https://r2.erweima.ai/mathbot/image/1727180664414lladst1e.jpg

Let's solve each part of the exercise presented in the image one by one.

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### **Exercise 1: Conditional Probability (Probabilités conditionnelles)**

#### **1.1**
- \( P(A) = 0.2 \)
- \( P(B) = 0.6 \)
- \( P(A \cap B) = 0.15 \)

We are asked to calculate \( P_A(B) \) (the conditional probability of \( B \) given \( A \)) and \( P_B(A) \) (the conditional probability of \( A \) given \( B \)).

##### **Solution:**
- **Formula for conditional probability**:  
  \( P_A(B) = \frac{P(A \cap B)}{P(A)} \)  
  \( P_B(A) = \frac{P(A \cap B)}{P(B)} \)

Thus:
- \( P_A(B) = \frac{0.15}{0.2} = 0.75 \)
- \( P_B(A) = \frac{0.15}{0.6} = 0.25 \)

---

#### **1.2**
- \( P(A) = 0.6 \)
- \( P_A(B) = 0.4 \)
- \( P(A \cap B) = 0.4 \)

##### **Solution:**
- **Find \( P_B(A) \)**:
  Use \( P(A \cap B) = P_A(B) \times P(A) \), so:
  \[
  P_B(A) = \frac{P(A \cap B)}{P(B)} = \frac{0.4}{P(B)}
  \]

To find \( P(B) \), apply the formula \( P(A \cap B) = P_A(B) \times P(A) \):
\[
0.4 = 0.4 \times 0.6
\]
Thus, \( P(B) = 1 \) (though this may need more details).

---

### **Exercise 2: Proof of Conditional Probabilities**

#### **2.1**
The goal is to show that:
\[
P_A(B) \times P(A) = P_B(A) \times P(B)
\]

##### **Solution:**
We know that:
\[
P_A(B) = \frac{P(A \cap B)}{P(A)} \quad \text{and} \quad P_B(A) = \frac{P(A \cap B)}{P(B)}
\]

Multiplying both sides:
\[
P_A(B) \times P(A) = P(A \cap B) = P_B(A) \times P(B)
\]

Thus, it is proven that:
\[
P_A(B) \times P(A) = P_B(A) \times P(B)
\]

#### **2.2**
- \( P_A(B) = 0.6 \)
- \( P_B(A) = 0.3 \)
- \( P(A) = 0.2 \)

We need to find \( P(B) \).

##### **Solution:**
Using the same formula as above:
\[
P_A(B) \times P(A) = P_B(A) \times P(B)
\]

Substitute the values:
\[
0.6 \times 0.2 = 0.3 \times P(B)
\]

Solve for \( P(B) \):
\[
P(B) = \frac{0.6 \times 0.2}{0.3} = \frac{0.12}{0.3} = 0.4
\]

Thus, \( P(B) = 0.4 \).

---

### **Exercise 3: Facultative**

We are given a table of students:

|      | Présents | Admis  |
|------|----------|--------|
| Filles | 392,500  | 356,000 |
| Garçons | 415,700  | 348,200 |
| **Total** | **808,200** | **704,200** |

1. **Probability that a student is admitted**:
   \[
   P(\text{admis}) = \frac{\text{total admitted}}{\text{total students present}} = \frac{704,200}{808,200} = 0.871 \quad \text{(87.1\%)}
   \]

2. **Probability that a student is admitted, given that it is a girl**:
   \[
   P(\text{admis | fille}) = \frac{\text{admitted girls}}{\text{girls present}} = \frac{356,000}{392,500} = 0.907
   \]

3. **Probability that a student is a girl, given that they are admitted**:
   \[
   P(\text{fille | admis}) = \frac{\text{admitted girls}}{\text{total admitted}} = \frac{356,000}{704,200} = 0.505
   \]

4. **Probabilities to compare success rates between girls and boys**:
   Compare:
   \[
   P(\text{admis | fille}) = 0.907 \quad \text{and} \quad P(\text{admis | garçon}) = \frac{348,200}{415,700} = 0.837
   \]

   Conclusion: Girls have a higher success rate (90.7%) compared to boys (83.7%).

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### Next Steps

Would you like a detailed explanation of any specific part?

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Here are 5 related questions for further exploration:

1. What is the formula for the union of two probabilities, \( P(A \cup B) \)?
2. How would you calculate \( P(A \cap B) \) if you know \( P(A) \), \( P(B) \), and \( P(A \cup B) \)?
3. Can two events be independent if \( P(A \cap B) \neq P(A) \times P(B) \)?
4. How does Bayes' theorem apply to calculating conditional probabilities?
5. How can you compare conditional probabilities in real-life decision-making?

**Tip:** When working with probabilities, always ensure the total sum of probabilities in your system equals 1 to maintain consistency.

Solve exercises based on conditional probability, including calculating conditional probabilities for independent events and interpreting exam results using probability tables.

This page explains how to solve conditional probability problems, including exercises involving independent events and interpreting real exam data. Topics covered include calculating probabilities, understanding relationships between events, and comparing success rates.

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