1.Table :

2.Variation :
Direct Variation
3. Equation
y = kx
\(k=\frac{y}{x}=\frac{1}{2}=\frac{2}{4}=\ldots . . \frac{1}{2}\) 
\(y=\frac{1}{2} x\) 
4. Points :
(2,1),(4,2),(6,3),(8,4),(1,5)
5. Solution
From the graph
(i) If x = 9 then, y = 4.5
(ii) if y = 7.5 then, x = 15

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle

Total number of students n(S) = 50.
Let A and B be the events of students opted for NCC and NSS respectively.
n(A) = 28, n(B) = 30, n(A⋂B) = 18
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 28 }{ 50 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 30 }{ 50 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 18 }{ 50 } \)
(i) Probability of the students opted for NCC but not NSS
P(A ∩ \(\bar { B } \)) = P(A) - P(A ∩ B) = \(\frac { 28 }{ 50 } -\frac { 18 }{ 50 } =\frac { 1 }{ 5 } \)
(ii) Probability of the students opted for NSS but not NCC.
P(A ∩ \(\bar { B } \)) = P(B) - P(A ∩ B) = \(\frac { 30 }{ 50 } -\frac { 18 }{ 50 } =\frac { 6 }{ 25 } \)
(iii) Probability of the students opted for exactly one of them
= P[(A ∩ \(\bar { B } \)) U (\(\bar { A } \) ∩ B)]
= P(A ∩ \(\bar { B } \)) + P(\(\bar { A } \) ∩ B) =\(\frac { 1 }{ 5 } +\frac { 6 }{ 25 } =\frac { 11 }{ 25 } \)
(Note that (A ∩ \(\bar { B } \)),(\(\bar { A } \) ∩ B) are mutually exclusive events)

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

When we roll two dice, the sample space is given by
S = \(\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \)
n(S) = 36
(i) Let A be the event of getting the sum of outcome values equal to 4.
Then A = {(1, 3),(2, 2),(3, 1)}; n(A) = 3.
Probability of getting the sum of outcomes equal to 4 is  P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(ii) Let B be the event of getting the sum of outcome values greater than 10.
Then B = {(5,6),(6,5),(6,6)}; n(B) = 3
Probability of getting the sum of outcomes greater than 10 is P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(iii) Let C be the event of getting the sum of outcomes less than 13. Here all the outcomes have the sum value less than 13. Hence C = S
Therefore, n(C) = n(S) = 36
Probability of getting the sum value less than 13 is P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 36 } \) = 1

Total number of possible outcomes n(S) = 5 + 4 = 9
(i) Let A be the event of getting a blue ball.
Number of favourable outcomes for the event A. Therefore, n(A) = 5
Probability that the ball drawn is blue. Therefore, P(A) = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 9 } \)
(ii) \(\bar { A } \) will be the event of not getting a blue ball. So P(\(\bar { A } \)) = 1 - P(A) = \(1-\frac { 5 }{ 9 } =\frac { 4 }{ 9 } \).

Since A and B are mutually exclusive events
\(\mathrm{P}(A \cap B)=0\)
P(not A) = 0.45
P(A) = 1 - P(not A)
P(A) = 1 - 0.45 = 0.55
\(\mathrm{P}(A \cup B)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
0.65 = 0.55 + P(B) - 0
P(B) = 0.65 - 0.55 = 0.1

(i) Given P(A) = 0.42
P(not A) = 1 - P(A)
\(\mathrm{P}(\bar{A})=1-0.42=0.58\)
(ii) Given P(B) = 0.48
P(not B) = 1 - P(B)
\(\mathrm{P}(\bar{B})=1-0.48=0.52\)
(iii) P(A or B) = \(P(A \cup B)\)
\(=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
= 0.42 + 0.48 - 015
= 0.90 - 0.16
P(A or B) = 0.74

\( \mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\frac{1}{3} =\frac{2}{3}+\frac{2}{5}-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cap B) =\frac{2}{3}+\frac{2}{5}-\frac{1}{3} \)
\(\mathrm{P}(A \cap B) =\frac{10+6-5}{15}=\frac{16-5}{15}=\frac{11}{15} \)

P(A) = 0.37, P(B) = 0.42, P(A∩B) = 0.09
P(AUB) = P(A) + P(B) - P(A∩B)
P(AUB) = 0.37 + 0.42 - 0.09 = 0.7

When a coin is tossed thrice, the outcome will be

The sample space s = {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
n(S) = 3
Let A be the event of getting two consecutive tails
4 = {HTT, TTH, TTT}
n(A) = 3
\(\Rightarrow P=\frac { n\left\{ F \right\} }{ n\{ O\} } =\frac { 3 }{ 8 } \)
Probability of getting two consecutive tails = \(\frac{3}{8}\)

leap year has 366 days. So it has 52 full weeks and 2 days. 52 Saturdays must be in 52 full weeks.
The possible chances for the remaining two days will be the sample space.
S = {(Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}
n(S) = 7
Let A be the event of getting 53^rd Saturday.
Then A = {Fri-Sat, Sat-Sun}; n(A) = 2
Probability of getting 53 Saturdays in a leap year is P(A0 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

When two coins are tossed together, the sample space is
S = {HH, HT, TH, TT} n(S) = 4
Let A be the event of getting different faces on the coins.
A = {HT, TH}; n(A) = 2
Probability of getting different faces on the coins is P(A) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).