For two events $\mathrm{A}$ and $\mathrm{B}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6},…
For two events $\mathrm{A}$ and $\mathrm{B}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3}$, then $\mathrm{A}$ and $\mathrm{B}$ are
independent
mutually exhaustive
mutually exclusive
complementary
Solution
$\begin{aligned}
& \text { We have, } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A})=\frac{1}{6}, \mathrm{P}(\mathrm{B})=\frac{2}{3} \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& \therefore \frac{5}{6}=\frac{1}{6}+\frac{3}{2}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0
\end{aligned}$
Thus A and B are mutually exclusive events.