Supply the missing quantities.
Figure
Description
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 0.7.
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.
Outcome
P(A ∩ E)
=
P(A ∩ F)
=
P(B ∩ G)
=
P(B ∩ H)
=

To solve this problem, we need to use the principles of probability, specifically the *multiplication rule of probability*. In a tree diagram, the probability of reaching a specific outcome is the product of the probabilities along the branches leading to that outcome. Let's work through the missing values step-by-step.

### **Stage 1**: 
- The sum of all probabilities leading out of the starting point must be equal to 1.
  - P(A) = 0.3 (already given)
  - P(B) = 0.1 (already given)
  - Let P(C) be the missing probability. So we have:
    \[
    P(A) + P(B) + P(C) = 1
    \]
    \[
    0.3 + 0.1 + P(C) = 1
    \]
    \[
    P(C) = 1 - 0.4 = 0.6
    \]
    Therefore, \( P(C) = 0.6 \).

### **Stage 2**:
- From node A:
  - P(A ∩ E): Let this branch probability be \( P(E \mid A) \). We know that the other branch (A → F) has a probability of 0.5, so:
    \[
    P(E \mid A) + P(F \mid A) = 1
    \]
    \[
    P(E \mid A) + 0.5 = 1
    \]
    \[
    P(E \mid A) = 0.5
    \]
  - Now, to get the joint probabilities:
    \[
    P(A ∩ E) = P(A) \times P(E \mid A) = 0.3 \times 0.5 = 0.15
    \]
    \[
    P(A ∩ F) = P(A) \times P(F \mid A) = 0.3 \times 0.5 = 0.15
    \]

- From node B:
  - P(B ∩ G): Let \( P(G \mid B) \) be the branch probability for B → G. Since B → H is 0.7:
    \[
    P(G \mid B) + P(H \mid B) = 1
    \]
    \[
    P(G \mid B) + 0.7 = 1
    \]
    \[
    P(G \mid B) = 0.3
    \]
  - Now, to get the joint probabilities:
    \[
    P(B ∩ G) = P(B) \times P(G \mid B) = 0.1 \times 0.3 = 0.03
    \]
    \[
    P(B ∩ H) = P(B) \times P(H \mid B) = 0.1 \times 0.7 = 0.07
    \]

- From node C:
  - We are already given the joint probabilities for outcomes I and J:
    \[
    P(I ∩ C) = 0.24, \quad P(J ∩ C) = 0.36
    \]
  - We can verify that these probabilities add up to the total probability of C:
    \[
    P(I \mid C) + P(J \mid C) = 1
    \]
    \[
    0.24 + 0.36 = 0.6 \quad \text{(which matches } P(C) = 0.6 \text{)}
    \]

### **Final Outcome**:
1. \( P(A ∩ E) = 0.15 \)
2. \( P(A ∩ F) = 0.15 \)
3. \( P(B ∩ G) = 0.03 \)
4. \( P(B ∩ H) = 0.07 \)

These are the missing probabilities based on the given tree structure.

### Do you want more details or have any questions?

Here are 5 questions related to this problem:

1. How would the probabilities change if the probability of A was different?
2. How does the multiplication rule of probability work in tree diagrams?
3. Can you explain the difference between conditional and joint probabilities?
4. How would you find the probability of reaching a particular node in a tree with more than two stages?
5. What would happen if the probabilities from each node did not sum to 1?

**Tip**: Always check that probabilities at each branching point sum to 1; this ensures the diagram reflects a complete set of outcomes.

Supply the missing quantities. 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 0.7. 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. Outcome: P(A ∩ E) = P(A ∩ F) = P(B ∩ G) = P(B ∩ H) =

Learn how to solve a probability tree diagram problem by filling in the missing probabilities for various branches. The problem covers key concepts like joint and conditional probability, including the application of the multiplication rule. Suitable for Grades 11-12.

Math Problem Statement

P(A ∩ F)

P(B ∩ G)

P(B ∩ H)

Solution

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