Given the probability that A will be able to solve the problem = \(\frac{90}{100}=\frac{9}{10}\) and the probability that B will be able to solve the problem = \(\frac{70}{100}=\frac{7}{10}\)
i.e., P(A) = \(\frac{9}{10}\) and P(B) = \(\frac{7}{10}\)
\(P\left( \overline { A } \right) \) = 1 - P(A) = \(\frac{9}{10}=\frac{1}{10}\)
\(P\left( \overline { B } \right) \) = 1 - P(B) = \(\frac{7}{10}=\frac{3}{10}\)
P(at least one solve the problem) = \(P\left( A\cup B \right) \)
= \(1-P\left( \overline { A\cup B } \right) =1-P\left( \overline { A } \cap \overline { B } \right) \)
= \(1-P\left( \overline { A } \right) .P\left( \overline { B } \right) \)
= \(1-\frac{3}{100}=\frac{97}{100}\)
Hence the probability that at least one of them will solve the problem = \(\frac{97}{100}\)

Let the events E₁, E₂ and A be defined as
E₁ - event that male subscriber is under 30
E₂ - event that male subscriber is above 30
A is event that subscriber is male
P(E₁) = 0.80 and P(E₂) = 1 - 0.8 = 0.2
P(B/E₁) = 0.20, P(B/E₂) = 0.30
\(P({ E }_{ 1 }/B)=\frac { P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) +P({ E }_{ 2 }).P\left( \frac { B }{ { E }_{ 2 } } \right) } \)
\(=\frac { 0.8\times 0.2 }{ 0.8\times 0.2+0.2\times 0.3 } \)
\(=\frac { 0.16 }{ 0.16+0.06 } =\frac { 16 }{ 22 } =0.727\)

Let E₁, E₂, E₃ and A be the events defined as
E₁ = Screw is manufactured by machine A
E₂ = Screw is manufactured by machine B
E₃ = Screw is manufactured by machine C
A → screws drawn is defective
\(P({ E }_{ 1 })=\frac { 30 }{ 100 } \), \(P({ E }_{ 2 })=\frac { 40 }{ 100 } \), \(P({ E }_{ 3 })=\frac { 30 }{ 100 } \)
P(A/E₁) =\(\frac { 20 }{ 100 } \), P(A/E₂) = \(\frac { 4 }{ 100 } \), P(A/E₃) \(=\frac { 6 }{ 100 } \)
\(P({ E }_{ 3 }/A)=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/E_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 30 }{ 100 } \times \frac { 6 }{ 100 } }{ \frac { 30 }{ 100 } \times \frac { 2 }{ 100 } +\frac { 40 }{ 100 } \times \frac { 4 }{ 100 } +\frac { 30 }{ 100 } \times \frac { 6 }{ 100 } } \)
\(=\frac{180}{60+160+180}\)
\(=\frac { 180 }{ 400 } \)
= 0.45

Given P(A) = \(\frac{3}{4}\), P(B) = \(\frac{1}{2}\), P(C) = \(\frac{2}{3}\)
Then \(P\left( \overline { A } \right) =1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } ;P\left( \overline { B } \right) =1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \) and \(P\left( \overline { C } \right) =1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
(i) \(P\left( \overline { A } \right) \) = (all the three hit the targets)
= \(P\left( A\cap B\cap C \right) =P\left( A \right) P\left( B \right) P\left( C \right) \) (Since A, B, C hits independently)
= \(\frac{3}{4}.\frac{1}{2}.\frac{2}{3}=\frac{1}{4}\)
(ii) P(only one of them hits the target)
= \(P\{ \left( A\cap \overline { B } \cap \overline { C } \right) \cup \left( \overline { A } \cap B\cap \overline { C } \right) \cup \left( \overline { A } \cap \overline { B } \cap C \right) \} \)
= \(P\{ \left( A\cap \overline { B } \cap \overline { C } \right) +P\left( \overline { A } \cap B\cap \overline { C } \right) +P\left( \overline { A } \cap \overline { B } \cap C \right) \} \)
= \((\frac{3}{4}.\frac{1}{2}.\frac{1}{3})+(\frac{1}{4}.\frac{1}{2}.\frac{1}{3})+(\frac{1}{4}.\frac{1}{2}.\frac{2}{3})=\frac{1}{4}\)
(iii) P(at least one of them hit the target)
= 1– P(none of them hit the target)
= 1 - \(P\left( \overline { A } \cap \overline { B } \cap \overline { C } \right) \)
= 1 - \(P (\overline { A }) P(\overline { B }) P(\overline { C } )\)
= 1 - \(\frac{1}{24}=\frac{23}{24}\)

Let the events A, E₁, E₂, E₃ be defined as
A - Bolt is defective
E₁ - Bolt is manufactured by Machine A
E₂ - Bolt is manufactured by Machine B
E₃ - Bolt is manufactured by Machine C
P(E₁) = \(\frac{20}{100}\), P(E₂) = \(\frac{30}{100}\), P(E₃) = \(\frac{50}{100}\)
P(A/E₁) = \(\frac{7}{100}\),
P(A/E₂) = \(\frac{3}{100}\),
P(A/E₃) =\(\frac{5}{100}\)
\(P({ E }_{ 3 }/A)=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 50 }{ 100 } \times \frac { 5 }{ 100 } }{ \frac { 20 }{ 100 } \times \frac { 7 }{ 100 } +\frac { 30 }{ 100 } \times \frac { 3 }{ 100 } +\frac { 50 }{ 100 } \times \frac { 5 }{ 100 } } \)
\(=\frac { 250 }{ 140+90+250 } = \frac { 250 }{ 480 }  = \frac { 25 }{ 48 }\)
= 0.5208