Question

the cafeteria creates pre - made boxed lunches with equal numbers of the following items:

• a sandwich made with either white or wheat bread and either roast beef or bologna
• a snack that is either chips, popcorn, or pretzels
• a drink that is either bottled water or juice

if gretchen randomly chooses one of the boxed lunches, what is the probability that she will get a roast beef sandwich on wheat bread and popcorn in her box?
options: $\frac{1}{12}$, $\frac{1}{6}$, $\frac{1}{24}$, $\frac{3}{4}$

Explanation:

Step1: Determine number of choices for each category

• Bread: 2 options (white, wheat)
• Meat: 2 options (roast beef, bologna)
• Snack: 3 options (chips, popcorn, pretzels)
• Drink: 2 options (water, juice)

Step2: Calculate total number of possible lunches

Using the multiplication principle, total combinations = \(2\times2\times3\times2 = 24\)

Step3: Determine favorable outcomes

• Bread: 1 (wheat)
• Meat: 1 (roast beef)
• Snack: 1 (popcorn)
• Drink: 2 (since drink choice doesn't affect the specific snack/sandwich, but we still consider all drink options? Wait, no—wait, the problem is about the sandwich (bread + meat) and snack. Wait, maybe I misread. Wait, the question is "roast beef sandwich on wheat bread and popcorn in her box". So:
• Bread: 1 (wheat) out of 2
• Meat: 1 (roast beef) out of 2
• Snack: 1 (popcorn) out of 3
• Drink: doesn't matter? Wait, no, the total number of lunches is all combinations, and the favorable is the number of lunches with wheat, roast beef, popcorn, and any drink. Wait, no—wait, the problem says "equal numbers of the following items", so each combination is equally likely. Let's re-express:

The sandwich has 2 (bread) * 2 (meat) = 4 options.

The snack has 3 options.

The drink has 2 options.

Total lunches: \(4\times3\times2 = 24\), which matches Step2.

Favorable: sandwich (wheat, roast beef) → 1 option for bread, 1 for meat → 11=1 sandwich type. Snack: popcorn → 1 option. Drink: 2 options (since drink can be either, but the problem is about the sandwich and snack; wait, no—wait, the question is "roast beef sandwich on wheat bread and popcorn in her box". So regardless of drink, as long as sandwich is wheat/roast beef and snack is popcorn. So number of favorable: 1 (sandwich) 1 (snack) * 2 (drink) = 2? Wait, no, that can't be. Wait, no—wait, maybe the drink is not a factor? Wait, no, the total number of lunches is all possible combinations, so each lunch is a unique combination of bread, meat, snack, drink. So to have roast beef on wheat, popcorn, and any drink:

Bread: wheat (1 choice)

Meat: roast beef (1 choice)

Snack: popcorn (1 choice)

Drink: 2 choices (water or juice)

So favorable outcomes: \(1\times1\times1\times2 = 2\)? Wait, but that would make probability \(2/24 = 1/12\)? Wait, no, wait—maybe I messed up. Wait, let's re-express:

Wait, the problem is "what is the probability that she will get a roast beef sandwich on wheat bread and popcorn in her box?"

So:

• Probability of wheat bread: \(1/2\) (since 2 breads, equal number)

• Probability of roast beef: \(1/2\) (2 meats)

• Probability of popcorn: \(1/3\) (3 snacks)

• Probability of any drink: 1 (since drink doesn't matter, but actually, the drink is part of the lunch, but the question is about the sandwich and snack. Wait, no—actually, the total number of possible lunches is (2 breads) (2 meats) (3 snacks) * (2 drinks) = 24.

The number of lunches with wheat bread, roast beef, popcorn, and any drink: (1 bread) (1 meat) (1 snack) * (2 drinks) = 2.

Wait, but that would be 2/24 = 1/12. Which is one of the options. Wait, let's check again.

Alternatively, maybe the drink is not considered? No, because the lunch includes a drink, so each lunch is a combination of all four: sandwich (bread+meat), snack, drink. So total combinations: 223*2=24.

Favorable: bread=wheat (1), meat=roast beef (1), snack=popcorn (1), drink=any (2). So 111*2=2. So probability is 2/24=1/12.

Yes, that makes sense.

Answer:

\(\frac{1}{12}\)