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
\(\therefore P(A_{ 1 }/B)=\frac { P({ A }_{ 1 }).P\left( B/{ A }_{ 1 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) +P({ A }_{ 3 }).P\left( B/{ A }_{ 3 } \right) } \)
\(=\frac { \frac { 1 }{ 6 } \times 0.01 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.015+\frac { 1 }{ 2 } \times 0.02 } =\frac { 1 }{ 600 } \times 60=\frac { 1 }{ 10 } \)
\(\therefore P(A_{ 1 }/B)=0.1\)

Q₁= Size of \({ \left( \frac { N }{ 4 } \right) }^{ th }\)value = size of\({ \left( \frac { 43 }{ 4 } \right) }^{ th }\) value
= Size of 10.75^th value
Q₁ lies in the interval (40-50) and its corresponding values are L = 40, f = 20, pcf = 8, c = 10
\({ Q }_{ 1 }=L+\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) \times c=40+\frac { 10.75-8 }{ 20 } \times 10\)
\(=40+\frac { 2.75 }{ 2 } =40+1.375\)
Q₁ = 43.38
Q₃=Size of \({ \left( \frac { 3N }{ 4 } \right) }^{ th }\) value = size of\({ \left( \frac { 3\times 43 }{ 4 } \right) }^{ th }\) value
= Size of (32.25)^th value
\(\therefore\) Q₃ lies in (50-60) and its corresponding values are L = 50, f = 10, pcf = 28 and c = 10
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) \times c=50+\frac { 32.25-28 }{ 10 } \times 10\)
\(\therefore\) Q₃ = 50 + 4.25 = 54.25
\(\therefore\) Quartile deviation QD = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })=\frac { 1 }{ 2 } (54.25-43.38)=\frac { 1 }{ 2 } (10.87)=5.435\)

Let A₁ and A₂ are two boxes.
P(A₁) = P(A₂) = 1/2
P(B/A₁) = P(White ball from I box) = 4/7
P(B/A₂) = P(White ball from II box) = 3/10
P(B) = P(White ball from A₁ or A₂)
= P(A₁). P(B/A₁)+P(A₂).P(B/A₂)
\(=\frac { 1 }{ 2 } \times \frac { 4 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 3 }{ 10 } =\frac { 2 }{ 7 } +\frac { 3 }{ 20 } =\frac { 40+21 }{ 140 } =\frac { 61 }{ 140 } \)
By Baye's theorem,
\(P({ B }_{ 1 }/A)=\frac { P({ A }_{ 1 }).P(B/{ A }_{ 1 }) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P(B/{ A }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } \times \frac { 4 }{ 7 } }{ \frac { 1 }{ 2 } \times \frac { 4 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 3 }{ 10 } } =\frac { \frac { 2 }{ 7 } }{ \frac { 61 }{ 140 } } =\frac { 2 }{ 7 } \times \frac { 140 }{ 61 } \)
\(\therefore P({ B }_{ 1 }/A)=\frac { 40 }{ 61 } \)

Given \(P(A)=\frac { 2 }{ 5 } ,P(B)=\frac { 1 }{ 3 } \) and \(P(C)=\frac { 1 }{ 2 } \)
\(P(A\cap C)=\frac { 1 }{ 5 } \) and \(P(B\cap C)=\frac { 1 }{ 4 } \)
\(\therefore P(C/B)=\frac { P(C\cap B) }{ P(B) } =\frac { \frac { 1 }{ 4 } }{ \frac { 1 }{ 3 } } =\frac { 3 }{ 4 } \)
and \(P(\bar { A } \cap \bar { C } )=P\left( \overline { A\cup C } \right) =1-P(A\cup C)\)
= 1 - [P(A) + P(C) - \(P(A\cap C)\)]
= 1-\(\left( \frac { 2 }{ 5 } +\frac { 1 }{ 2 } -\frac { 1 }{ 5 } \right) =1-\left( \frac { 4+5-2 }{ 10 } \right) =1-\frac { 7 }{ 10 } \)
\(=\frac { 10-7 }{ 10 } =\frac { 3 }{ 10 } \)

The events favourable to A are
{(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)} and to B are {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)}
\(\Rightarrow\) n(A) = 6, n(B) = 6
\(\therefore\) \(A\cap B=\) {(6, 2)}
\(\Rightarrow\) \(n(A\cap B)\) = 1
\(P(A)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
\(P(B)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)
and \(P(A\cap B)=\frac { 1 }{ 36 } =\frac { 1 }{ 6 } \times \frac { 1 }{ 6 } \)
= P(A). P(B)
\(\therefore P(A\cap B)\)  = P(A).P(B)
Hence, A and B are independent events.