两只债券的至少一次违约概率范围 Bond Default Probability Range 专题 Probability / 概率 难度 L4 来源 QuantQuestion 题目详情 两只债券各自的违约概率分别为 30% 与 50%。问：至少有一只债券违约的概率可能落在什么范围内？ Two bonds have individual default probabilities of 30% and 50%. What is the possible range for the probability that at least one of the two bonds defaults? 解析 设事件 AAA、BBB 分别表示两只债券违约，P(A)=0.3, P(B)=0.5\mathbb{P}(A)=0.3,\ \mathbb{P}(B)=0.5P(A)=0.3, P(B)=0.5。 由容斥： P(A∪B)=P(A)+P(B)−P(A∩B)=0.8−P(A∩B).\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)=0.8-\mathbb{P}(A\cap B).P(A∪B)=P(A)+P(B)−P(A∩B)=0.8−P(A∩B). 而交集概率满足 max{0, P(A)+P(B)−1}≤P(A∩B)≤min{P(A),P(B)},\max\{0,\,\mathbb{P}(A)+\mathbb{P}(B)-1\}\le \mathbb{P}(A\cap B)\le \min\{\mathbb{P}(A),\mathbb{P}(B)\},max{0,P(A)+P(B)−1}≤P(A∩B)≤min{P(A),P(B)}, 即 0≤P(A∩B)≤0.30\le \mathbb{P}(A\cap B)\le 0.30≤P(A∩B)≤0.3。 因此 0.8−0.3≤P(A∪B)≤0.8−0⇒0.5≤P(A∪B)≤0.8.0.8-0.3\le \mathbb{P}(A\cup B)\le 0.8-0 \Rightarrow \boxed{0.5\le \mathbb{P}(A\cup B)\le 0.8}.0.8−0.3≤P(A∪B)≤0.8−0⇒0.5≤P(A∪B)≤0.8.