At a museum cafe you can get a pre-made boxed lunch with a sandwich, fruit, and drink for only $3

Question

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The sandwiches are made with either turkey or ham.
The fruit is either an apple or an orange.
The drink is either bottled water or juice.
The number of boxes they make for every possible combination is the same. If you randomly choose one of the boxed lunches without knowing the contents, what is the probability you will get an orange and not get juice in your box?

Answer 1

To find the probability of getting a boxed lunch that contains an orange and does not contain juice, we first need to determine all the possible combinations of the boxed lunches.

1. Sandwich options: There are 2 types of sandwiches: turkey and ham.
2. Fruit options: There are 2 types of fruit: apple and orange.
3. Drink options: There are 2 types of drinks: bottled water and juice.

Now, we calculate the total number of combinations:

\[ \text{Total combinations} = (\text{number of sandwiches}) \times (\text{number of fruits}) \times (\text{number of drinks}) = 2 \times 2 \times 2 = 8 \]

The 8 combinations are:

1. Turkey, Apple, Water
2. Turkey, Apple, Juice
3. Turkey, Orange, Water
4. Turkey, Orange, Juice
5. Ham, Apple, Water
6. Ham, Apple, Juice
7. Ham, Orange, Water
8. Ham, Orange, Juice

Next, we focus on the combinations that meet our criteria: we want an orange and not juice.

The relevant combinations are:

1. Turkey, Orange, Water
2. Ham, Orange, Water

There are 2 combinations that meet the criteria (orange and not juice).

Now, we can calculate the probability:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4} \]

Thus, the probability of getting a boxed lunch that contains an orange and does not contain juice is \(\boxed{\frac{1}{4}}\).