King spade, Queen spade, Jack spade are removed
∴ total number of cards = 52 - 3 = 49.
(i) Probability (diamond)= \(\frac { 13 }{ 49 } \)
(ii) Probability (queen) \(\frac { 4-1 }{ 49 } =\frac { 3 }{ 49 } \)
(iii) Probability (spade) = \(\frac { 13-3 }{ 49 } =\frac { 10 }{ 49 } \)
(iv) Probability (heart bearing number 5) = \(\frac { 13-5 }{ 49 } =\frac { 8 }{ 49 } \)

Range = L- S = 20
Co-efficient of range = \(\frac { L-S }{ L+S } \) = 0.2
L - S = 20 ...(1)
L - S = 0.2(L + S)
(L + S) 0.2 = 20   ...(2)

L = \(\frac{24}{0.4}\) = 60
Substitute L = 60 in (1)
60 - S = 20
-S = 20 - 60 =-40
S = 40
∴ The largest is 60, the smallest is 40°.

Total students = 35
n(S) = 35
Boys : girls = 4 : 3
Let the number of boys = 4x
and number of girls = 3x
4x + 3x = 35
7x = 35
\(x=\frac{35}{7}=5\)
Number of boys = 4 x 5 = 20
Number of girls = 3 x 5 = 15
Boys are numbered = {1, 2,3,4,5,6,7,8, 9,10, 11, 12,13, 14, 15,16, 17, 18,19, 20}
Girls are numbered = {21, 22,23, 24, 25, 26,27 , 29, 29, 30, 31, 32, 33, 34, 35}
Let A be the event of getting a boy with prime roll number.
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
\(\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{8}{35}\)
Let B be the event of getting a girl with composite roll number
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34,35}
n(B) = 12
\(\mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{12}{35}\)
Let C be the event of getting an even roll number
C = {2, 4, 6,8, 10, 12, 14, 16, 19, 20, 22, 24, 26, 29, 34, 32, 34}
n(C) = 17
\(\mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{17}{35}\)
Since A and B are mutually exclusive events
P(A n B) = 0
B n C = {22, 24, 26, 29, 30, 32, 34}
n ( B n C ) = 7
\(P(B \cap C) =\frac{n(B \cap C)}{n(\mathrm{~S})}=\frac{7}{35} \)
\(A \cap C =\{2\} \)
\(\mathrm{n}(A \cap C) =1 \)
\(P(A \cap C) =\frac{n(A \cap C)}{n(S)}=\frac{1}{35} \)
\(P(A \cap B \cap C) =0[\because n(A \cap B)=0] \)
\(P(A \cup B \cup C)=P(A)+P(B)+P(C)- P(A \cap B)-P(B \cap C)-P(A \cap C) + P(A \cap B \cap C) \)
\(=\frac{8}{35}+\frac{12}{35}+\frac{17}{35}-0-\frac{7}{35}-\frac{1}{35}+0 \)
\(=\frac{8+12+17-7-1}{35}=\frac{29}{35} \)
Required probabiliry = \(\frac{29}{35}\)

Total number of people = 8000
n(S) = 8000
Let A be the event of selecting a female
Number of females n(A) = 3000
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Let 'B' be the event of selecting an individual as over 50 years.
Number of people who are over 50 years = 1300
n(B) = 1300
\(\mathrm{P}(\mathrm{B}) =\frac{n(\mathrm{~B})}{n(S)}=\frac{1300}{8000} \)
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B}) =30 \% \text { of } 3000 \)
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B}) =\frac{30}{100} \times 3000=900 \)
\(\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =\frac{n(A \cap B)}{n(S)}=\frac{900}{8000} \)
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \)
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\frac{3000}{8000}+\frac{1300}{8000}-\frac{900}{8000} \)
\(\mathrm{P}(\mathrm{A} \cup \mathrm{B}) =\frac{3000+1300-900}{8000} \)
\(=\frac{3400}{8000}=\frac{17}{40} \)
Probability that a chosen individual is either female or over 50 years is \(\frac{17}{40}\)

When two dice are rolled once, the Sample space
S = \(\left\{ \begin{matrix} (1,1) \\ (2,1) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,2) \\ (2,2) \\ \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,3) \\ (2,3) \\ \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ 6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) \\ (2,4) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,5) \\ (2,5) \\ \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,6) \\ (2,6) \\ \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
n(S) = 36
Let 'A' be the event of getting an even number on the first die.
\(A=\left\{ \begin{matrix} (2,1), & (2,2), & \begin{matrix} (2,3), & (2,4), & \begin{matrix} (2,5), & (2,6) \end{matrix} \end{matrix} \\ (4,1), & (4,2), & \begin{matrix} (4,3), & (4,4), & \begin{matrix} (4,5), & (4,6) \end{matrix} \end{matrix} \\ (6,1), & (6,2), & \begin{matrix} (6,3), & (6,4), & \begin{matrix} (6,5), & (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)
n(A) = 18
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 18 }{ 36 } \)
Let B be the event of getting total face sum 8.
B = {(2, 6), (3,5), (4, 4), (5, 3), (6, 2)}
n(B) = 5
\(P(A)=\frac { n(B) }{ n(S) } =\frac { 5 }{ 36 } \)
(A ∩ B) = {(2, 6), (4,4), (6, 2)}
n(A ∩ B) = 3
p(A ∩ B) = \(\frac{(A∩B)}{n(S)}=\frac{3}{36}\)
∴ P(A U B) = P(A) + P(B)+ P(A ∩ B)
\( =\frac{18}{36}+\frac{5}{36}-\frac{3}{36} \)
\(\mathrm{P}(A \cup B)=\frac{18+5-3}{36}=\frac{23-3}{36}=\frac{20}{36}=\frac{5}{9} \)
Probability of getting even number in the first die or a total face sum 8 is \(\frac{5}{9} .\)

Area of the rectangle = (length x breadth) square units
= 4 x 3 square feet
n(S) = 12 square feet
Area of the circle = \(\pi r^{2}\) square units
= \(\pi(1)^{2}\) square feet
= \(\pi\) square feet
Let A be the event of winning the game, then n(A) = \(\pi\)
Probability of winning the game = \(\frac{n(A)}{n(S)}=\frac{\pi}{12}\)
\(=\frac{3.14}{12}=\frac{1.57}{6}=\frac{157}{600}\)

When three fair coins are tossed together, the sample space
S = {(HHH), (THH), (HTH),(HHT), (TTH), (THT), (HTT), (TTT)}
N(s) = 8
(i) Let A be the event of getting all heads
A = {HHH}
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{7}{8}\)
(iii) Let C be the event of getting at most one head
C = {HTT, THT, TTH, TTT}
n(C) = 4
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let D be the event of getting at most two tails
P = {HHH, HHT, HTH, HTT, THH, THT, TTH}
n(D) = 7
\(\mathrm{P}(\mathrm{D})=\frac{n(D)}{n(S)}=\frac{7}{8}\)

When two unbiased dice are rolled, the Sample Space
s = {(1, 1) (1, 2) (r,3) (1,4) (1,5) (1,6)
(2, 1) (2,2) (2, 3) (2, 4) (2,5) (6, 6)
(3, 1) (3,2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4,2) (4,3) (4,4) (4, 5) (4,6)
(5, 1) (5,2) (5,3) (5,4) (5,5) (6,6)
(6, 1) (6,2) (6, 3) (6,4) (6, 5) (6, 6)}
n(S) = 36
(i) Let A be the event of getting a doublet
A = {( 1, 1) (2,2) (3,3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting the product as a prime number.
B = {(1,2) (1,3) (1, 5) (2,1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting the sum as a prime number.
c = {(1, 1) ( 1, 2) (1, 4) ( 1, 6) (2, 1) (2, 3) (2, 5) (3,2) (3, 4) (4, 1) (4,3) (5,2) (5,6) (6, 1) (6,5)}
n(C) = 15
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting the sum as 1. Since it is an impossible event.
n(D) = 0 and P(D) = g

\(\bar { { x }_{ 1 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 110 }{ 5 } \)
= 22
\({ \sigma }_{ 1 }=\sqrt { \frac { { \Sigma d }_{ 1 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 40 }{ 5 } } \)
= \(\sqrt { 8 } \)
\(\bar { { x }_{ 2 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 75 }{ 5 } \)
= 15
\({ \sigma }_{ 2 }=\sqrt { \frac { { \Sigma d }_{ 2 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 20 }{ 5 } } \)
= \(\sqrt { 4 } \)
= \(2\sqrt { 2 } \)
\(CV_{ 1 }=\frac { { { \sigma }_{ 1 } } }{ { x }_{ 1 } } \times 100\)
= \(\frac { 2.83 }{ 22 } \times 100\)
= 12.86
\({ CV }_{ 2 }=\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \times 100\)
= \(\frac { 2 }{ 15 } \times 100\)
= 13.33
\(\therefore\)Co-efficient of variation of City A is less than C.V of City B.
\(\therefore\) City A is more consistent.

First we find the coefficient of variation for guavas and oranges separately.
Number of guavas, n=7

Mean \(\bar { { x }_{ 1 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₁=\(\sqrt { \frac { 136 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 19.43-18.40 } \) ≃ 1.01
Coefficient of variation for guavas
C.V₁=\(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \times 100\)% = \(\frac { 1.01 }{ 4.29 } \) x 100% =23.54%

 Number of oranges n=7
Mean \(\bar { { x }_{ 2 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₂=\(\\ \sqrt { \frac { 184 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 26.29-18.40 } \)=2.81
Coefficient of variation for oranges
C.V₁=\(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100% = \(\frac { 2.81 }{ 4.29 } \) x 100% = 65.50%
CV₁=23.54%, CV₂=65.50%. Since, C.V₁ < C.V₂,  we can conclude that the consumption of guavas is more consistent than oranges.