Answer :
To solve this problem, let's break it down into several logical steps.
### Step 1: Determine Total Possible Combinations
First, we need to calculate the total number of possible combinations that can be created for the boxed lunches.
1. Bread Options:
- White Bread
- Wheat Bread
This gives us [tex]\(2\)[/tex] options.
2. Meat Options:
- Roast Beef
- Bologna
This also gives us [tex]\(2\)[/tex] options.
3. Snack Options:
- Chips
- Popcorn
- Pretzels
This gives us [tex]\(3\)[/tex] options.
4. Drink Options:
- Bottled Water
- Juice
This gives us [tex]\(2\)[/tex] options.
To find the total number of combinations, multiply the number of options for each category together:
[tex]\[ \text{Total Combinations} = 2 \text{ (Bread)} \times 2 \text{ (Meat)} \times 3 \text{ (Snack)} \times 2 \text{ (Drink)} = 24 \][/tex]
### Step 2: Determine Favorable Combinations
Next, we need to calculate the number of favorable combinations that meet the condition of having a roast beef sandwich and popcorn.
1. Condition for Sandwich (Roast Beef):
- White Bread with Roast Beef
- Wheat Bread with Roast Beef
This provides [tex]\(2\)[/tex] favorable options for the sandwich.
2. Condition for Snack (Popcorn):
- Only [tex]\(1\)[/tex] option for popcorn as a snack.
3. Drink Options:
- Bottled Water
- Juice
The drinks can be either, so there are [tex]\(2\)[/tex] options.
Therefore, for roast beef and popcorn, the number of favorable combinations is:
[tex]\[ \text{Favorable Combinations} = 2 \text{ (Sandwich)} \times 1 \text{ (Snack)} \times 2 \text{ (Drink)} = 4 \][/tex]
### Step 3: Calculate the Probability
Finally, we calculate the probability that Gretchen will randomly choose a boxed lunch with a roast beef sandwich and popcorn. Probability is given by:
[tex]\[ \text{Probability} = \frac{\text{Number of Favorable Combinations}}{\text{Total Number of Combinations}} = \frac{4}{24} = \frac{1}{6} \][/tex]
Therefore, the probability that Gretchen will get a boxed lunch with a roast beef sandwich and popcorn is [tex]\(\frac{1}{6}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{\frac{1}{6}} \][/tex]
### Step 1: Determine Total Possible Combinations
First, we need to calculate the total number of possible combinations that can be created for the boxed lunches.
1. Bread Options:
- White Bread
- Wheat Bread
This gives us [tex]\(2\)[/tex] options.
2. Meat Options:
- Roast Beef
- Bologna
This also gives us [tex]\(2\)[/tex] options.
3. Snack Options:
- Chips
- Popcorn
- Pretzels
This gives us [tex]\(3\)[/tex] options.
4. Drink Options:
- Bottled Water
- Juice
This gives us [tex]\(2\)[/tex] options.
To find the total number of combinations, multiply the number of options for each category together:
[tex]\[ \text{Total Combinations} = 2 \text{ (Bread)} \times 2 \text{ (Meat)} \times 3 \text{ (Snack)} \times 2 \text{ (Drink)} = 24 \][/tex]
### Step 2: Determine Favorable Combinations
Next, we need to calculate the number of favorable combinations that meet the condition of having a roast beef sandwich and popcorn.
1. Condition for Sandwich (Roast Beef):
- White Bread with Roast Beef
- Wheat Bread with Roast Beef
This provides [tex]\(2\)[/tex] favorable options for the sandwich.
2. Condition for Snack (Popcorn):
- Only [tex]\(1\)[/tex] option for popcorn as a snack.
3. Drink Options:
- Bottled Water
- Juice
The drinks can be either, so there are [tex]\(2\)[/tex] options.
Therefore, for roast beef and popcorn, the number of favorable combinations is:
[tex]\[ \text{Favorable Combinations} = 2 \text{ (Sandwich)} \times 1 \text{ (Snack)} \times 2 \text{ (Drink)} = 4 \][/tex]
### Step 3: Calculate the Probability
Finally, we calculate the probability that Gretchen will randomly choose a boxed lunch with a roast beef sandwich and popcorn. Probability is given by:
[tex]\[ \text{Probability} = \frac{\text{Number of Favorable Combinations}}{\text{Total Number of Combinations}} = \frac{4}{24} = \frac{1}{6} \][/tex]
Therefore, the probability that Gretchen will get a boxed lunch with a roast beef sandwich and popcorn is [tex]\(\frac{1}{6}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{\frac{1}{6}} \][/tex]