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, pretzels, and juice in her box?
(\boldsymbol{\frac{1}{6}})
(\boldsymbol{\frac{1}{12}})
(\boldsymbol{\frac{3}{4}})
(\boldsymbol{\frac{1}{24}})

Explanation:

Step1: Calculate sandwich options

Bread: 2 choices (white/wheat), meat: 2 choices (roast beef/bologna). So sandwich options: \(2\times2 = 4\). Probability of roast beef: \(\frac{1}{2}\) (since 2 meat choices, 1 is roast beef? Wait, no: meat has 2 options, so probability of roast beef is \(\frac{1}{2}\)? Wait, bread: 2, meat: 2. So total sandwich combinations: \(2\times2 = 4\). The number of roast beef sandwiches: for each bread (2), roast beef (1), so 2. Wait, no: probability of roast beef is \(\frac{2}{4}=\frac{1}{2}\)? Wait, no, the problem is about the probability of getting roast beef sandwich (regardless of bread), pretzels, and juice. So first, sandwich: bread (2) and meat (2). So total sandwich types: \(2\times2 = 4\). The number of roast beef sandwiches: for each bread (2), meat is roast beef (1), so 2. Wait, no, the probability of roast beef (given bread and meat) is \(\frac{1}{2}\) (since meat has 2 options: roast beef or bologna). So probability of roast beef sandwich: \(\frac{1}{2}\) (because meat is 2 options, 1 is roast beef; bread doesn't affect the meat choice? Wait, no, the sandwich is made with either white or wheat bread AND either roast beef or bologna. So the sandwich combinations are: white+roast, white+bologna, wheat+roast, wheat+bologna. So 4 total. The number of roast beef sandwiches: 2 (white+roast, wheat+roast). So probability of roast beef sandwich: \(\frac{2}{4}=\frac{1}{2}\).

Step2: Calculate snack options

Snack: chips, popcorn, pretzels. 3 options. Probability of pretzels: \(\frac{1}{3}\).

Step3: Calculate drink options

Drink: bottled water or juice. 2 options. Probability of juice: \(\frac{1}{2}\).

Step4: Multiply the probabilities

Since the choices are independent, we multiply the probabilities: \(\frac{1}{2}\times\frac{1}{3}\times\frac{1}{2}=\frac{1}{12}\). Wait, wait, let's recheck. Wait, the sandwich: total sandwich combinations: 2 (bread) 2 (meat) = 4. The desired sandwich: roast beef (1 meat choice) and any bread (2), so 2. Wait, no, the problem is "roast beef sandwich" – so meat is roast beef, bread is either. So number of roast beef sandwiches: 2 (white+roast, wheat+roast). Total sandwiches: 4. So probability of roast beef sandwich: 2/4 = 1/2. Snack: 3 options, pretzels is 1, so 1/3. Drink: 2 options, juice is 1, so 1/2. So total probability: (1/2) (1/3) (1/2) = 1/12. Wait, but let's think of total possible boxed lunches. Sandwich: 4, snack: 3, drink: 2. Total combinations: 432 = 24. Desired combinations: roast beef sandwich (2) pretzels (1) juice (1) = 211 = 2? Wait, no, wait: sandwich: roast beef (meat) and any bread (2), so 2 sandwich types. Snack: pretzels (1). Drink: juice (1). So desired combinations: 211 = 2. Total combinations: 43*2 = 24. So probability is 2/24 = 1/12. Yes, that matches. So the probability is 1/12.

Answer:

\(\frac{1}{12}\) (corresponding to the option \(\frac{1}{12}\))