Answer
It is not verified that \[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=\frac{1}{2}\].
Work Step by Step
Suppose that three fair dice are tossed.
Let ${{A}_{i}}$ be the event that a 6 shows on the i th die, i = 1, 2, 3.
Hence,
\[\begin{align}
& P({{A}_{1}})=\frac{1}{6} \\
& P({{A}_{2}})=\frac{1}{6} \\
& P({{A}_{3}})=\frac{1}{6} \\
\end{align}\]
Consider,
\[\begin{align}
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=P(Atleast\text{ one 6 appears}) \\
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=1-P(no\text{ 6 appears}) \\
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=1-\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6} \\
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=1-\frac{125}{216} \\
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=\frac{216-125}{216} \\
& P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=\frac{91}{216} \\
\end{align}\]
Since, we see that \[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})=\frac{91}{216}\ne \frac{1}{2}\]
Therefore, we can say that ${{A}_{1}},\text{ }{{A}_{2}}\text{ and }{{A}_{3}}$are not mutually exclusive events, so
\[P({{A}_{1}}\cup {{A}_{2}}\cup {{A}_{3}})\ne P({{A}_{1}})+P({{A}_{2}})+P({{A}_{3}})\]
