https://file.aiquickdraw.com/mathbot/file/172963401581503jh538i.jpg

Let's analyze the problem step by step.

### 1. Total Ice Cream Combinations
- The supermarket offers **6 Ben & Jerry flavors** and **8 Eddy flavors**.
- Your roommate is bringing back **2 cartons**.
- The total number of combinations is calculated using the formula for combinations of two different types of objects:
  
  \[
  \text{Total combinations} = 6 \times 8 = 48
  \]

Thus, 91 seems to have been calculated as the total number of possible combinations (but should be **incorrect**).

---

### 2. Scenario: 1 Ben & Jerry, 1 Eddy Carton
- You are selecting **1 Ben & Jerry carton** from 6 flavors. The number of ways to do that is **6**.
- You are selecting **1 Eddy carton** from 8 flavors. The number of ways to do that is **8**.
- The total number of different bags of ice cream cartons in this case is:

  \[
  6 \times 8 = 48
  \]

Thus, the answer for "how many different bags of ice cream does this yield?" is **48**.

The probability of this selection is calculated by dividing the number of favorable outcomes (48) by the total number of possible combinations (**91**, assuming that's the correct number):

  \[
  \text{Probability} = \frac{48}{91}
  \]

---

### 3. Scenario: All Ben & Jerry Cartons
- You are selecting **2 Ben & Jerry cartons** from the 6 flavors. The number of ways to do this is calculated as:

  \[
  \binom{6}{2} = \frac{6 \times 5}{2} = 15
  \]

Thus, there are **15** different ways to select all Ben & Jerry cartons.

The probability of this type of bag (all Ben & Jerry) is:

  \[
  \text{Probability} = \frac{15}{91}
  \]

---

### 4. Probability of At Least 1 Ben & Jerry Carton
- The probability of getting **at least 1 Ben & Jerry** carton is the complement of the probability of getting **only Eddy** cartons. 
- The probability of getting **only Eddy** cartons is calculated next.

### 5. Probability of Only Eddy Cartons
- You are selecting **2 Eddy cartons** from 8 flavors. The number of ways to do this is:

  \[
  \binom{8}{2} = \frac{8 \times 7}{2} = 28
  \]

Thus, the number of different bags of only Eddy cartons is **28**.

The probability is:

  \[
  \text{Probability} = \frac{28}{91}
  \]

### 6. Probability of At Least 1 Eddy Carton
- The probability of getting **at least 1 Eddy carton** is the complement of getting **only Ben & Jerry** cartons, which is already calculated as:

  \[
  \text{Probability} = 1 - \frac{15}{91} = \frac{76}{91}
  \]

---

Let me know if you would like further clarification or help filling in the probabilities!

---

**Questions:**
1. How do you calculate the total combinations when selecting multiple objects from different groups?
2. What is the formula for calculating combinations (n choose r)?
3. How do we find the probability of mutually exclusive events?
4. Why is the complement rule useful in probability calculations?
5. How can we calculate the probability for more complex events, such as "at least one of an item"?

**Tip**: Always use the complement rule to simplify calculations when finding "at least" probabilities!

Your roommate buys 2 ice cream cartons, and the supermarket offers 6 Ben & Jerry and 8 Eddy flavors. How many total combinations are possible, and what are the probabilities of different combinations?

This problem explores probabilities of selecting ice cream flavors. With 6 Ben & Jerry and 8 Eddy flavors, learn how to calculate combinations and probabilities when picking two cartons. Ideal for high school students studying probability and combinatorics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation