Question

A jar contains four coins: a nickel, a dime, a quarter, and a half-dollar. Three coins are randomly selected from the jar. a. What is the probability that the selection will contain the half-dollar? b. What is the probability that the total amount drawn will equal \(60 \phi\) or more?

Short answer

Answer: The probability of selecting the half-dollar coin is \(\frac{3}{4}\), and the probability of selecting a group of coins with a total amount equal to or more than \(60 \phi\) is also \(\frac{3}{4}\).

Step-by-step solution

Calculate the total number of groups of coins

We are selecting three coins out of four. To find the number of possible groups, we'll use the combination formula: \(_{n}C_{k} = \dfrac{n!}{k!(n-k)!}\). In this case, n=4 and k=3. So the number of groups is: \(_{4}C_{3} = \dfrac{4!}{3!(4-3)!} = \dfrac{4!}{3!1!} = 4\). The possible groups are: (nickel, dime, quarter), (nickel, dime, half-dollar), (nickel, quarter, half-dollar), and (dime, quarter, half-dollar).

a. Probability of selecting the half-dollar

We will now find out how many groups have the half-dollar. We can see from the possible groups that there are 3 such combinations: (nickel, dime, half-dollar), (nickel, quarter, half-dollar), and (dime, quarter, half-dollar). P(half-dollar) = (Number of groups containing half-dollar) / (Total number of groups) P(half-dollar) = 3/4 Thus, the probability of selecting the half-dollar is \(\frac{3}{4}\).

b. Probability of total amount equal to or more than \(60 \phi\)

We will first find the total amount for each possible group: 1. (nickel, dime, quarter): 5 + 10 + 25 = 40 phi 2. (nickel, dime, half-dollar): 5 + 10 + 50 = 65 phi 3. (nickel, quarter, half-dollar): 5 + 25 + 50 = 80 phi 4. (dime, quarter, half-dollar): 10 + 25 + 50 = 85 phi We can see that Groups 2, 3, and 4 have a total of \(60 \phi\) or more. P(total amount >= 60 phi) = (Number of suitable groups) / (Total number of groups) P(total amount >= 60 phi) = 3/4 Thus, the probability of selecting a group with a total amount of \(60 \phi\) or more is \(\frac{3}{4}\).

Key concepts

Probability Calculation

Probability calculation is the process of determining the likelihood of a particular outcome happening in a random event. It forms the basis of probability theory and is a fundamental concept in statistics, helping us to make informed predictions about events in various fields like gaming, finance, and science.

In our exercise, we're determining the probability of certain coin combinations being drawn from a jar. The formula used here is \[ P( ext{event}) = \frac{ ext{Number of favorable outcomes}}{ ext{Total number of possible outcomes}} \]This formula allows us to calculate the chance of drawing the half-dollar or achieving a total coin value of at least \(60 \phi\).

The total number of possible outcomes is a key element in this calculation. For example, the exercise involves drawing three coins from a total of four. The total combinations can be calculated using the combination formula which is a concept in combinatorics. With 3 favorable scenarios in our exercise, we find the probability of selecting the half-dollar or reaching at least \(60 \phi\) both at \(\frac{3}{4}\).
This simplicity of calculation is what makes probability an accessible and extremely useful tool for everyday decisions.

Combinatorics

Combinatorics is an area of mathematics concerned with counting, arranging, and finding patterns within sets. It provides the tools needed to solve problems related to the arrangement and combination of objects without deeper focus on their actual order.

In our exercise, combinatorics is used to identify the number of ways to select three coins from a set of four coins. This uses the formula for combinations:\[_{n}C_{k} = \frac{n!}{k!(n-k)!} \]Where \( n \) is the total number of items to choose from, and \( k \) is the number of items to be chosen. "!" symbolizes a factorial, which is the product of an integer and all the integers below it.

Applied to our coin problem, we set \( n=4 \) and \( k=3 \) to calculate \(_{4}C_{3} = 4\). This calculation gives us four possible combinations of coins. These specific combinations make it easy to determine how many include the half-dollar or reach the required \(60 \phi\) value.

• A clear understanding of this principle allows you to solve many similar problems where order doesn’t matter.
• It's a powerful tool in both statistics and real-life application.

Conditional Probability

Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. It’s a key concept in probability theory because it allows us to refine probabilities based on additional information.

In the context of our exercise, we indirectly explore conditional probability when considering the selection groups. If we know a half-dollar is included in a selection, our probability of achieving \(60 \phi\) or more changes accordingly. The conditional probability formula is given as:\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]Where \( P(A|B) \) is the probability of event A occurring given that B has already occurred.

Although not directly calculated in our exercise, understanding this concept helps you see how having additional information (e.g., which coins are drawn) can affect the likelihood of certain outcomes.

• This understanding can apply to many fields like risk assessment, where knowing certain conditions can significantly refine predictions.
• In real life, it’s all about using available information to make the best possible predictions.