A and B are independent events with $\mathrm{P}(\mathrm{A})=\frac{3}{10}$,…
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
$\frac{41}{50}$
$\frac{41}{125}$
$\frac{7}{25}$
$\frac{7}{50}$
Solution
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]
$\begin{aligned}
& =\frac{7}{10}+\frac{2}{5}-\frac{7}{10} \times \frac{2}{5} \\
& =\frac{41}{50}
\end{aligned}$