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Let's go through the problem step by step:

### Problem Overview:
- **Children**: Oldest (O), Middle (M), and Youngest (Y)
- **Jobs**: Kitchen (K), Living room (L), and Bedroom (B)
- Each child is randomly assigned one of the jobs.

#### Part A: **List the possible outcomes.**
Each child can be assigned any of the three jobs, so the total possible outcomes are the combinations of children and jobs. The form of the outcomes will be like: \( (Child, Job) \).

- O (Oldest) can be assigned to K, L, or B
- M (Middle) can be assigned to K, L, or B
- Y (Youngest) can be assigned to K, L, or B

So, the possible outcomes are:
- \( O-K, O-L, O-B \)
- \( M-K, M-L, M-B \)
- \( Y-K, Y-L, Y-B \)

These are all the possible combinations, totaling 9 outcomes.

---

#### Part C: **P(D) - Probability that the oldest child is chosen.**

There are 9 possible outcomes in total, and the event \( D \) (Oldest child is chosen) corresponds to the three outcomes where O is selected:
- \( O-K, O-L, O-B \)

Thus, the probability of \( D \) is:
\[
P(D) = \frac{\text{Number of favorable outcomes for D}}{\text{Total outcomes}} = \frac{3}{9} = \frac{1}{3}
\]

---

#### Part E: **P(N) - Probability that the bedroom is not cleaned.**

The event \( N \) corresponds to the outcomes where no one is assigned to clean the bedroom (B). To count these, we exclude the outcomes where B is assigned:
- Remaining outcomes: \( O-K, O-L, M-K, M-L, Y-K, Y-L \)

There are 6 outcomes where the bedroom is not cleaned, so the probability is:
\[
P(N) = \frac{6}{9} = \frac{2}{3}
\]

---

#### Part F: **P(D ∪ N) - Probability that either the oldest child is chosen or the bedroom is not cleaned (or both).**

The union \( D \cup N \) corresponds to all outcomes where either the oldest child is chosen, or the bedroom is not cleaned, or both. These outcomes include:
- \( O-K, O-L, O-B \) (Oldest chosen)
- \( M-K, M-L, Y-K, Y-L \) (Bedroom not cleaned)

These give 7 favorable outcomes in total, so:
\[
P(D \cup N) = \frac{7}{9}
\]

---

#### Part G: **P(D ∩ N) - Probability that the oldest child is chosen and the bedroom is not cleaned (both occur).**

This intersection corresponds to the outcomes where both the oldest child is chosen, and the bedroom is not cleaned. The favorable outcomes are:
- \( O-K, O-L \)

There are 2 such outcomes, so:
\[
P(D \cap N) = \frac{2}{9}
\]

---

Would you like further details on any step? Also, here are some related questions you might find useful:

1. What would be the probability if the youngest child is chosen?
2. How does the probability change if there are 4 jobs instead of 3?
3. What is the complementary event of \( N \), and how do we calculate its probability?
4. How does the intersection of two independent events like \( D \) and \( N \) work mathematically?
5. How do we interpret the union and intersection of events in probability theory?

**Tip:** Always ensure that the total number of possible outcomes is clearly understood before calculating any probability.

Beth randomly picks one of her 3 children (O, M, Y) to help clean the house, assigning them to Kitchen, Living Room, or Bedroom. What are the possible outcomes, and what are the probabilities of specific events like the oldest child being chosen and the bedroom not being cleaned?

Solve a probability problem involving 3 children assigned to different cleaning jobs. Learn how to calculate the probability of events like choosing the oldest child or ensuring the bedroom is not cleaned.

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