if the

It is given that $\mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11}$, and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$

(i) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}$

$\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{7}{11}$

$\Rightarrow \frac{6}{11}+\frac{5}{11}-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{7}{11}$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{11}{11}-\frac{7}{11}=\frac{4}{11}$

(ii) It is known that, $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

$\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\frac{4}{11}}{\frac{5}{11}}=\frac{4}{5}$

(iii) It is known that, $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$

$\Rightarrow P(B \mid A)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}$