Mahesh

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(b)

I.

Question 1.
Locate the position of the point P with respect to the circle S = 0 when
i) P(3, 4) and S ≡ x² + y² – 4x – 6y – 12 = 0
Solution:
S ≡ x² + y² – 4x – 6y – 12
P(3, 4) = (x₁, y₁)
S₁₁ = 3² + 4² – 4.3 – 6.4 – 12
= 9 + 16 – 12 – 24 – 12
= – 23 < 0
P (3, 4) lies inside the circle

ii) P(1, 5) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
S₁₁ = (1)² + (5)² – 2(-1) – 4(5) + 3 = 7
S₁₁ > 0 [∴ P is outside the circle]

iii) P (4, 2) and S ≡ 2x² + 2y² – 5x – 4y – 3 = 0
Solution:
S₁₁ = 2(4)² + 2(2)² – 5(4) – 4(2) – 3 = 9
S₁₁ > 0 (P lies outside the circle)

iv) P(2, -1) and S ≡ x² + y² – 2x – 4y + 3 = 0
Solution:
S₁₁ = (2)² + (-1)² – 2(2) – 4 (-1) + 3 = 8
S₁₁ > 0 [P is outside the circle]

Question 2.
Find the power of the point P with respect to the circle S = 0 when
i) P = (5, -6), and S ≡ x² + y² + 8x + 12y + 15
Solution:
S₁₁ = power of the point
= 25 + 36 + 40 – 72 + 15
= 116 – 72 = 44

ii) P = (-1, 1) and S ≡ x² + y² -6x + 4y – 12
Power of the point = S₁₁
= 1+1+6 + 4-12 = 0

iii) P = (2, 3) and S ≡ x² + y² – 2x + 8y – 23
Power of the point = S₁₁
= 4 + 9-4 + 24-23 = 10

iv) P = (2, 4) and S ≡ x² + y² – 4x – 6y – 12
Power of the point = 4 + 16 – 8 – 24 – 12
= -24.

Question 3.
Find the length of tangent from P to the circle S = 0 when
i) P = (-2, 5) and S = x² + y² – 25
Solution:
Length of tangent = \(\sqrt{s_{11}}\)
= \(\sqrt{(-2)^{2}+(5)^{2}-25}\) = 2 units

ii) P = (0, 0), S = x² + y² – 14x + 2y + 25
Solution:
Length of the tangent = \(\sqrt{s_{11}}\)
= \(\sqrt{0+0-0+0+25}\) = 5 units

iii) P = (2, 5) and S ≡ x² + y² – 5x + 4y – 5
Solution:
Length of the tangent = \(\sqrt{s_{11}}\)
= \(\sqrt{4+25-10+20-5}\)
= \(\sqrt{34}\) units

II.

Question 1.
If the length of the tangent from (5, 4) to the circle x² + y² + 2ky = 0 is 1 then find k.
Solution:
Length of tangent
\(\sqrt{s_{11}}=\sqrt{(5)^{2}+(4)^{2}+8k}\)
But length of tangent = 1
∴ 1 = \(\sqrt{25+16+8k}\)
Squaring both sides we get 1 = 41 + 8k
k = – 5 units.

Question 2.
If the length of the tangent from (2, 5) to the circle x² + y² – 5x + 4y + k = 0 is √37 then find k.
Solution:
Length of tangent = \(\sqrt{s_{11}}\)
= \(\sqrt{(2)^2+(5)^2-5 \times 2+4 \times 5+k}\)
= 37 = 39 + k
k = -2 units.

III.

Question 1.
If a point P is moving such that the lengths of tangents drawn from P to the circles x² + y² – 4x – 6y – 12 = 0 and x² + y² + 6x + 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P.
Solution:
P(x, y) is any point on the locus
S ≡ x² + y² – 4x – 6y – 12

Locus of P is
5x² + 5y² – 60x – 126y – 212 = 0

Question 2.
If a point P is moving such that the lengths of the tangents drawn from P to the circles x² + y² + 8x + 12y + 15 = 0 and x² + y² – 4x – 6y – 12 = 0 are equal then find the equation of the locus of P.

Solution:
Equations of the circles are
S ≡ x² + y² + 8x + 12y + 15 = 0
S¹ ≡ x² + y² – 4x – 6y – 12 = 0
P (x₁, y₁) is any point on the locus and PT₁ PT₂ are the tangents from P to the two circles.
Given condition is PT₁ = PT₂ ⇒ P₁T₁² = PT₂²
x²₁ + y²₁ + 8x₁ + 12y₁ + 15
= x²₁ + y²₁ – 4x₁ – 6y₁ – 12
12x₁ + 18y₁ + 27 = 0
(or) 4x₁ + 6y₁ + 9 = 0
Locus of P(x₁, y₁) is 4x + 6y + 9 = 0

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(a)

I.

