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) =\(\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 } \)

We know that the sample space is S = {1,2,3,4,5,6}
Now, A = {3,6} ; B = { 2,4,6} then (A∩B) = {6}
P(A) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
P(B) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) and P(A∩B) = \(\frac{1}{6}\)
Clearly P(A∩B) = P(A) P(B)
Hence A and B are independent events.

\(n(S)=52{ C }_{ 2 }=\frac { 52\times 51 }{ 2\times 1 } =1326\)
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 = {(1, 1) (1, 2) (1,3), ..... (6, 6)}
(2, 1), (2, 2) ...... (2, 6)
(6, 1), (6, 2)  ..... (6, 6)}
n(S) = 36
Let B be the event that sum of numbers appearing is 6
B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\(P(B)=\frac{5}{36}\)
Let A be the event that 4 has appeared atleast once
A = {(2, 4), (4, 2)}
\(A\cap B\)  = {(2, 4),(4, 2)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

Arranging in ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
n = 11
 Q₁ = size of \((\frac { { n+1} }{ 4 } )^{ th }\)  value
= size of \({ \left( \frac { 11+1 }{ 4 } \right) }^{ th }\) value
= size of 3^rd value = 6
Q₃ = Size of \({ \left( \frac { 3(n+1) }{ 4 } \right) }^{ th }\) value
= size of \({ \left( 3\left( \frac { 11+1 }{ 4 } \right) \right) }^{ th }\) value
=size of 3(3)^th value
= size of 9^th value = 18

(i) S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5}
n(A) = 3
\(P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event that the number is greater than or equal to 3.
B = {3, 4, 5, 6}    
n(B) = 4
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Let n(s) denote the number of men psychologist.
\(\therefore\) n(s) = 50
Let n(A) denote the number of men psychologist
n(A) = 15
P(A) = \(\frac { 15 }{ 50 } =\frac { 3 }{ 10 } \)

Harmonic mean is useful in calculating average speed
HM = \(\frac { n }{ 1/a+1/b+1/c+1/d } =\frac { 4 }{ 1/100+1/200+1/300+1/400 } \)
\(=\frac { 4 }{ \frac { 12+6+4+3 }{ 1200 } } =\frac { 4 \times 1200}{ 25 } \)
= 192 km/hr

n{S) = 1000
Let A be the event that the student selected is from XI std and B be the event that the selected student is a girl.
n(B) = 450
\(P(B)=\frac { 450 }{ 1000 } \)
n(A\(\cap \)B) = \(\frac { 20 }{ 100 } \times 450=90\)
P(A\(\cap \)B) \(=\frac { 90 }{ 1000 } \)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 90 }{ 1000 } }{ \frac {450 }{ 1000} } =\frac { 1 }{ 5 } =0.20\)