations, we refer to the likelihood of achieving a specific outcome. Here, our goal is to find the number of different ways to make change for a quarter using various coin denominations. In probability terms, this means calculating the fraction of successful outcomes over all possible outcomes. For coin combinations, each distinct way to form 25 cents can be considered a unique outcome. Therefore, if there are 5 distinct ways to make a quarter, the probability of selecting any one specific combination randomly would be \(\frac{1}{5}\).
Understanding Coin Denominations
Coin denominations are the different values that coins can have. In the problem, we are interested in U.S. coins, specifically pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents). Understanding these values is essential to solving the problem because it helps identify possible combinations to make a quarter. Each combination must sum exactly to 25 cents, without considering the sequence of coins.
Applying Combinatorics
Combinatorics is the branch of mathematics dealing with combinations and arrangements of objects. In this exercise, we use combinatorics to identify all possible ways to combine the given coin denominations to total 25 cents. Key to this approach is recognizing that different arrangements of the same coins do not count separately. For example, one dime and fifteen pennies is the same combination regardless of the order the coins are used in. By systematically considering each possible highest coin denomination (starting with quarters and moving down to pennies), we ensure that no combinations are missed.
Using Fractions to Express Probabilities
In probability, fractions are used to represent the ratios of favorable outcomes to total possible outcomes. For this problem, expressing our probability as a fraction helps to clearly see the relationship between the number of ways to make change for a quarter and the overall problem space. If there are 5 unique combinations to make a quarter, the probability of selecting any specific combination is calculated as \(\frac{1}{5}\). Using fractions in this way provides a clear and concise method to understand probabilistic outcomes.

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Most popular questions from this chapter

Express all probabilities as fractions. In the game of blackjack played with one deck, a player is initially dealt 2 different cards from the 52 different cards in the deck. A winning "blackjack" hand is won by getting 1 of the 4 aces and 1 of 16 other cards worth 10 points. The two cards can be in any order. Find the probability of being dealt a blackjack hand. What approximate percentage of hands are winning blackjack hands?

Find the probability and answer the questions. When Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. Based on those results, estimate the probability of getting an offspring pea that is green. Is the result reasonably close to the expected value of \(3 / 4\), as Mendel claimed?

Answer the given questions that involve odds. When the horse California Chrome won the 140th Kentucky Derby, a \(\$ 2\) bet on a California Chrome win resulted in a winning ticket worth \(\$ 7\). a. How much net profit was made from a \(\$ 2\) win bet on California Chrome? b. What were the payoff odds against a California Chrome win? c. Based on preliminary wagering before the race, bettors collectively believed that California Chrome had a \(0.228\) probability of winning. Assuming that \(0.228\) was the true probability of a California Chrome victory, what were the actual odds against his winning? d. If the payoff odds were the actual odds found in part (c), what would be the worth of a \(\$ 2\) win ticket after the California Chrome win?

Express all probabilities as fractions. The author owns a safe in which he stores all of his great ideas for the next edition of this book. The safe "combination" consists of four numbers between 0 and 99 , and the safe is designed so that numbers can be repeated. If another author breaks in and tries to steal these ideas, what is the probability that he or she will get the correct combination on the first attempt? Assume that the numbers are randomly selected. Given the number of possibilities, does it seem feasible to try opening the safe by making random guesses for the combination?

What does the symbol ! represent? Six different people can stand in a line \(6 !\) different ways, so what is the actual number of ways that six people can stand in a line?
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