Question 1.
Find the equation of the circle with centre C and radius r where.
i) C = (2, -3), r = 4
Solution:
Equation of the circle is
⇒ (x – h)² + (y – k)² = a²
⇒ (x – 2)² + (y + 3)² = 4²
x² – 4x + 4 + y² + 6y + 9 = 16
x² + y² – 4x + 6y – 3 = 0

ii) C = (-1, 2), r = 5
Solution:
Equation of the circle is
(x + 1)² + (y – 2)² = 5²
⇒ x² + 2x + 1 + y² – 4y + 4 = 25
⇒ x² + y² + 2x – 4y – 20 = 0

iii) C = (a, -b); r = a + b
Solution:
Equation of the circle is
(x – a)² + (y + b)² = r²
⇒ x² – 2xa + a² + y² + 2by + b² = (a + b)²
⇒ x² + y² – 2xa + 2by – 2ab = 0

iv) C = (-a, -b); r = \(\sqrt{a^{2}-b^{2}}=2\) (|a| > |b|)
Solution:
Equation of the circle is
(x + a)² + (y + b)² = [latex]\sqrt{a^{2}-b^{2}}=2[/latex]²
⇒ x² + y² + 2xa + 2yb + a² + b² = a² – b²
⇒ x² + y² + 2xa + 2yb + 2b² = 0

v) C = (cos α, sin α); r = 1
Solution:
Equation of the circle is
(x – cos α)² + (y – sin α)² = 1
x² + y² – 2x cos α – 2y sin α + sin² α + cos² α = 1
x² + y² – 2x cos α – 2y sin α = 0

vi) C = (-7, -3);r = 4
Solution:
Equation of the circle is
(x + 7)² + (y + 3)² = 4²
x² + y² + 14x + 6y +49 + 9=16
⇒ x² + y² + 14x + 6y + 42 = 0

vii) C = (-\(\frac{1}{2}\), -9), r = 5
Solution:
Equation of the circle is
(x + \(\frac{1}{2}\))² + (y + 9)² = 5²
x² + x + \(\frac{1}{4}\) + y² + 18y + 81 =25
x² + y² + x + 18y + 56 + \(\frac{1}{4}\) = 0
4x² + 4y² + 4x + 72y + 225 = 0

viii) C = (\(\frac{5}{2}\), \(\frac{4}{3}\)), r = 6
Solution:
Equation of the circle is
(x – \(\frac{5}{2}\))² + (y + \(\frac{4}{3}\))² = 6²

Multiplying with 36
36x² + 36y² – 180x + 96y + 225 + 64 – 1296 = 0
⇒ 36x² + 36y² – 180x + 96y – 1007 = 0

ix) C = (1, 7), r = \(\frac{5}{2}\)
Solution:
Equation of the circle is
(x – 1)² + (y – 7)² = (\(\frac{5}{2}\))²
⇒ x² – 2x + 1 + y² – 14y + 49 = \(\frac{25}{4}\)
⇒ x² + y² – 2x – 14y + \(\frac{175}{4}\) = 0
4x² + 4y² – 8x – 56y + 175 = 0

x) C = (0, 0); r = 9
Solution:
Equation of the circle is
(x – 0)² + (y – 0)² = (9)²
x² + y² = 81

Question 2.
Find the equation of the circle passing through the origin and having the centre at (-4, -3).
Solution:
Equation of the circle is
(x – h)² + (y – k)² = r²; (h, k) = (-4, -3)
(x + 4)² + (y + 3)² = r²
Circle passes through origin.
∴ (0 + 4)² + (0 + 3)² = r² ⇒ 25 = r²
Then required equation of circle be
(x + 4)² + (y + 3)² = 25
x² + y² + 8x + 6y = 0

Question 3.
Find the equation of the circle passing through (2, -1) having the centre at (2, 3).
Solution:
C = (2, 3), P = (2, -1)
Radius CP = \(\sqrt{(2 – 2)^{2} + (3 + 1)^{2}}\) = 4
Equation of circle be
(x – 2)² + (y – 3)² = 42
x² + y² – 4x – 6y – 3 = 0

