Question

Events a, b, and c in a sample space have p(a)=0.2, p(b)=0.4, and p(c)=0.1. find p(a ∪ b ∪

Answer 1

If $A$ and $B$ are mutually independent, then $\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)$ and $\mathbb P(A\cap B)=0$.

If $A$ and $C$ are independent, then $\mathbb P(A\cap C)=\mathbb P(A)\cdot\mathbb P(C)$. Ditto for $B$ and $C$.

The inclusion/exclusion principle says that

$\mathbb P(A\cup B\cup C)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\mathbb P(A\cap B)-\mathbb P(A\cap C)-\mathbb P(B\cap C)+\mathbb P(A\cap B\cap C)$

Since $A$ and $B$ are mutually exclusive, their intersection is the empty set. It then follows that $A\cap B\cap C$ also is the empty set, so the last probability is also 0.

Plug in what we know:

$\mathbb P(A\cup B\cup C)=0.2+0.4+0.1-0-0.2\cdot0.1-0.4\cdot0.1+0$

$\implies\mathbb P(A\cup B\cup C)=0.64$