Probability of solving specific problem independently by

Probability of solving the problem by $A, P(A)=\frac{1}{2}$

Probability of solving the problem by $B, P(B)=\frac{1}{3}$

Since the problem is solved independently by A and B,

$\therefore \mathrm{P}(\mathrm{AB})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$

$\mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=1-\frac{1}{2}=\frac{1}{2}$

$\mathrm{P}\left(\mathrm{B}^{\prime}\right)=1-\mathrm{P}(\mathrm{B})=1-\frac{1}{3}=\frac{2}{3}$

i. Probability that the problem is solved $=P(A \cup B)$

$=P(A)+P(B)-P(A B)$

$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$

$=\frac{4}{6}$

$=\frac{2}{3}$

(ii) Probability that exactly one of them solves the problem is given by, $\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\mathrm{A}^{\prime}\right)$

$=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}$

$=\frac{1}{3}+\frac{1}{6}$

$=\frac{1}{2}$