Question 4.
Find the equation of the circle passing the through (- 2, 3) centre at (0, 0).
Solution:
C = (0, 0), P = (- 2, 3)
Radius = \(\sqrt{(0 + 2)^{2} + (0 – 3)^{2}}\)
= √13
Equation of circle be
(x – 0)² + (y – 0)² = (√13)²
x² + y² = 13

Question 5.
Find the equation of the circle passing through (3, 4) having and the centre at (-3, 4).
Solution:
Let the equation of the circle be
(x – h)² + (y – k)² – r²
Centre (h, k) = (-3, 4)
(x + 3)² + (y – 4)² = r²
Circle passes through (3, 4)
(3 + 3)² + (4 – 4)² = r²
r² = 36
Equation of the circle is
(x + 3)² + (y – 4)² = 36
x² + 6x + 9 + y² – 8y + 18 – 36 = 0
x² + y² + 6x – 8y – 11 =0

Question 6.
Find the value of ‘a’ if 2x² + ay² – 3x + 2y – 1 =0 represents a circle and also find its radius.
Solution:
General equation of second degree
ax² + 2hxy + by² + 2gx + 2fy + c = 0
Represents a circle, when
a = b, h= 0, g² + f² – c > 0 In
2x² + ay² – 3x + 2y -1 = 0
a = 2, above equation represents circle.
x² + y² – \(\frac{3}{2}\) x + y – \(\frac{1}{2}\) = 0
2g = –\(\frac{3}{2}\); 2f = 1; c = –\(\frac{1}{2}\)
c = (-g, -f) = (\(\frac{+3}{4}\), \(\frac{-1}{2}\))
Radius = \(\sqrt{g^{2} + f^{2} – c}\) = \(\sqrt{\frac{9}{16}+\frac{1}{4}+\frac{1}{2}}\)
= \(\frac{\sqrt{21}}{4}\) units

Question 7.
Find the values of a, b if ax² + bxy + 3y² – 5x + 2y – 3 = 0 represents a circle. Also find the radius and centre of the circle.
Solution:
General equation of second degree
ax² + 2hxy + by² + 2gx + 2fy + c = 0
represents a circle a = b, h = 0
∴ ax² + bxy + 3y² – 5x + 2y – 3 = 0
represents, a circle.
When b = 0, a = 3
3x² + 3y² – 5x + 2y – 3 = 0

Question 8.
If x² + y² + 2gx + 2fy -12 = 0 represents a circle with centre (2, 3), find g, f and its radius.
Solution:
Circle is x² + y² + 2gx + 2fy – 12 = 0
C = (-g, -f) C = (2, 3)
∴ g = -2, f = -3, c = – 12
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{4 + 9 + 12}\)
= 5 units

Question 9.
If x² + y² + 2gx + 2fy = 0 represents a circle with Centre (-4, -3) then find g,f and the radius of the circle.
Solution:
Circle is x² + y² + 2gx + 2fy – 12 = 0
⇒ C = (-g, -f) C = (-4, -3)
∴ g = 4, f = 3
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{16 + 9}\) = 5 units

Question 10.
If x² + y² – 4x + 6y + c = 0 represents a circle with radius 6 then find the value of c.
Solution:
Circle is x² + y² – 4x + 6y + c = 0
r = \(\sqrt{g^{2} + f^{2} – c}\) ; g = – 2, f = 3
6 = \(\sqrt{4 + 9 – c}\)
36 = 13 – c
or c = -23

Question 11.
Find the centre and radius of each of the circles whose equations are given below.
i) x² + y² – 4x – 8y – 41 = 0
Solution:
x² + y² – 4x – 8y – 41 = 0 ……….. (i)
x² + y² + 2gx + 2fy + c = 0 ……….. (ii)
Comparing (i) and (ii) we get
g = – 2, f = – 4, c = – 41
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{4 + 16 + 41}\)
= \(\sqrt{61}\) units
Centre = (-g, -f) = (2, 4)

ii) 3x² + 3y² – 5x- 6y + 4 = 0
Solution:
Equation of the circle is
3x² + 3y² – 5x – 6y + 4 = 0
x² + y² – \(\frac{5}{3}\)x – \(\frac{6}{3}\)y + \(\frac{4}{3}\) = 0 …………. (i)
x² + y² + 2gx + 2fy + c = 0 …………. (ii)
Comparing equations (i) and (ii) we get

