\(\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)\)
Because \(A\) and \(B\) are independent, you have \(\mathbb P(A\cap B)=\mathbb P(A)\times\mathbb P(B)\), so
\(\mathbb P(A\cup B)=0.60+0.39-0.60\times0.39=0.756\)
If \(A\) and \(B\) are mutually exclusive, you have
\(\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)=0.60+0.39=0.99\)
