两只债券的至少一次违约概率范围

设事件 $A$、$B$ 分别表示两只债券违约，$\mathbb{P}(A)=0.3,\ \mathbb{P}(B)=0.5$。

由容斥：

$$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)=0.8-\mathbb{P}(A\cap B).$$

而交集概率满足

$$\max\{0,\,\mathbb{P}(A)+\mathbb{P}(B)-1\}\le \mathbb{P}(A\cap B)\le \min\{\mathbb{P}(A),\mathbb{P}(B)\},$$

即 $0\le \mathbb{P}(A\cap B)\le 0.3$。

因此

$$0.8-0.3\le \mathbb{P}(A\cup B)\le 0.8-0 \Rightarrow \boxed{0.5\le \mathbb{P}(A\cup B)\le 0.8}.$$

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英文解析

Let events $A$ and $B$ represent the default of the two bonds, with $\mathbb{P}(A)=0.3$ and $\mathbb{P}(B)=0.5$.

By the inclusion-exclusion principle:

$$\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)=0.8-\mathbb{P}(A\cap B).$$

The probability of the intersection satisfies

$$\max\{0,\,\mathbb{P}(A)+\mathbb{P}(B)-1\}\le \mathbb{P}(A\cap B)\le \min\{\mathbb{P}(A),\mathbb{P}(B)\},$$

which implies $0\le \mathbb{P}(A\cap B)\le 0.3$.

Therefore

$$
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}.