Probability of solving the problem by \\\\(A\\\\), \\\\(P\\left( A \\right) = \\displaystyle\\frac { 1 }{ 2 }\\\\)Probability of solving the problem by \\\\(B\\\\), \\\\(P\\left( B \\right) = \\displaystyle\\frac { 1 }{ 3 }\\\\)Since the problem is solved independently by \\\\(A\\\\) and \\\\(B\\\\),\\\\(\ herefore P\\left( AB \\right) = P\\left( A \\right) \\cdot P\\left( B \\right) = \\displaystyle\\frac { 1 }{ 2 } \ imes \\displaystyle\\frac { 1 }{ 3 } = \\displaystyle\\frac { 1 }{ 6 }\\\\)\\\\( P\\left( A\\prime \\right) = 1 - P\\left( A \\right) = 1 - \\displaystyle\\frac { 1 }{ 2 } = \\displaystyle\\frac { 1 }{ 2 }\\\\)\\\\( P\\left( B\\prime \\right) = 1 - P\\left( B \\right) = 1 - \\displaystyle\\frac { 1 }{ 3 } = \\displaystyle\\frac { 2 }{ 3 }\\\\)(i) Probability that the problem is solved \\\\( = P\\left( A\\cup B \\right)\\\\)\\\\( = P\\left( A \\right) + P\\left( B \\right) - P\\left( AB \\right)\\\\)\\\\( = \\displaystyle\\frac { 1 }{ 2 } + \\displaystyle\\frac { 1 }{ 3 } - \\displaystyle\\frac { 1 }{ 6 }\\\\)\\\\( = \\displaystyle\\frac { 4 }{ 6 }\\\\)\\\\( = \\displaystyle\\frac { 2 }{ 3 }=0.66\\\\)(ii) Probability that exactly one of them solves the problem is given by,\\\\( P\\left( A \\right) \\cdot P\\left( B\\prime \\right) + P\\left( B \\right) \\cdot P\\left( A\\prime \\right)\\\\)\\\\( = \\displaystyle\\frac { 1 }{ 2 } \ imes \\displaystyle\\frac { 2 }{ 3 } + \\displaystyle\\frac { 1 }{ 2 } \ imes \\displaystyle\\frac { 1 }{ 3 }\\\\)\\\\( = \\displaystyle\\frac { 1 }{ 3 } + \\displaystyle\\frac { 1 }{ 6 }\\\\)\\\\( = \\displaystyle\\frac { 1 }{ 2 }=0.50 \\\\)