To find the average prices for shares we find harmonic mean:
\(\therefore\) Harmonic mean = \(\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } } =\frac { 4 }{ \frac { 1 }{ 10 } +\frac { 1 }{ 15 } +\frac { 1 }{ 20 } +\frac { 1 }{ 30 } } \)
\(=\frac { 4 }{ \frac { 6+4+3+2 }{ 60 } } =\frac { 4 \times 60}{ 15 } \)
= 16

Since A and B are independent events then P(A∩B) = P(A) P(B)
Given that P(A) =\(\frac{3}{5}\) and P(B) =\(\frac{1}{5}\),
then P(A∩B) =\(\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 } \)

\(n(S)=52{ C }_{ 2 }=\frac { 52\times 51 }{ 2\times 1 } =1326\)
Let A be the event that one king and one queen card is drawn
n(A) = 4C₁ \(\times\) 4C₁ = 16
\(P(A)=\frac { 16 }{ 1326 } =0.012\)

S {(G₁, G₂), (G₁, B₂), (B₁, G₂), (B₁, B₂)}
n(S) = 4
Let A be the event that both are girls
A= {(G₁ G₂)}
n(A) = 1
\(P(A)=\frac{1}{4}\)
Let B to the event that atleast one of the children is a girl.
B = {(G₁ G₂), (G₁ B₁), (G₂ B₂)}
\(P(B)=\frac{3}{4}\)
\( A\cap B=\{ ({ G }_{ 1 }{ G }_{ 2 })\}, P(A\cap B)=\frac{1}{4}\)
\(P(A / B)=\frac { P(A\cap B) }{ P(B) } =\frac{\frac{1}{4}}{\frac{3}{4}}=\frac { 1 }{ 3 } \)

Since we are given rate per rupee, harmonic mean will give the correct answer.
\(HM=\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } } \)
\(=\frac { 4 }{ \frac { 1 }{ 1 } +\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } } \)
\(=\frac { 4\times 12 }{ 25 } \)
= 1.92 kg per rupee.