A and B are independent events with \(\mathrm{P}(\mathrm{A})=\frac{3}{10}\), \(\mathrm{P}(\mathrm{B})=\frac{2}{5}\), then \(\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)\) has the value

Given that \(\mathrm{P}(\mathrm{A})=\frac{3}{10}, \mathrm{P}(\mathrm{B})=\frac{2}{5}\)
\(\begin{aligned}
\therefore \quad & \mathrm{P}\left(\mathrm{~A}^{\prime}\right)=\frac{7}{10} \\
& \mathrm{P}\left(\mathrm{~A}^{\prime} \cup \mathrm{B}\right) \\
& =\mathrm{P}\left(\mathrm{~A}^{\prime}\right)+\mathrm{P}(\mathrm{~B})-\mathrm{P}\left(\mathrm{~A}^{\prime} \cap \mathrm{B}\right) \\
& =\mathrm{P}\left(\mathrm{~A}^{\prime}\right)+\mathrm{P}(\mathrm{~B})-\mathrm{P}\left(\mathrm{~A}^{\prime}\right)(\mathrm{B})
\end{aligned}\)
\(\ldots\left[\because A\right.\) and \(B\) are independent, \(A^{\prime}\) and \(B\) are also independent]
\(\begin{aligned}
& =\frac{7}{10}+\frac{2}{5}-\frac{7}{10} \times \frac{2}{5} \\
& =\frac{41}{50}
\end{aligned}\)