A tree diagram has two stages. Stage 1 has three nodes and stage 2 has six nodes.
In stage 1, the branch from the starting position to node A is labeled 0.3. The branch from the starting position to node B is labeled 0.1. The branch from the starting position to node C is an answer blank.
In stage 2, the branch from node A to node E is an answer blank. The branch from node A to node F is labeled 0.5.
In stage 2, the branch from node B to node G is an answer blank. The branch from node B to node H is labeled 0.9.
In stage 2, the branch from node C to node I is an answer blank. The branch from node C to node J is an answer blank.
Node I is labeled P(I ∩ C) = 0.24.
Node J is labeled P(J ∩ C) = 0.36.

Let's break down the information from the tree diagram step by step and fill in the missing values based on probability principles. We'll assume that each branch represents conditional probabilities and use the fact that the sum of probabilities at any node must be 1.

### Stage 1:
- **From starting position to Node A**: Labeled \( P(A) = 0.3 \)
- **From starting position to Node B**: Labeled \( P(B) = 0.1 \)
- **From starting position to Node C**: Since probabilities must sum to 1, \( P(C) = 1 - P(A) - P(B) = 1 - 0.3 - 0.1 = 0.6 \).

### Stage 2:
We have branches from nodes \( A \), \( B \), and \( C \).

1. **From Node A**:
   - The branch from \( A \) to \( F \) is labeled \( P(F \mid A) = 0.5 \).
   - The remaining branch from \( A \) to \( E \) must be \( P(E \mid A) = 1 - P(F \mid A) = 1 - 0.5 = 0.5 \).

2. **From Node B**:
   - The branch from \( B \) to \( H \) is labeled \( P(H \mid B) = 0.9 \).
   - The remaining branch from \( B \) to \( G \) must be \( P(G \mid B) = 1 - P(H \mid B) = 1 - 0.9 = 0.1 \).

3. **From Node C**:
   - We are given the joint probabilities for nodes \( I \) and \( J \) as:
     \[
     P(I \cap C) = 0.24, \quad P(J \cap C) = 0.36
     \]
     To find the conditional probabilities \( P(I \mid C) \) and \( P(J \mid C) \), we use the formula:
     \[
     P(I \mid C) = \frac{P(I \cap C)}{P(C)}, \quad P(J \mid C) = \frac{P(J \cap C)}{P(C)}
     \]
     Substituting the values:
     \[
     P(I \mid C) = \frac{0.24}{0.6} = 0.4, \quad P(J \mid C) = \frac{0.36}{0.6} = 0.6
     \]

### Summary of Results:
- \( P(C) = 0.6 \) (Stage 1)
- \( P(E \mid A) = 0.5 \), \( P(F \mid A) = 0.5 \) (Stage 2, from A)
- \( P(G \mid B) = 0.1 \), \( P(H \mid B) = 0.9 \) (Stage 2, from B)
- \( P(I \mid C) = 0.4 \), \( P(J \mid C) = 0.6 \) (Stage 2, from C)

Would you like a more detailed breakdown of how I computed the missing values or have any questions?

Here are 5 related questions:
1. How do we calculate the total probability of reaching any node in stage 2?
2. What is the probability of reaching node \( J \) from the starting position?
3. How does the tree diagram change if one of the probabilities is conditional on an external event?
4. Can we apply Bayes' Theorem to this tree diagram?
5. How would adding a third stage impact the overall probabilities?

**Tip:** Always ensure that the sum of the probabilities at each branching node equals 1 to maintain consistency in tree diagrams.

A tree diagram has two stages. Stage 1 has three nodes and stage 2 has six nodes. In stage 1, the branch from the starting position to node A is labeled 0.3. The branch from the starting position to node B is labeled 0.1. The branch from the starting position to node C is an answer blank. In stage 2, the branch from node A to node E is an answer blank. The branch from node A to node F is labeled 0.5. In stage 2, the branch from node B to node G is an answer blank. The branch from node B to node H is labeled 0.9. In stage 2, the branch from node C to node I is an answer blank. The branch from node C to node J is an answer blank. Node I is labeled P(I ∩ C) = 0.24. Node J is labeled P(J ∩ C) = 0.36.

This detailed solution breaks down a two-stage probability tree diagram with missing values. Learn how to apply conditional probability formulas to calculate probabilities for each branch, including nodes A, B, C, E, F, G, H, I, and J. Suitable for Grades 11-12 students studying advanced probability.

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