iii) 3x² + 3y² + 6x – 12y – 1 = 0
Find the radius and centre of the circle.
Solution:
Equation of the circle is
3x² + 3y² + 6x – 12y – 1 = 0
x² + y² + \(\frac{6}{3}\)x – \(\frac{12}{3}\)y – \(\frac{1}{3}\) = 0
C = (-g, -f) = (-1, 2)
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{1 + 4 + \frac{1}{3}}\)
= \(\frac{4}{\sqrt{3}}\)

iv) x² + y² + 6x + 8y – 96 = 0
Solution:
Equation of the circle is
x² + y² + 6x + 8y – 96 = 0
C = (-g, -f) = (-3, -4)
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{9 + 16 + 96}\)
= \(\sqrt{121}\) = 11 units

v) 2x² + 2y² – 4x + 6y – 3 = 0 Sol. Equation of the circle is
Solution:
x² + y² – 2x + 3y-|- = 0 ………….. (i)
x² + y² + 2gx + 2fy + c = 0 …………… (ii)
Comparing (i) and (ii) we get
C = (1, –\(\frac{3}{2}\))
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{1 + \frac{9}{4} + \frac{3}{2}}=\frac{\sqrt{19}}{2}\) units

vi) 2x² + 2y² – 3x + 2y – 1 = 0
Solution:
Equation of the circle is
x + y – \(\frac{3}{2}\)x + y – \(\frac{1}{2}\) = 0
x² + y² + 2gx + 2fy + c = 0
Comparing we get
C = (-g, -f) = (\(\frac{3}{2}\), –\(\frac{1}{2}\))
r = \(\sqrt{g^{2} + f^{2} – c}\)
\(\sqrt{\frac{9}{16} + \frac{1}{4} + \frac{3}{2}}=\frac{\sqrt{21}}{4}\) units

vii) \(\sqrt{1 + m^{2}}\) (x² + y²) – 2cx – 2mcy = 0
Solution:
Equation of the circle is

viii) x² + y² + 2ax – 2by + b2 = 0
Solution:
Equation of the circle is
x² + y² + 2gx + 2fy + c = 0
C = (-g, -f) = (-a, b)
Radius = \(\sqrt{g^{2} + f^{2} – c}\)
= \(\sqrt{a^{2} + b^{2} – b^{2}}\) = a units

Question 12.
Find the equations of the circles for which the points given below are the end points of a diameter.
i) (1, 2), (4, 6)
Solution:
Equation of the circle with (x₁, y₁), (x₂, y₂) as ends of a diameter is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 1) (x – 4) + (y – 2) (y – 6) = 0
⇒ x² – 5x + 4 + y² – 8y + 12 = 0
⇒ x² + y² – 5x – 8y + 16 = 0

ii) (-4, 3); (3, -4)
Solution:
Equation of circle with (x₁, y₁) and (x₂, y₂) are end points of diameter is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = o
Required equation of circle be
(x + 4) (x – 3) + (y – 3) ( y + 4) = 0
x² + y² + x + y – 24 = 0

iii) (1, 2); (8, 6)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
(x – 1) (x – 8) + (y – 2) (y – 6) = 0
x² + y² – 9x – 8y + 20 = 0

iv) (4, 2); (1, 5)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 4) (x – 1) + (y – 2) (y – 5) = 0
x² + y² – 5x – 7y + 14 = 0

v) (7, -3); (3, 5)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 7)(x – 3) + (y + 3) (y – 5) = 0
x² + y² – 10x – 2y + 6 = 0

vi) (1, 1); (2, -1)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 1) (x – 2) + (y – 1) (y + 1) = 0
x² + y² – 3x +1=0

vii) (0, 0); (8, 5)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 0) (x – 8) + (y – 0) (y – 5) = 0
⇒ x² – 8x + y² – 5y = 0
x² + y² – 8x – 5y = 0

viii) (3, 1); (2, 7)
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
(x – 3) (x – 2) + (y – 1) (y – 7) = 0
x²+ y² – 5x – 8y + 13 = 0

