To determine the probability that Libby will have oatmeal and apple juice, we first need to identify the total number of food and drink options she has:
- Food options: Oatmeal, pancakes, eggs (3 options)
- Drink options: Apple juice, orange juice (2 options)
Next, we calculate the total number of possible combinations of food and drink:
\[ \text{Total combinations} = \text{Number of food options} \times \text{Number of drink options} = 3 \times 2 = 6 \]
The combinations can be listed as follows:
- Oatmeal and apple juice
- Oatmeal and orange juice
- Pancakes and apple juice
- Pancakes and orange juice
- Eggs and apple juice
- Eggs and orange juice
Now, we identify the favorable outcome, which is Libby having oatmeal and apple juice. There is only 1 favorable outcome (oatmeal and apple juice) in the 6 total combinations.
Now, we can calculate the probability:
\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{6} \]
The options given (A, B, C, D) seem to be numerical values rather than probability fractions, so it's likely there's a misunderstanding or the wrong context for these options.
However, the correct probability that Libby will have oatmeal and apple juice is \( \frac{1}{6} \), which does not match the options provided.
If the question is asking for a numeric representation based on a different scale (e.g., out of 18), we'd need that context to adjust the calculation accordingly. If you meant to ask about a scaled format, please clarify! As it stands, based purely on random selection of food and drink combinations, the answer is:
\[ \text{Probability of oatmeal and apple juice} = \frac{1}{6} \]