To solve the problem, we will follow the steps outlined in the video transcript, ensuring to clarify each step with detailed explanations.
### Given:
- A fair die is rolled.
- Events:
- \( E = \{1, 3, 5\} \)
- \( F = \{2, 3\} \)
- \( G = \{2, 3, 4, 5\} \)
### Total Outcomes:
When a die is rolled, the total number of outcomes is \( 6 \) (i.e., \( \{1, 2, 3, 4, 5, 6\} \)).
### Step 1: Calculate Probabilities of Individual Events
1. **Probability of \( E \)**:
\[
P(E) = \frac{\text{Number of elements in } E}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2}
\]
2. **Probability of \( F \)**:
\[
P(F) = \frac{\text{Number of elements in } F}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}
\]
3. **Probability of \( G \)**:
\[
P(G) = \frac{\text{Number of elements in } G}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3}
\]
### Step 2: Calculate Intersection of Events
1. **Intersection of \( E \) and \( F \)**:
\[
E \cap F = \{3\} \quad \text{(common element)}
\]
\[
P(E \cap F) = \frac{1}{6}
\]
2. **Intersection of \( E \) and \( G \)**:
\[
E \cap G = \{3\} \quad \text{(common element)}
\]
\[
P(E \cap G) = \frac{1}{6}
\]
3. **Intersection of \( F \) and \( G \)**:
\[
F \cap G = \{2, 3\} \quad \text{(common elements)}
\]
\[
P(F \cap G) = \frac{2}{6} = \frac{1}{3}
\]
### Step 3: Calculate Conditional Probabilities
1. **Calculate \( P(E|F) \)**:
\[
P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \times \frac{3}{1} = \frac{1}{2}
\]
2. **Calculate \( P(F|E) \)**:
\[
P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times \frac{2}{1} = \frac{1}{3}
\]
3. **Calculate \( P(E|G) \)**:
\[
P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}
\]
4. **Calculate \( P(G|E) \)**:
\[
P(G|E) = \frac{P(E \cap G)}{P(E)} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{6} \times \frac{2}{1} = \frac{1}{3}
\]
### Step 4: Calculate Union and Intersection Probabilities
1. **Calculate \( P(E \cup F | G) \)**:
- First, find \( E \cup F = \{1, 2, 3, 5\} \).
- Intersection with \( G \):
\[
E \cup F \cap G = \{2, 3\} \quad \text{(common elements)}
\]
\[
P(E \cup F \cap G) = \frac{2}{6} = \frac{1}{3}
\]
\[
P(E \cup F | G) = \frac{P(E \cup F \cap G)}{P(G)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}
\]
2. **Calculate \( P(E \cap F | G) \)**:
\[
P(E \cap F | G) = \frac{P(E \cap F \cap G)}{P(G)} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}
\]
### Final Answers:
(i) \( P(E|F) = \frac{1}{2}, P(F|E) = \frac{1}{3} \)
(ii) \( P(E|G) = \frac{1}{4}, P(G|E) = \frac{1}{3} \)
(iii) \( P(E \cup F | G) = \frac{1}{2}, P(E \cap F | G) = \frac{1}{4} \)