Question 13.
Obtain the parametric equation of each of the following circles.
i) x² + y² = 4
Solution:
Equation of the circle is x² + y² = 4
C (0, 0), r = 2
Parametric equations are
x = – g + r cos θ = 2 cos θ
y = – b + r sin θ = 2 sin θ, 0 ≤ θ < 2π

ii) 4(x² + y²) = 9
Solution:
Equation of the circle is 4(x² + y²) = 9
x² + y² = \(\frac{9}{4}\)
C(0, 0), r = \(\frac{3}{2}\)
Parametric equations are
x = \(\frac{3}{2}\) cos θ, y = \(\frac{3}{2}\) sin θ, 0 ≤ θ < 2π

iii) 2x² + 2y² = 7
Solution:
Equation of the circle is 2x² + 2y² = 7
x² + y² = \(\frac{7}{2}\)
C(0, 0), r = \(\sqrt{\frac{7}{2}}\)
Parametric equations are
x = \(\sqrt{\frac{7}{2}}\) cos θ, y = \(\sqrt{\frac{7}{2}}\) sin θ, 0 ≤ θ < 2π

iv) (x – 3)² + (y – 4)² = 8²
Solution:
Equation of the circle is (x – 3)² + (y – 4)² = 8²
Centre (3, 4), r = 8
Parametric equations are
x = 3 + 8 cos θ, y = 4 + 8 sin θ, 0 ≤ θ < 2π

v) x² + y² – 4x – 6y – 12 = 0
Solution:
Equation of the circle is
x² + y² – 4x – 6y – 12 = 0
Centre (2, 3), r = \(\sqrt{4 + 9 + 12}\) = 5
Parametric equations are
x = 2 + 5 cos θ, y = 3 + 5 sin θ, 0 ≤ θ < 2π

vi) x² + y² – 6x + 4y – 12 = 0
Solution:
Equation of the circle is x² + y² – 6x + 4y – 12 = 0
Centre (3, – 2), r = \(\sqrt{9 + 4 + 12}\) = 5
Parametric equations are
x = 3 + 5 cos θ, y = -2 + 5 sin θ, 0 ≤ θ < 2π

II.

Question 1.
If the abscissae of points A, B are the roots of the equation, x² + 2ax – b² = 0 and ordinates of A, B are roots of y² + 2py – q²= 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter.
Solution:
Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
x² – x(x₁ + x₂) + x₁x₂ + y² – y (y₁ + y₂) + y₁y₂ = 0
x₁, x₂ are roots of x² + 2ax – b² = 0
y₁, y₂ are roots of y² + 2py – q² = 0
x₁x₂ = -2a,    y₁y₂ = -2p
x₁x₂ = -b², y₁y₂ = -q²

Equation of circle be
x² – x(- 2a) – b² + y² – y(- 2p) – q² = 0
x² + 2xa + y² + 2py – b² – q² = 0

Question 2.
i) Show that A (3, -1) lies on the circle x² + y² – 2x + 4y = 0. Also find the other end of the diameter through A.
Solution:
Equation of the circle is
x² + y² – 2x + 4y = 0 …………… (i)
A (3, -1); Let B (x₁, y₁)
Substituting A in equation (i)
(3)² + (-1)² – 2(3) + 4 (-1) = 0
∴ A lies on the circle
C (-g, -f)
C = (1, -2)

C is the centre of circle
C is midpoint of AB
\(\frac{x_{1}+3}{2}\) = -1 \(\frac{y_{1}-1}{2}\) = -2
x₁ = -1 y₁ = -3
B(x₁, y₁) = (-1, -3)

ii) Show that A(-3, 0) lies on x² + y² + 8x + 12y + 15 = 0 and find the other end of diameter through A.
Solution:
If A (-3, 0) satisfy
x² + y² + 8x + 12y + 15 = 0
then (-3)² + (0)² – 8 × 3 + 12 × 0 +15
9 – 24 + 1 5 = 0
∴ (-3, 0) is one end of diameter.
A (-3, 0), C(-4, -6), B(x₁, y₁)
\(\frac{x_{1}+(-3)}{2}\) = -4 \(\frac{y_{1}+0}{2}\) = -6
x₁ = -5 y₁ = -12
∴ Other end of diameter is (-5, -12)

Question 3.
Find the equation of a circle which passes through (2, -3) and (-4, 5) and having the centre on 4x + 3y + 1 =0
Solution:
x² + y² +2gx + 2fy + c = 0 ……….. (i)
Equation (i) passes through (2, -3), (-4, 5)
∴ 4 + 9 + 4g – 6f + c = 0 …………. (ii)
16 + 25 – 8g + 10f + c = 0 …………. (iii)
Equation (iii) – (ii) we get
28 – 12g + 16f = 0
(or) 3g – 4f = 7
Centre lies on (-g, -f) lies on 4x + 3y + 1 = 0
then 4(-g) + 3(-f) + 1 = 0
3g – 4f – 7 = 0
Solving we get f = -1
g = 1
Now substituting f, g values in equation (ii) we get
4 + 9 + 4(1) – 6 (-1) + c = 0, c = -23
x² + y² + 2x – 2y – 23 = 0 is required equation of circle.

