Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. 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 and popcorn in her box? (\frac{1}{12}) (\frac{1}{2}) (\frac{1}{6}) (\frac{1}{3})

Answer:

each drink, there are 2 (bread) × 1 (meat) × 1 (snack) = 2 boxes per drink. There are 2 drinks, so 2×2 = 4? Wait, no, 2 (bread) × 1 (meat) × 1 (snack) × 2 (drink) = 4. Total boxes: 4×3×2 = 24. So probability is $\frac{4}{24}=\frac{1}{6}$? Wait, I must have made a mistake.

Wait, let's start over.

Sandwich: bread (2) × meat (2) = 4 possible sandwiches (white - roast, white - bologna, wheat - roast, wheat - bologna).

Snack: 3 possible (chips, popcorn, pretzels).

Drink: 2 possible (water, juice).

Total number of boxed lunches: $4\times3\times2 = 24$.

Now, number of boxed lunches with roast beef sandwich and popcorn:

Roast beef sandwich: 2 types (white - roast, wheat - roast).

Popcorn: 1 type.

Drink: 2 types (water, juice).

So number of favorable boxes: $2\times1\times2 = 4$. Wait, no, 2 (sandwiches) × 1 (snack) × 2 (drinks) = 4.

Wait, but $4\div24=\frac{1}{6}$? But the option $\frac{1}{12}$ is there. Wait, maybe I messed up the sandwich count. Wait, the sandwich is "either white or wheat bread and either roast beef or bologna"—so bread: 2, meat: 2, so 2×2 = 4 sandwiches. Correct.

Snack: 3, drink: 2. Total: 24.

Favorable: roast beef (meat: 1 out of 2) and popcorn (snack: 1 out of 3). The drink doesn't matter, but actually, in the total, drink is included. Wait, maybe the problem is considering only sandwich and snack, ignoring drink? But no, the box has all three. Wait, maybe the question is "roast beef sandwich (any bread) and popcorn (any drink)"? Wait, no, the problem says "roast beef sandwich and popcorn in her box"—so the drink is part of the box, but we are only concerned with the sandwich (roast beef) and the snack (popcorn). So the probability of roast beef sandwich is (number of roast beef sandwiches)/(total sandwiches) = 2/4 = 1/2. Probability of popcorn is 1/3. Since the choices are independent (sandwich, snack, drink are independent components), the probability of roast beef sandwich AND popcorn is (1/2)×(1/3) = 1/6? But that's one of the options. Wait, but earlier when I calculated total boxes, I got 4 favorable out of 24, which is 1/6. But the first option is 1/12. Wait, maybe I made a mistake in the number of sandwiches. Wait, no—wait, maybe the sandwich is defined as (bread, meat), so 2 breads × 2 meats = 4. Snack: 3, drink: 2. Total: 24. Favorable: (roast beef sandwiches: 2) × (popcorn: 1) × (drink: 2) = 4. 4/24 = 1/6. So the answer should be 1/6? But let's check again.

Wait, probability of roast beef sandwich: meat has 2 options, so 1/2. Probability of popcorn: snack has 3 options, so 1/3. The drink has 2 options, but since we are not restricting the drink, the probability of any drink is 1. So the combined probability is (1/2)×(1/3)×1 = 1/6. So the correct answer is $\frac{1}{6}$? But the first option is 1/12. Wait, maybe I misread the problem. Let me check again.

Wait, the problem says "equal numbers of the following items"—maybe each item (sandwich, snack, drink) is in equal number, not the combinations. Wait, that would be different. If there are equal numbers of each sandwich, each snack, and each drink, then the number of boxes is the least common multiple or something, but the problem says "pre - made boxed lunches with equal numbers of the following items"—so each box has one sandwich, one snack, one drink, and the number of boxes with each sandwich type is equal, each snack type is equal, each drink type is equal. So let's say there are N boxes with each sandwich, N with each snack, N with each drink. Then total number of boxes: for sandwiches: 4 types, so 4N; for snacks: 3 types, so 3N; for drinks: 2 types, so 2N. So total boxes: LCM(4N, 3N, 2N) = 12N (since LCM(4,3,2)=12). So total boxes: 12N.

Number of boxes with roast beef sandwich: there are 2 roast beef sandwich types (white - roast, wheat - roast), each with N boxes? No, wait, if equal numbers of each item, then number of boxes with white - roast beef sandwich: N, white - bologna: N, wheat - roast: N, wheat - bologna: N. So roast beef sandwiches: 2N.

Number of boxes with popcorn: N (since 3 snack types, each with N boxes).

Number of boxes with roast beef sandwich and popcorn: for each roast beef sandwich type (2) and each drink type (2), but wait, no—if each sandwich type, snack type, and drink type has N boxes, then the number of boxes with roast beef sandwich (2 types) and popcorn (1 type) and any drink (2 types) is 2×1×2×N? No, this is getting confusing.

Wait, let's use the multiplication principle for probability.

Probability of roast beef sandwich: meat is roast beef or bologna, so 2 options, so P(roast beef) = $\frac{1}{2}$.

Probability of popcorn: snack is chips, popcorn, pretzels, 3 options, so P(popcorn) = $\frac{1}{3}$.

The drink has 2 options, but since we don't care about the drink, the probability of any drink is 1. Also, the sandwich and snack choices are independent. So the probability of roast beef sandwich AND popcorn is P(roast beef) × P(popcorn) = $\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$. So the answer should be $\frac{1}{6}$. But let's check the options. The options are $\frac{1}{12}$, $\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{3}$. So $\frac{1}{6}$ is an option. Wait, maybe I made a mistake earlier when I thought 4/24, but 4/24 is 1/6. Yes, 4 divided by 24 is 1/6. So the correct answer is $\frac{1}{6}$.

Wait, but let's re - calculate the total number of possible boxed lunches.

Sandwich: bread (2) × meat (2) = 4.

Snack: 3.

Drink: 2.

Total: 4×3×2 = 24.

Favorable: roast beef sandwich (2 types: white - roast, wheat - roast) × popcorn (1 type) × drink (2 types: water, juice). So 2×1×2 = 4.

4/24 = 1/6. So the answer is $\frac{1}{6}$.