(i) If A is a subset of B, then
A\(\cap \)B = A
\(\Rightarrow\) n(A\(\cap \)B) = n(A)
\(\Rightarrow\) P(A\(\cap \)B) = P(A)
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { P(A) }{ P(A) } =1\)
(ii) If \(A\cap B=\phi \) then n(A\(\cap \)B) = 0 \(\Rightarrow\) P(A\(\cap \)B) = 0
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 0 }{ P(A) } =0\)

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

Consider the following events.
A: Person gets an electric contract
B: Person gets plumbing contract
Given P(A) = \(\frac { 2 }{ 5 } ,P(\bar { B } )=\frac { 4 }{ 7 } \)and P(AUB) = \(\frac { 2 }{ 3 } \)
We know, P(AUB) = P(A) + P(B) - P(\(A\cap B\))
\(\Rightarrow \frac { 2 }{ 3 } =\frac { 2 }{ 5 } +\left( 1-\frac { 4 }{ 7 } \right) -P(A\cap B)\)
\(\Rightarrow \frac { 2 }{ 3 } -\frac { 2 }{ 5 } =\frac { 3 }{ 7 } -P(A\cap B)\)
\(\Rightarrow P(A\cap B)=\frac { 2 }{ 5 } +\frac { 3 }{ 7 } -\frac { 2 }{ 3 } =\frac { 17 }{ 105 } \)

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 \(P(E\cap F)=\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
\(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { \frac { 1 }{ 13 } }{ \frac { 2 }{ 13 } } =\frac { 1 }{ 13 } \times \frac { 13 }{ 2 } =\frac { 1 }{ 2 } \)

Here n(s) = 36
Let A be the event of getting 3 on first die and B be the event of getting the sum of 8 or greater.
\(\therefore\) A = {(3, 1) (3, 2) (3, 3)(3, 4) (3, 5) (3, 6)}
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2) (3, 6) (4, 5) (5, 4) (6, 3) (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}
and \(A\cap B\) = {(3, 5), (3, 6)}
\(\therefore P(A)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)   \(P(B)=\frac { 15 }{ 36 } =\frac { 5 }{ 12 } \) and \(P(A\cap B)=\frac { 2 }{ 36 } =\frac { 1 }{ 18 } \)
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 18 } }{ \frac { 1 }{ 6 } } =\frac { 1 }{ 18 } \times \frac { 6 }{ 1 } =\frac { 1 }{ 3 } \)

Sample space S = {1, 2, 3, 4, 5, 6}.
Let A be the event of die shows an odd number.
Let B be the event of getting 5.
Then, A = {1, 3, 5}, B = {5}, and A\(\cap \)B = {5}.
Therefore, P(A) = \(\frac{3}{6}\) and P(A\(\cap \)B) = \(\frac{1}{6}\)
P(getting 5 / die shows an odd number) = P(B / A)
= \(\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 6 } }{ \frac { 3 }{ 6 } } \)
P(B/A) = \(\frac{1}{3}\) .

The sample space is
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, n(S) = 10
Let A be the event of choosing an even number and B be the event of choosing an integer multiple of three.
A = {2, 4, 6, 8, 10}, n(A) = 5,
B = {3, 6, 9} , n(B) = 3
P(choosing an even integer) = P(A) = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 10 } =\frac { 1 }{ 2 } \)
P(choosing an integer multiple of three) = P(B) =\(\frac { n(B) }{ n(S) } =\frac { 3 }{ 10 } \) 

When'a die is thrown twice, then the sanmple space
\(S=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)
n(S) = 36
A be the event, first die shows 5 and
B be the event, second die shows 5 then,
\(A =\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\)
\(n(A) =6\)
\(B =\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)\} \)
\(n(B) =6 \)
\(A \cap B =\{(5,5)\}, \quad n(A \cap B)=1\)
Now, \(P(A \cup B) =P(A)+P(B)-P(A \cap B) \)
\(=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)} \)
\(=\frac{6}{36}+\frac{6}{36}-\frac{1}{36} \)
\(=\frac{11}{36}\)

Given \(P(A)=0.35\)
\(P(A \text { or } B)=0.85 \text { and } P(A \text { and } B)=0.15\)
\((i) P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.85 =0.35+P(B)-0.15 \)
\(P(B) =0.85-0.20\)
\(=0.65\)
\((ii) P(\bar{B})=1-P(B)=1-0.65=0.35\)
\((iii) P( only\ A)=P(A)-P(A \cap B)\)
\(=0.35-0.15=0.20\)

Let S be the sample space,
A be the event of getting an even number,
B be the event of getting multiple of three.
Therefore,
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) =6 
A = {2, 4, 6} ⇒ n(A) = 3
B = {3, 6} ⇒ n(B) = 2
The required probabilities are
(i) P (getting an even number) = P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) P (getting multiple of three) = P(B) = \(\frac { n(B) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)