Question 4.
Find the equation of a circle which passes through (4, 1) (6, 5) and having the centre on 4x + 3y – 24 = 0
Solution:
Equation of circle be
x² + y² + 2gx + 2fy + c = 0 passes through (4, 1) and (6, 5) then
4² + 1² + 2g(4) + 2f(1) + c = 0 ………… (i)
6² + 5² + 2g(6) + 2f(5) + c = 0 ………… (ii)
Centre lie on 4x + 3y – 24 = 0
∴ 4(-g) + 3 (-f) – 24 = 0 ………… (iii)
(ii) – (i) we get
44 + 4g + 8f = 0 …………… (iv)
Solving (iii) and (iv) we get
f = -4, g = -3, c = 15
∴ Required equation of circle is
x² + y² – 6x – 8y +15 = 0

Question 5.
Find the equation of a circle which is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (-2, 14).
Solution:
x² + y² – 6x – 4y – 12 = 0 ……… (i)
C = (-g, -f)
= (3, 2)
Equation of circle concentric with (i) be
(x – 3)² + (y – 2)² = r²
Passes through (-2, 14)
∴ (-2 – 3)² + (14 – 2)² = r²
169 = r²
Required equation of circle be
(x – 3)² + (y – 2)² = 169
x² + y² – 6x – 4y – 156 = 0

Question 6.
Find the equation of the circle whose centre lies on the X – axis and passing through (-2, 3) and (4, 5).
Solution:
x² + y² + 2gx + 2fy + c = 0 ………… (i)
(-2, 3) and (4, 5) passes through (i)
4 + 9 – 4g + 6f + c = 0 ………….. (ii)
16 + 25 + ,8g + 10f + c = 0 ………… (iii)
(iii) – (ii) we get
28 + 12g + 4f = 0
f + 3g = -7
Centre lies on X – axis then f = 0
g = -, \(\frac{7}{3}\) f = 0, c = \(\frac{67}{3}\) -, we get by substituting g, f in equation (ii)
Required equation will be
3(x² + y²) – 14x – 67 = 0

Question 7.
If ABCD is a square then show that the points A, B, C and D are concyclic.
Solution:
AB = a, AD = a
A (0, 0), B(0, a), D (a, 0)
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
passes through A, B, D we get
A : 0 + 0 + 2g(0) + 2f(0) + c = 0
C = 0
B : 0 + a2 + 2g(0) + 2fa + 0 = 0
f = \(\frac{a}{2}\)
Similarly g = –\(\frac{a}{2}\)
Required equation of circle be
x² + y² – ax – ay = 0
Co-ordinates of C are (a, a)
a² + a² – a² – a² = 0
⇒ C lies on the circle passing through A, B, D.
∴ A, B, C, D are concyclic.

III.

Question 1.
Find the equation of circle passing through each of the following three points.
i) (3, 4); (3, 2); (1, 4)
Solution:
Equation of circle
x² + y² + 2gx + 2fy + c = 0
Given points satisfy above equation then
9 +16 + 6g + 8f + c + = 0 ………….. (i)
9 + 4 + 6g + 4f + c = 0 ………….. (ii)
1 + 16 + 2g + 8f + c = 0 ………… (iii)
Subtracting (ii) – (i) we get
-12 – 4f = 0 (or) f = – 3
(ii) – (iii) we get
– 4 + 4g – 4f = 0
g – f = 1
9 = – 2
Now substituting g, f in equation (i) we get
25 + 6 (- 2) + 8 (- 3) + c = 0
we get c = 11
Required equation of circle be
x² + y² – 4x – 6y + 11 = 0

ii) (1, 2); (3, -4); (5, -6)
Solution:
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
1 + 4 + 2g + 4f + c = 0 …………. (i)
9 + 16 + 6g — 8f + c = 0 …………. (ii)
25 + 36 + 10g – 12f + c = 0 …………. (iii)
Subtracting (ii) – (i) we get
20 + 4g – 12f = 0
(or) 5 + g – 3f = 0 …………. (iv)
Similarly (iii) – (ii), we get
36 + 4g – 4f = 0
(or) 9 + g – f = 0 …………. (v)
Solving (v) and (iv) we get
f = -2, g = -11, c = 25
Required equation of circle be
x² + y² – 22x – 4y + 25 = 0

