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)
=

Let's work through this problem by identifying the missing probabilities and calculating the specified outcomes using the principles of probability trees.

### Step 1: Find the Missing Probabilities in Stage 1
In Stage 1, we have three branches:
1. From the starting position to node A: Probability = \( 0.3 \)
2. From the starting position to node B: Probability = \( 0.1 \)
3. From the starting position to node C: Probability = ?

Since the total probability from the starting point must be 1, we can find the probability of node C as follows:
\[
P(C) = 1 - (P(A) + P(B)) = 1 - (0.3 + 0.1) = 1 - 0.4 = 0.6
\]

So, **the missing probability for the branch from the starting position to node C is 0.6.**

### Step 2: Find the Missing Probabilities in Stage 2
We have the following probabilities labeled in Stage 2:

- **Branches from Node A:**
  - From A to F: Probability = \( 0.5 \)
  - From A to E: Probability = ?
  
  Since the total probability from node A must be 1:
  \[
  P(E \mid A) = 1 - P(F \mid A) = 1 - 0.5 = 0.5
  \]

- **Branches from Node B:**
  - From B to H: Probability = \( 0.7 \)
  - From B to G: Probability = ?
  
  Similarly, since the total probability from node B must be 1:
  \[
  P(G \mid B) = 1 - P(H \mid B) = 1 - 0.7 = 0.3
  \]

- **Branches from Node C:**
  - We are given the joint probabilities:
    - \( P(I \cap C) = 0.24 \)
    - \( P(J \cap C) = 0.36 \)
  
  The probability of node C is \( P(C) = 0.6 \), so we can find the conditional probabilities:
  \[
  P(I \mid C) = \frac{P(I \cap C)}{P(C)} = \frac{0.24}{0.6} = 0.4
  \]
  \[
  P(J \mid C) = \frac{P(J \cap C)}{P(C)} = \frac{0.36}{0.6} = 0.6
  \]

### Step 3: Calculate the Requested Joint Probabilities
Now, let's calculate the outcomes using \( P(X \cap Y) = P(X) \times P(Y \mid X) \):

1. **\( P(A \cap E) \):**
   \[
   P(A \cap E) = P(A) \times P(E \mid A) = 0.3 \times 0.5 = 0.15
   \]

2. **\( P(A \cap F) \):**
   \[
   P(A \cap F) = P(A) \times P(F \mid A) = 0.3 \times 0.5 = 0.15
   \]

3. **\( P(B \cap G) \):**
   \[
   P(B \cap G) = P(B) \times P(G \mid B) = 0.1 \times 0.3 = 0.03
   \]

4. **\( P(B \cap H) \):**
   \[
   P(B \cap H) = P(B) \times P(H \mid B) = 0.1 \times 0.7 = 0.07
   \]

### Summary of Solutions
- **Missing Probability for \( P(C) \):** \( 0.6 \)
- **Missing Probability for \( P(E \mid A) \):** \( 0.5 \)
- **Missing Probability for \( P(G \mid B) \):** \( 0.3 \)
- **Missing Probability for \( P(I \mid C) \):** \( 0.4 \)
- **Missing Probability for \( P(J \mid C) \):** \( 0.6 \)
  
- **Outcomes:**
  - \( P(A \cap E) = 0.15 \)
  - \( P(A \cap F) = 0.15 \)
  - \( P(B \cap G) = 0.03 \)
  - \( P(B \cap H) = 0.07 \)

Would you like further explanation on any step? 

---

Here are some questions to deepen your understanding:

1. How do probability tree diagrams help visualize joint probabilities?
2. Why must probabilities from a single point add up to 1 in a tree diagram?
3. How would you approach finding conditional probabilities from joint probabilities?
4. What would change if we altered one of the initial probabilities?
5. How does understanding joint and conditional probabilities aid in complex problem-solving?

**Tip:** Always check that the probabilities for branches from the same point add up to 1 to ensure accuracy.

Discover how to calculate missing probabilities in a two-stage probability tree diagram. This example demonstrates joint and conditional probabilities and applies the law of total probability. Ideal for students in Grades 10-12, with detailed calculations of P(A ∩ E), P(A ∩ F), P(B ∩ G), and P(B ∩ H).

Math Problem Statement

P(A ∩ F)

P(B ∩ G)

P(B ∩ H)

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation