The odds against A being selected for a job are 3: 5 imply that the probability of the rejection of A for a job is $\(\frac{3}{3+5}=\frac{3}{8}(=\overline{\mathrm{A)\)$ whereas the probability of A's selection is $\(\frac{5}{3+5}=\frac{5}{8}(=\mathrm{A})\)$

Similarly the odds in favor of B being selected are 5: 6 imply that the probability of selection and rejection of $B$ is $\(\frac{5}{5+6}=\frac{5}{11}(=\mathrm{B}) \& \frac{6}{5+6}=\frac{6}{11}(=\vec{B})\)$ respectively.

Now, the probability of A \& B both being selected \(= Probability (A selected)\) $\(\times$ Probability (B selected $)=\mathrm{P}(\mathrm{A}) \times \mathrm{P}(\mathrm{B})=\frac{5}{8} \times \frac{5}{11}=\frac{25}{88}\)$

Next the probability that exactly one of $\(A \& B\)$ is selected $\(=P(\bar{A}) \times P(B)+P(A) \times P(\bar{B})\)$ $\(=\frac{3}{8} \times \frac{5}{11}+\frac{5}{8} \times \frac{6}{11}=\frac{15+30}{88}=\frac{45}{88}\)$

Hence options (B) \& (D) are correct. .