Given observations arranged in ascending order are
38, 42, 57, 58, 61, 65, 66, 72 and n = 8
D₆ = Size of 6\({ \left( \frac { n+1 }{ 10 } \right) }^{ th }\) value = Size of \(6{ \left( \frac { 8+1 }{ 10 } \right) }^{ th }\) value
= Size of 5.4^th value \(\simeq \) Size of 5^th value
= 61
D₈ = Size of \(8{ \left( \frac { n+1 }{ 10 } \right) }^{ th }\) value = Size of \(8{ \left( \frac { 8+1 }{ 10 } \right) }^{ th }\) value
=size of 7.2^th value \(\simeq \) size of 7^th value
= 66
P₇ = Size of \(7{ \left( \frac { n+1 }{ 100 } \right) }^{ th }\) value = size of \(7{ \left( \frac { 8+1 }{ 100 } \right) }^{ th }\) value
= size of (0.6)^th value \(\simeq \) size of first value
=38
P₂₀=Size of 20\({ \left( \frac { n+1 }{ 100 } \right) }^{ th }\) value = Size of \({ 20\left( \frac { 9 }{ 100 } \right) }^{ th }\) value
= Size of (1.8^th value) \(\simeq \) size of 2^nd value
= 42

Given \(P(A)=\frac { 1 }{ 2 } \Rightarrow p(\bar { A } )=1-P(A)=1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
\(P(B)=\frac { 2 }{ 3 } \Rightarrow P(\bar { B } )=1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
\(P(C)=\frac { 3 }{ 4 } \Rightarrow P(\bar { C } )=1-P(C)=1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
P(Only one of them hits the target)
\(=P(A\cap \bar { B } \cap \bar { C } )+P(\bar { A } \cap B\cap \bar { C } )+P(\bar { A } \cap \bar { B } \cap C)\)
\(=P(A).P(\bar { B) } .P(\bar { C) } +P(\bar { A } ).P(B).P(\bar { C } )+P(\bar { A } ).P(\bar { B } ).P(C)\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 3 }{ 4 } \)
\(=\frac { 1 }{ 24 } +\frac { 2 }{ 24 } +\frac { 3 }{ 24 } =\frac { 6 }{ 24 } =\frac { 1 }{ 4 } \)

Total Number of bolts produced = 1000 + 2000 + 3000 = 6000
\(P({ A }_{ 1 })=\frac { 1000 }{ 6000 } =\frac { 1 }{ 6 } \)
\(P({ A }_{ 2 })=\frac { 2000 }{ 6000 } =\frac { 1 }{ 3 } \)
\(P({ A }_{ 3 })=\frac { 3000 }{ 6000 } =\frac { 1 }{ 2 } \)
Let B be the event of selecting defective bolts.
\(\therefore \) P(B/A₁) = 1% = \(\frac { 1 }{ 100 } \) = 0.01
P(B/A₂) = 1.5% = 0.015
and (P(B/A₃) = 2% = 0.02
\(\therefore P(A_{ 1 }/B)=\frac { P({ A }_{ 1 }).P\left( B/{ A }_{ 1 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) +P({ A }_{ 3 }).P\left( B/{ A }_{ 3 } \right) } \)
\(=\frac { \frac { 1 }{ 6 } \times 0.01 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.015+\frac { 1 }{ 2 } \times 0.02 } =\frac { 1 }{ 600 } \times 60=\frac { 1 }{ 10 } \)
\(\therefore P(A_{ 1 }/B)=0.1\)

Let A₁, A₂ and A₃ be the three boxes, then
P(A₁) = P(A₂) = P(A₃) = 1/3
let B be the event of getting 1 white and 1 red
P(B/A₁) = P(getting 1 white and 1 red ball from A₁)
\(=\frac { 1{ C }_{ 1 }\times 2{ C }_{ 1 } }{ 6{ C }_{ 2 } } =\frac { 2 }{ \frac { 6\times 5 }{ 2 } } =\frac { 4 }{ 6\times 5 } =\frac { 2 }{ 15 } \)
P(B/A₂)=P(getting 1 white and 1 red ball from A₂)
\(=\frac { 2{ C }_{ 1 }\times 3{ C }_{ 1 } }{ 6{ C }_{ 2 } } =\frac { 2\times 3 }{ \frac { 6\times 5 }{ 2 } } =\frac { 6\times 2 }{ 6\times 5 } =\frac { 2 }{ 5 } \)
PB/A₃) = P(getting 1 white and 1 red ball from A₃)
\(=\frac { 2{ C }_{ 1 }\times 1{ C }_{ 1 } }{ 5{ C }_{ 2 } } =\frac { 2\times 1 }{ \frac { 5\times 4 }{ 2\times 1 } } =\frac { 4 }{ 5\times 4 } =\frac { 1 }{ 5 } \)
By Baye's theorem,
\(P(A_{ 2 }/B)=\frac { P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 })P(B/{ A }_{ 2 })+P({ A }_{ 3 }).P(B/{ A }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } \times \frac { 2 }{ 5 } }{ \frac { 1 }{ 3 } \times \frac { 2 }{ 15 } +\frac { 1 }{ 3 } \times \frac { 2 }{ 5 } +\frac { 1 }{ 3 } \times \frac { 1 }{ 5 } } =\frac { \frac { 2 }{ 15 } }{ \frac { 2 }{ 45 } +\frac { 2 }{ 15 } +\frac { 1 }{ 15 } } =\frac { \frac { 2 }{ 15 } }{ \frac { 2+6+3 }{ 45 } } \)
\(=\frac { \frac { 2 }{ 15 } }{ \frac { 11 }{ 45 } } =\frac { 2 }{ 15 } \times \frac { 45 }{ 11 } =\frac { 6 }{ 11 } \)
\(P(A_{ 2 }/B)=\frac { 6 }{ 11 } \)

Let A = {First ball drawn is green}
B = {Second ball drawn is green}
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 6 }{ 10 } \)   [\(\because\) Total number of balls = 4+6 = 10]
Let C = {getting green ball after taking out first green ball}
\(P(C)=\frac { 5{ C }_{ 1 } }{ 9{ C }_{ 1 } } =\frac { 5 }{ 9 } \)   [\(\because\) First ball is not replaced]
\(\therefore\) P(getting green ball) = P(A).P(C) = \(\frac { 6 }{ 10 } \times \frac { 5 }{ 9 } \)
\(\therefore P(A\cap B)=\frac { 1 }{ 3 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 3 } }{ \frac { 6 }{ 10 } } \)
\(P(B/A)=\frac { 1 }{ 3 } \times \frac { 10 }{ 6 } =\frac { 5 }{ 9 } \)