(i) p(A) = 0.5, P(B)= 0.6,\(P(A\cap B)=0.24\)
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) 
(i.e.,) \(P(A\cup B)\) = 0.5 + 0.6 - 0.24
= 1.1- 0.24 = 0.86
\(\therefore P(A\cup B)=0.86\) 
(ii) \(P(\bar { A } \cap B)=P(B)-P(A\cap B)\) 
= 0.6 - 0.24 = 0.36
(iii) \(P(A\cap \bar { B } )=P(A)-P(A\cap B)\) 
= 0.5 - 0.24 = 0.26
(iv) \(P(\bar { A } \cup \bar { B } )=P\{ A\cap B'\} =1-P(A\cap B)\) 
= 1 - 0.24 = 0.76
(v) \(P(\bar { A } \cap \bar { B } )=P\left\{ A\cup B' \right\} =1-P(A\cup B)\) 
= 1- 0.86 = 0.14.

n(S) = 6 \(\times\) 6 \(\times\) 6 = 216
A = {(1, 1, 4) (1, 2, 4) (1, 3, 4) (1, 4, 4) (1, 5, 4)(1, 6, 4) (2, 1, 4) (2, 2, 4) (2, 3, 4) (2, 4, 4) (2, 5, 4) (2, 6, 4) (6, 1, 4) (6, 2, 4) (6, 3, 4) (6, 4, 4) (6, 5, 4) (6, 6, 4)}
B = { (6, 5, 1) (6, 5, 2) (6, 5, 3) (6, 5, 4) (6, 5, 5) (6, 5, 6)}
\(\Rightarrow\) n(B) = 6
\(\Rightarrow P(B)=\frac { 6 }{ 216 } \)
\(A\cap B=\{ (6,\ 5,\ 4)\} \)
\(\Rightarrow \quad n(A\cap B)=1\)
\(\therefore P(A\cap B)=\frac { 1 }{ 216 } \)
\(\therefore P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 216 } \times \frac { 216 }{ 6 } =\frac { 1 }{ 6 } \)
\(\therefore P(A/B)=\frac { 1 }{ 6 } \)

n(S) = 6
Given n(A) = 3, n(B) = 2, n(C) = 4
A\(\cap \)B = {3}
\(\Rightarrow n(A\cap B)=1\)
\(\Rightarrow P(A\cap B)=1\)
\(P(A\cap B\cap C)=\frac { 1 }{ 6 } \)
\(A\cap B\cap C=\{ 3\} \)
\(\Rightarrow \quad n(A\cap B\cap C)=\frac { 1 }{ 6 } \)
(i) \(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 6 } }{ \frac { 2 }{ 6 } } =\frac { 1 }{ 6 } \times \frac { 6 }{ 2 } =\frac { 1 }{ 2 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 6 } }{ \frac { 3 }{ 6 } } =\frac { 1 }{ 6 } \times \frac { 6 }{ 3 } =\frac { 1 }{ 3 } \)
\(\Rightarrow \quad P(A\cap B/C)=\frac { P(A\cap B\cap C) }{ P(C) } =\frac { \frac { 1 }{ 6 } }{ \frac { 4 }{ 6 } } =\frac { 1 }{ 6 } \times \frac { 6 }{ 4 } =\frac { 1 }{ 4 } \)

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

n(S) = 52
There are 4 kings and 4 queens in a pack of cards
\(\therefore\) n(E) = 8
There are 4 queens and 4 aces in a pack of cards
\(\therefore\) n(F) = 8
\(\therefore P(E)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(F)=\frac { 8 }{ 52 } =\frac { 2 }{ 13 } \) and \(P(E\cap F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 13 } }{ \frac { 2 }{ 13 } } =\frac { 1 }{ 13 } \times \frac { 13 }{ 2 } =\frac { 1 }{ 2 } \)