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

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)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

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.

Here n = 11 observations are arranged into ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
D₂ = size of 2 \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₆ = size of 6  \(\left( \frac { n+1 }{ 10 } \right) \)^th value
D₂ = size of 2.4^th value ≈ size of 2^nd value = 4
D₆ = size of 7.2^th value ≈ size of 7^th value = 14

Arrange the values in ascending order 20, 30, 40, 45, 55, 60, 80.
Median = size of \( \left( \frac { \left( n+1 \right) }{ 2 } \right) ^{ th }\) value when n is odd
= size of \(\left( \frac { \left( 7+1 \right) }{ 2 } \right) ^{ th }\) value
= size of 4^th item = 45

MD about Median = \(\frac { \sum { |X – Median| } }{ n } =\frac { 105 }{ 7 } =15.0\)
Coefficient of MD about median \(=\frac { 15}{ 45 } =0.33\)

HM = \(\frac { N }{ \sum { \left( \frac { f }{ x } \right) } } =\frac { 37 }{ 0.9771 } \) = 37.86 km/hr

GM = Antilog\(\left( \frac { \sum { logx } }{ n } \right) =Anitlog\left( \frac { 6.3026 }{ 3 } \right) =Antilog(2.1009)\)
GM = 126.2
\(\therefore\) Average increase of the commodity from 2004 to 2007
= 126.1 - 100 = 26.1%

GM = Antilog\(\left( \frac { \sum { f\ log\ x } }{ N } \right) \)
\(=Anitlog\left( \frac { 66.771 }{ 31 } \right) \)
= 142.5 Ibs

Calculation of Harmonic Mean

\(HM=\frac { N }{ \sum { \left( \frac { f }{ X } \right) } } =\frac {50 }{ 3.46 } =14.45\)

Q₁ = Size of  \(\left( \frac { N+1 }{ 4 } \right) ^{th}\) value
= Size of \(\frac{43+1}{4}\) = 11^th value = 20
D₂ = Size of  \(\left( \frac { 2(N+1) }{ 100 } \right) ^{th}\) value
= Size of \(\frac{88}{10}\) = 8.8^th value = 20
P₉₀ = Size of \(\left( \frac { 90(N+1) }{ 100 } \right) ^{th}\) value
= Size of \(\frac{3960}{100}\)  = 39.6^th value = 50

Here the sample space of the experiment is
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
A = {Three heads or Three tails}
= {HHH, TTT}
B = {at least two heads}
= {HHH, HHT, HTH, THH} and
C = {at most two heads} = {HHT, HTH, HTT, THH, TTH, THT, TTT}
Also (A∩B) = {HHH}; (A∩C) = {TTT} and (B∩C) ={HHT, HTH, THH}
∴ P(A) = \(\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \); P(B) =\(\frac{1}{2}\); P(C) = \(\frac{7}{8}\) and
P(A∩B)= \(\frac{1}{8}\), P(A∩C)=\(\frac{1}{8}\), P(B∩C)=\(\frac{3}{8}\)
Also P(A). P(B)=\(\frac { 1 }{ 4 } .\frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
P(A). P(C) =\(\frac { 1 }{ 4 } .\frac { 7 }{ 8 } =\frac { 7 }{ 32 } \)
and P(B). P(C) =\(\frac { 1 }{ 2 } .\frac { 7 }{ 8 } =\frac { 7 }{ 16 } \)
Thus, P(A∩B) = P(A). P(B)
P(A∩C) ≠ P(A) P(C) and
P(B∩C) ≠ P(B). P(C)
Hence, the events (A and B) are independent, and the events (A and C) and (B and C) are dependent.

Let the events E₁, E₂ and A be defined as
E₁ - event that male subscriber is under 30
E₂ - event that male subscriber is above 30
A is event that subscriber is male
P(E₁) = 0.80 and P(E₂) = 1 - 0.8 = 0.2
P(B/E₁) = 0.20, P(B/E₂) = 0.30
\(P({ E }_{ 1 }/B)=\frac { P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) +P({ E }_{ 2 }).P\left( \frac { B }{ { E }_{ 2 } } \right) } \)
\(=\frac { 0.8\times 0.2 }{ 0.8\times 0.2+0.2\times 0.3 } \)
\(=\frac { 0.16 }{ 0.16+0.06 } =\frac { 16 }{ 22 } =0.727\)

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) \(=\frac { 90 }{ 1000 } \)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 90 }{ 1000 } }{ \frac {450 }{ 1000} } =\frac { 1 }{ 5 } =0.20\)

Q₁ = size of\({ \left( \frac { N }{ 4 } \right) }^{ th }\)value
= size of\({ \left( \frac { 68 }{ 4 } \right) }^{ th }\) value
= size of 17^th  value
Q₁ Lies in the C.I 20 - 30
L = 20, f = 13, pcf = 15, c = 10, \(\frac { N }{ 4 } =17\)
\({ Q }_{ 1 }=L+\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) c\)
\(=20+\frac { 17-15 }{ 13 } \times 10=20+\frac { 20 }{ 13 } = 20 + 1.54\)
= 21.54
Q₃ = Size of \({ \left( \frac { 3N }{ 4 } \right) }^{ th }\) value
= size of 3(17)^th value
= size of 51^st value
Q₃ lies in the C.I 40 - 50
\(L=40,\frac { 3N }{ 4 } =51\), pcf = 46, f = 14, c = 10
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) c\)
\(=40+\frac { 51-46 }{ 14 } \times 10=40+\frac { 50 }{ 14 } =43.57\)
Q.D = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })= \frac{43.57-21.54}{2}\)
\(=\frac { 22.03 }{ 2 } =11.02\)
Co-efficient of Q.D  \(=\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 43.57-21.54 }{ 43.57+21.54 } \)
\(=\frac { 22.03 }{ 65.11 } =0.3384\)

\(AM=\frac { \sum { fx } }{ N } =\frac { 2357 }{ 140 } =16.84\)
GM = Anti \(\log { \left( \frac { \sum { flogX } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 167.209 }{ 140 } \right) } \)
= Anti log(1.1944) = 15.65
\(HM=\frac { N }{ \sum { \left( \frac { f }{ X } \right) } } =\frac { 140 }{ 9.6852 } =14.46\)
i.e.,16.84 >15.65 >14.46
\(\therefore\) AM > GM > HM

6, 8, 9, 10, 11, 12, 14
\(\therefore \) Median = 10

GM = \(\sqrt{8 \times 18} = \sqrt{144} = 12\)