$\begin{array}{1 1}(i) 2/3 (ii) 1/2 \\(i) 1/2 (ii) 2/3\\ (i) 1/3 (ii) 1/2 \\ (i) 1/3 (ii) 2/3 \end{array} $

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- If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
- P (\(\bar{A})\) = 1 - P(A)
- P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)

Given P (A solving the problem) = P (A) = $\large\frac{1}{2}$ and P (B solving the problem) = P(B) = $\large\frac{1}{3}$.

(i) P (the problem is solved)

P (problem is solved) = 1 - P (problem is not solved).

P (problem is not solved by A) = 1 - P (A solving the problem) = 1 - $\large\frac{1}{2} = \large\frac{1}{2}$

P (problem is not solved by B) = 1 - P (B solving the problem) = 1 - $\large\frac{1}{3} = \large\frac{2}{3}$

P (neither A nor B solved the problem) = P (problem not solved by A) $\times$ P (problem not solved by B) = $\large \frac{1}{2}$ $\times$ $\large \frac{2}{3}$ = $\large\frac{1}{3}$

Therefore P (problem is solved) = 1 - $\large\frac{1}{3} = \frac{2}{3}$

(ii) P (exactly one of them solves the problem)

P (exactly one of them solve the problem) = P (A solved the problem but B doesnt) + P (B solved the problem but A doesnt)

P (A solved the problem but B doesnt) = P(A) $\times$ P ($\bar{B}$) = $\large\frac{1}{2}\;$$\times$ $\large\frac{2}{3} = \frac{1}{3}$

P (B solved the problem but A doesnt) = P(B) $\times$ P ($\bar{A}$) = $\large\frac{1}{3}\;$ $\times$ $\large\frac{1}{2} = \frac{1}{6}$

Therefore P (exactly one of them solves the problem) = $\large\frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{1}{2}$

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