Nous savons déjà $P_A(B) = \frac{P(A \cap B)}{P(A)}$ et $P_B(A) = \frac{P(A \cap B)}{P(B)}$, alors
$$\begin{align*}P_A(B) \times \left( \frac{1}{P_B(A)} - 1 \right) &= \frac{P(A \cap B)}{P(A)} \times \left(\frac{P(B)}{P(A \cap B)} - 1\right) \\ &= \frac{P(A \cap B)}{P(A)} \times \left(\frac{P(B) - P(A \cap B)}{P(A \cap B)} \right) \\ &= \frac{P(A \cap B)}{P(A)} \times \frac{P(B) - P(A \cap B)}{P(A \cap B)} \\ &= \frac{P(B) - P(A \cap B)}{P(A)} \end{align*}$$
C'est à dire $$P_A(B) \times \left( \frac{1}{P_B(A)} - 1 \right) = \frac{P(B) - P(A \cap B)}{P(A)}$$
Nous savons aussi $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, alors $$P(A \cup B) - P(A) = P(B) - P(A \cap B)$$D'où $$\frac{P(A \cup B) - P(A)}{P(A)} = \frac{P(B) - P(A \cap B)}{P(A)}$$
Or $\frac{P(A \cup B) - P(A)}{P(A)} = \frac{P(A \cup B)}{P(A)} - 1$, alors $$\frac{P(B) - P(A \cap B)}{P(A)} = \frac{P(A \cup B)}{P(A)} - 1$$
Il s'ensuit que $$P_A(B) \times \left( \frac{1}{P_B(A)} - 1 \right) = \frac{P(B) - P(A \cap B)}{P(A)} = \frac{P(A \cup B)}{P(A)} - 1$$
Bài viết đã được chỉnh sửa nội dung bởi Konstante: 27-11-2023 - 01:37