iii) (2, 1); (5, 5); (-6, 7)
Solution:
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
4 + 1 + 4g + 2f + c = 0 …………. (i)
25 + 25 + 10g + 10f + c = 0 …………. (ii)
36 + 49 – 12g + 14f + c = 0 …………. (iii)
Subtracting (ii) – (i)
45 + 6g + 8f = 0 …………. (iv)
Subtracting (iii) – (ii)
35 – 22g + 4f = 0 …………. (v)
Solving (iv) and (v) we get
g = \(\frac{1}{2}\); f = -6; c = 5
Required equation of circle be
x² + y² + x – 12y + 5 = 0

iv) (5, 7); (8, 1); (1, 3)
Solution:
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
25 + 49 + 10g + 14f + c = 0 …………. (i)
64 + 1 + 16g + 2f + c = 0 …………. (ii)
1 + 9 + 2g + 6f + c = 0 …………. (iii)
Subtracting (ii) – (i)
-9 + 6g – 12f = 0 …………. (iv)
(or) 2g – 4f – 3 = 0
Subtracting (iii) – (ii)
– 55 – 14g + 4f = 0 ……….. (v)
Solving (v) and (iv) we get
g = \(\frac{-29}{6}\), f = \(\frac{-19}{6}\), c = \(\frac{56}{3}\)
∴ Required equation of circle is
x² + y² – \(\frac{-29}{3}\)x – \(\frac{-19}{3}\)y + \(\frac{56}{3}\) = 0
3(x² + y²) – 29x – 19y + 56 = 0

Question 2.
i) Find the equation of the circle passing through (0, 0) and making intercepts 4,3 on X- axis and Y – axis respectively.
Solution:
x² + y² + 2gx + 2fy + c = 0
Circle is passing through
(0, 0), (4, 0) and (0, 3)
0 + 0 + 2g(0) + 2f(0) + c = 0
c = 0 ………… (i)

16 + 0 + 8g + 2f . 0 + c = 0
g = -2 as c = 0
Similarly 0 + 9 + 2g. 0 + 6f + c = 0
f = –\(\frac{3}{2}\) as c = 0
Required equation of circle will be
x² + y² – 4x – 3y = 0
If intercepts are negative when circle passes through the points (0, 0) (-4, 0) (0, -3). Required equation of circle is x² + y² + 4x + 3y = 0

ii) Find the equation of the circle passing through (0, 0) and making intercept 6 units on X – axis and intercept 4 units on Y – axis.
Solution:
OA = 6 units,
OB = 4 units,
OD = 3 units, OE = 2 units.
∴ Co-ordinates of centre b(3, 2)
Radius OC = \(\sqrt{(0+3)^{2}+(0-2)^{2}}\) = √13
Equation of circle with (h, k) as centre r be radius is (x – h)² + (y – k)² = r²
∴ Required equation of circle be
(x – 3)² + (y – 2)² = 13
x² + y² – 6x – 4y = 0
If intercepts are negative when circle passes through the points (0, 0) (-6, 0) (0, -4)
Required equation of circle is x² + y² + 6x + 4y = 0

Question 3.
Show that the following four points in each of the following are concyclic and find the equation of the circle on which they lie.
i) (1, 1), (-6, 0), (-2, 2), (-2, -8)
Solution:
Suppose the equation of the required circle is
x² + y² + 2gx + 2fy + c = 0 ………… (i)
This circle passes through A (1, 1)
1 + 1 + 2g + 2f + c = 0
⇒ 2g + 2f + c = -2 …………. (ii)
The circle passes through B (-6, 0)
36 + 0 – 12g + 0 + c = 0
– 12g + c = – 36 …………. (iii)
This circle passes through C (-2, 2)
4 + 4 – 4g + 4f + c = 0
– 4g + 4f + c = – 8 ………… (iv)
(iii) – (iv) gives – 8g – 4f = 0
⇒ 2g + f = 7
(i) – (ii) gives 14g + 2f = 34
7g + f = 17 ………….. (v)
Solving (iv) and (v) we get g = 2, f = 3
Putting g = 2, f = 3 in
4 + 6 + c = -2
c = – 12
Required circle is x² + y² + 4x + 6y – 12 = 0
Substituting (-2, -8) in this above equation, we get
4 + 64 – 8- 48 -12 = 68 – 68 = 0
(- 2, -8) satisfies the above equation
∴ A, B, C, D are concyclic.
Equation of the circle is
x² + y² + 4x + 6y – 12 = 0

