(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
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

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 \(=\frac { 15}{ 45 } =0.33\)

GM = Anti \(\log { \left( \frac { \sum { logX } }{ n } \right) } ;\) where n = 10
GM = Anti log \(\left( \frac { 17.6373 }{ 10 } \right) \)
= Anti log(1.7637)
GM = 58.03

Find out the coefficient of mean deviation about median

Mean deviation about median = \(\frac { \sum { f|D| } }{ \sum f } =\frac { 2225.04 }{ 150 } =14.83\)
Median = 45.14
MD = 14.83

P(A₁) = P(that the machine A₁ produces screws) = \(\frac{1000}{6000}=\frac{1}{6}\)
P(A₂) = P(that the machine A₂ produces screws) = \(\frac{2000}{6000}=\frac{1}{3}\)
P(A₃) = P(that the machine A₃ produces screws) = \(\frac{3000}{6000}=\frac{1}{2}\)
Let B be the event that the chosen screw is defective
\(\therefore\) P(B/A₁) = P(that defective screw from the machine A₁) = 0.01
P(B/A₂) = P(that defective screw from the machine A₂) = 0.015 and
P(B/A₃) = P(that defective screw from the machine A₃) = 0.02
We have to find P(A₁/B)
Hence by Baye’s theorem, we get
\(P(A_1/B)=\frac{P(A_1)P(B/A_1)}{P(A_1)P(B/A_1)+P(A_2)P(B/A_2)+P(A_3)P(B/A_3)}\)
\(=\frac{(\frac{1}{6})(0.01)}{(\frac{1}{6})(0.01)+(\frac{1}{3})(0.015)+(\frac{1}{2})(0.02)}\)
= \(\frac{0.01}{0.01+0.03+0.06}=\frac{0.01}{0.1}=\frac{1}{10}\)

Q₁ = Size of \(\left( \frac { N }{ 4 } \right) ^{ th }\)value =\(\left( \frac { 67 }{ 4 } \right) ^{ th }\)= 16.75^th value Thus Q₁ lies in the class 20 – 30; and the corresponding values are
L = 20,\(\frac { N }{ 4 } \) = 16.75 ; pcf = 12, f = 19, c = 10
\({ Q }_{ 1 }=L+\left( \frac { \frac { N }{ 4 } -pcf }{ f } \right) \times c\)
\({ Q }_{ 1 }=20+\left( \frac { 16.75-12 }{ 19 } \right) \times 10\)= 20 + 2.5 = 22.5
Q₃ = Size of \(\left( \frac { 3N }{ 4 } \right) ^{th}\) value = 50.25^th value
Thus Q₃ lies in the class 50 – 60 and corresponding values are 
L = 50 ; \(\frac { 3N }{ 4 } \)= 50.25 ; pcf = 46, f = 9, c = 10
\({ Q }_{ 3 }=L+\left( \frac { \frac { 3N }{ 4 } -pcf }{ f } \right) \times c\)
\({ Q }_{ 3 }=20+\left[ \frac { 50.25-46 }{ 9 } \right] \times 10=54.72\)
\(QD=\frac{1}{2}(Q_3-Q_1)\)
\(=\frac { 54.72-22.5 }{ 2 } =16.11\)
\(\therefore\)  QD = 16.11

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

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