Ricky Takes 2 Coins At Random From 3 Quarters, 5 Dimes, And Two Nickels In His Pocket.1) What Is The

Ricky Takes 2 Coins At Random From 3 Quarters, 5 Dimes, And Two Nickels In His Pocket.1) What Is The

Introduction

Imagine Ricky reaching into his pocket, where he has a mixture of coins—specifically, 3 quarters, 5 dimes, and 2 nickels. Without looking, he randomly picks out two coins. This simple act might seem straightforward, but it opens the door to a fascinating exploration of probability, combinations, and coin values.

Understanding the likelihood of different outcomes when selecting coins at random is a common problem in probability theory. It’s a fundamental concept that applies not only in theoretical mathematics but also in real-world scenarios such as games of chance, financial calculations, and decision-making processes.

In this article, we will analyze the problem step-by-step, exploring the possible combinations Ricky might pick, calculating the probability of each, and understanding the implications of these outcomes. This detailed examination will help clarify concepts such as probability distribution, combinations, and expected value, all within the context of coins and everyday transactions.

The Coin Collection: Details and Counts

Before diving into the calculations, it’s essential to understand the composition of Ricky’s coin collection:

• Quarters: 3 coins (each worth 25 cents)


• Dimes: 5 coins (each worth 10 cents)


• Nickels: 2 coins (each worth 5 cents)

Total coins in Ricky’s pocket:
3 (quarters) + 5 (dimes) + 2 (nickels) = 10 coins

Ricky randomly selects 2 coins from these 10 without replacement, meaning once a coin is chosen, it’s not put back into the pocket before the second pick.

The goal is to determine:

• The total number of possible outcomes when selecting 2 coins at random.


• The probability of each type of outcome (e.g., both coins being quarters, one quarter and one dime, etc.).


• The expected value of the sum of the two coins selected.


• The possible scenarios and their likelihoods.

Total Number of Ways to Pick 2 Coins

When choosing 2 coins at random from 10 without replacement, the total number of combinations is calculated using the combination formula:

\[ C(n, k) = \frac{n!}{k!(n - k)!} \]

where:

• \( n \) is the total number of items (coins, in this case),


• \( k \) is the number of items to choose.

Applying this:

\[ C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]

So, there are 45 possible outcomes when selecting any 2 coins at random from Ricky’s collection.

Possible Types of Coin Pairs

Next, we categorize the possible pairs based on the types of coins selected:

• Both coins are quarters


• Both coins are dimes


• Both coins are nickels


• One quarter and one dime


• One quarter and one nickel


• One dime and one nickel

These categories are mutually exclusive and cover all possible combinations.

Calculating the Number of Favorable Outcomes for Each Category

To find the probability of each category, we need to determine the number of favorable outcomes for each case (i.e., the number of ways to pick such pairs).

1. Both Quarters

Number of ways to choose 2 quarters from 3:

\[ C(3, 2) = \frac{3!}{2!(3-2)!} = 3 \]

1. Both Dimes

Number of ways to choose 2 dimes from 5:

\[ C(5, 2) = \frac{5!}{2!(5-2)!} = 10 \]

1. Both Nickels

Number of ways to choose 2 nickels from 2:

\[ C(2, 2) = 1 \]

1. One Quarter and One Dime

Number of ways to choose 1 quarter from 3 and 1 dime from 5:

\[ C(3, 1) \times C(5, 1) = 3 \times 5 = 15 \]

1. One Quarter and One Nickel

Number of ways to choose 1 quarter from 3 and 1 nickel from 2:

\[ C(3, 1) \times C(2, 1) = 3 \times 2 = 6 \]

1. One Dime and One Nickel

Number of ways to choose 1 dime from 5 and 1 nickel from 2:

\[ C(5, 1) \times C(2, 1) = 5 \times 2 = 10 \]

Summarizing the Outcomes

| Category | Number of Ways | Total Possible Outcomes |
|------------|----------------|-------------------------|
| Both Quarters | 3 | 3 |
| Both Dimes | 10 | 10 |
| Both Nickels | 1 | 1 |
| Quarter & Dime | 15 | 15 |
| Quarter & Nickel | 6 | 6 |
| Dime & Nickel | 10 | 10 |
| Total | — | 45 |

This table confirms that the sum of all favorable outcomes equals the total possible combinations:

\[ 3 + 10 + 1 + 15 + 6 + 10 = 45 \]

which matches the total number of combinations.

Calculating Probabilities for Each Outcome

The probability of each category is the number of favorable outcomes divided by the total number of outcomes (45):

1. Both Quarters: \( \frac{3}{45} = \frac{1}{15} \approx 6.67\% \)


2. Both Dimes: \( \frac{10}{45} = \frac{2}{9} \approx 22.22\% \)


3. Both Nickels: \( \frac{1}{45} \approx 2.22\% \)


4. Quarter & Dime: \( \frac{15}{45} = \frac{1}{3} \approx 33.33\% \)


5. Quarter & Nickel: \( \frac{6}{45} = \frac{2}{15} \approx 13.33\% \)


6. Dime & Nickel: \( \frac{10}{45} = \frac{2}{9} \approx 22.22\% \)

These probabilities help us understand the likelihood of each type of coin pair being selected at random.

Calculating the Expected Value of the Sum of Selected Coins

Beyond just counting combinations, a common question is: What is the expected monetary value of the two coins Ricky picks?

Step 1: Assign values to each coin type

• Quarters: 25 cents


• Dimes: 10 cents


• Nickels: 5 cents

Step 2: Compute the sum for each combination

For each category, calculate the sum of the coin values:

| Category | Number of Ways | Sample Pair | Sum of Values (in cents) |
|------------|----------------|--------------|--------------------------|
| Both Quarters | 3 | (Q, Q) | 25 + 25 = 50 |
| Both Dimes | 10 | (D, D) | 10 + 10 = 20 |
| Both Nickels | 1 | (N, N) | 5 + 5 = 10 |
| Quarter & Dime | 15 | (Q, D) | 25 + 10 = 35 |
| Quarter & Nickel | 6 | (Q, N) | 25 + 5 = 30 |
| Dime & Nickel | 10 | (D, N) | 10 + 5 = 15 |

Step 3: Calculate the expected value

The expected value (EV) is computed as:

\[
EV = \sum (\text{probability of category} \times \text{average sum in that category})
\]

Using the probabilities and sums:

\[
EV = \left(\frac{3}{45} \times 50\right) + \left(\frac{10}{45} \times 20\right) + \left(\frac{1}{45} \times 10\right) + \left(\frac{15}{45} \times 35\right) + \left(\frac{6}{45} \times 30\right) + \left(\frac{10}{45} \times 15\right)
\]

Calculating each term:

• \( \frac{3}{45} \times 50 = \frac{1}{15} \times 50 \approx 3.33 \)


• \( \frac{10}{45} \times 20 = \frac{2}{9} \times 20 \approx 4.44 \)


• \( \frac{1}{45} \times 10 \approx 0.22 \)


• \( \frac{15}{45} \times 35 = \frac{1}{3} \times 35 \approx 11.67 \)


• \( \frac{6}{45} \times 30 = \frac{2}{15} \times 30 = 4 \)


• \( \frac{