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

From the table, we found that as x increases, y also increases. Thus, the variation is a direct variation.
Let y = kx, where k is a constant of proportionality.
From the given values, we have
\(k=\frac{3.1}{1}=\frac{6.2}{2}=\frac{9.3}{3}=\frac{12.4}{4}=\ldots=3.1\)
When you plot the points (1, 3.1) (2, 6.2) (3, 9.3), (4, 12.4), (5, 15.5), you find the relation y = (3.1)x forms a straight-line graph.
Clearly, from the graph, when diameter is 6 cm, its circumference is 18.6 cm.

 Given a triangle \(\triangle\)PQR. We have to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the given \(\triangle\)PQR.

Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
3. Joined Q₃ to R and drawn a line through Q₇ parallel to Q₃R, intersecting the extended line segment QR at R'
4. Drawn a line through R' parallel to RP intersecting the extended line segment QP at P.
5. Then PQR' is the required triangle each of whose sides is seven-thirds of the corresponding sides of PQR.

Given a triangle PQR we are required to construct another triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of the triangle PQR.



Steps of construction
1. Construct a \(\triangle\) PQR with any measurement
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 5 (the greater of 3 and 5 in \(\frac { 3 }{ 5 } \)) points.
Q₁Q_2, Q₃, Q₄ and Q₅ on QX so that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅
4. Join Q₅R  and draw a line through Q₃ (the third point, 3 being smaller of 3 and 5 in \(\frac { 3 }{ 5 } \)) parallel to Q₅R to intersect QR at R'.
5. Draw line through R' parallel to the line RP to intersect QP at P'.
Then, \(\triangle\)P'QR' is the required triangle each of whose sides is three-fifths of the corresponding sides of \(\triangle\) PQR.

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) = \(1-\frac { 5 }{ 9 } =\frac { 4 }{ 9 } \).

(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

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) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).