Let E₁ = the first compartment of the purse is chosen
E₂ = the second compartment of the purse is chosen
A = a rupee coin is drawn from the purse.
Since one of the two compartments is chosen randomly
\(P({ E }_{ 1 })=\frac { 1 }{ 2 } =P({ E }_{ 2 })\)
Also, P(A/E₁) = P(Drawing a rupee coin from the I compartment) = \(\frac { 2 }{ 5 } \)
P(A/E₂) = P(Drawing a rupee coin from the II compartment) = \(\frac { 3 }{ 5 } \)
\(\therefore\) By the law of total probability,
P(drawing a one-rupee coin) = P(A) = P(E₁).P(A/E₁)+P(E₂).P(A/E₂)
\(=\frac { 1 }{ 2 } \times \frac { 2 }{ 5 } +\frac { 1 }{ 2 } \times \frac { 3 }{ 5 } =\frac { 2 }{ 10 } +\frac { 3 }{ 10 } =\frac { 5 }{ 10 } =\frac { 1 }{ 2 } \)

We haveP(not A or not B) = 1/4
\(P(\overline { A } U\overline { B } )=\frac { 1 }{ 4 } \) \(\Rightarrow P(\overline { A\cap B } )=\frac { 1 }{ 4 } \)
\(\Rightarrow 1-P(A\cap B)=\frac { 1 }{ 4 } \)   \(\Rightarrow P(A\cap B)=1-\frac { 1 }{ 4 } =\frac { 3 }{ 4 } \)
\(\therefore P(A\cap B)=\frac { 3 }{ 4 } \)
Now, P(A). P(B) = \(\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } =\frac { 7 }{ 24 } \)
\(\therefore P(A\cap B)\neq P(A).P(B)\)
So, A and B are not independent events.

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

Total number of ways in which mother (M), Father (F) and Son (S) can be lined up at random in one of the following ways:
MFS, MSF, FMS, FSM, SFM, SMF
\(\therefore\) A = {SMF, SFM, MFS, FMS}
and B = {MFS, SFM}
\(\therefore A\cap B=\){MFS, SFM}
Clearly, n(A\(\cap \)B) = 2 and n(B) = 2
\(\therefore\) Required probability
\(P(A/B)=\frac { P(A\cap B) }{ n(B) } =\frac { n(A\cap B) }{ n(B) } =\frac { 2 }{ 2 } =1\)

Given n(A) = 3, n(B) = 2 and n(C) = 4
A\(\cap \)B = {3}
A\(\cap \)C = {2, 5}
\(\therefore\) n(A\(\cap \)B) = 1
and n(A\(\cap \)C) = 2
(i) P(A/B) = \(\frac { P(A\cap B) }{ P(B) } =\frac { n(A\cap B) }{ n(B) } =\frac { 1 }{ 2 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { n(A\cap B) }{ n(A) } =\frac { 1 }{ 3 } \)
(ii) \(P(A/C)=\frac { P(A\cap C) }{ P(C) } =\frac { n(A\cap C) }{ n(C) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \)
\(P(C/A)=\frac { P(A\cap C) }{ P(A) } =\frac { n(A\cap C) }{ n(A) } =\frac { 2 }{ 3 } \)