ii) (1, 2); (3, -4); (5, -6); (19, 8)
Solution:
Equation of circle
x² + y² + 2gx + 2fy + c = 0
1 + 4 + 2g + 4f + c = 0 …………. (i)
9 + 16 + 6g – 8f + c = 0 …………. (ii)
25 + 36 + 10g – 12f + c = 0 …………. (iii)
Subtracting (ii) – (i) we get
20 + 4g – 12f = 0
5 + g – 3f = 0 …………. (iv)
Subtracting (iii) – (ii) we get
36 + 4g – 4f = 0
(or)
9 + g – f = 0 …………. (v)
Solving (iv) and (v) we get
f = -2, g = – 11, c = 25
Equation of circle will be
x² + y² – 22x – 4y + 25 = 0 ……… (vi)
(19, 8) substituting in equation (vi)
(19)² + 8² – 22 × 19 – 4 × 8 + 25 = 0
Hence (19, 8) lie on circle and four points are concyclic.

iii) (1, -6); (5, 2); (7, 0); (-1, -4)
Solution:
Equation of circle
x² + y² + 2gx + 2fy + c =0
1 + 36 + 2g- 12f + c = 0 ……….. (i)
25 + 4 + 10g + 4f + c = 0 …………. (ii)
49 + 14g + c = 0 …………… (iii)
Subtracting (ii) – (i) we get
-8 + 8g + 16f = 0
(or)
2f+ g – 1 = 0 ……… (iv)
Subtracting (iii) – (ii) we get
20 + 4g – 4f = 0 ………… (v)
(or)
5 + g – f = 0
Solving (iv) and (v) we get
f = 2, g = -3, c = -7
Equation of circle be
x² + y² – 6x + 4y- 7 = 0 …….. (vi)
(-1, -4) satisfies equation (vi)
∴ Four points are concyclic.

iv) (9, 1), (7, 9), (-2, 12), (6, 10)
Solution:
Equation of the circle is
x² + y² + 2gx + 2fy + c = 0
This circle passes through
A(9, 1), B(7, 9), C (-2, 12)
81 + 1 + 18g + 2f + c = 0 ………….. (i)
49 + 81 14g + 18f + c = 0 ………….. (ii)
4 + 144 – 4g + 24f + c = 0 ………….. (iii)
(ii) – (i) gives – 4g + 16f + 48 = 0
4g – 16f = 48
g – 4f = 12 ……….. (iv)
(ii) – (iii) gives 18g – 6f – 18 = 0
18g – 6f = 18 ………… (v)
36g – 12f = 36 ……….. (v) × 2
3g – 12f = 36 ………… (iv) × 3
Subtracting 33g = 0 ⇒ g = 0
Substituting in (iv) we get – 4f = 12
f = \(\frac{12}{-4}\) = -3
Substituting the values of g, f in (i)
18(0) + 2 (-3) + c + 82 = 0
c = 6 – 82 = -76
Equation of the required circle is
x² + y² – 6y – 76 = 0
x² + y² – 6y – 76 = 6² + 10² – 6(10) – 76
= 36 + 100 – 60 – 76
= 136 – 136 = 0
D(6, 10) lies on the circle passing through A, B, C.
∴ A, B, C and D are concylic.
Equation of the circle is x² + y² – 6y – 76 = 0

Question 4.
If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic, and then find c.
Solution:
x² + y² + 2gx + 2fy + c = 0
Satisfies (2, 0), (0, 1) (4, 5) we get
4 + 0 + 4g + c₁ = 0 ………… (i)
0 + 1 + 2g. 0 + 2f + c₁ = 0 ………….. (ii)
16 + 25 + 8g + 10f + c₁ = 0 ……….. (iii)
(ii) – (i) we get
-3 – 4g + 2f = 0
4g – 2f = -3 ………. (iv)
(ii) – (iii) we get
-40 – 8g – 8f =0 (or)
g + f = – 5 ……… (v)
Solving(iv) and (v) we get
g = –\(\frac{13}{6}\), f = \(\frac{17}{6}\)
Substituting g and f values in equation (i) we get
4 + 4(-\(\frac{13}{6}\)) + c₁ = 0
c₁ = \(\frac{14}{3}\)
Now equation x² + y² – \(\frac{13}{3}\)x – \(\frac{17}{3}\)y + \(\frac{14}{3}\) = 0
Now circle passes through (0, c) then
c² – \(\frac{17}{3}\)c + \(\frac{14}{3}\)
3c² – 17c + 14 = 0
⇒ (3c – 14) (c – 1) = 0
(or)
c = 1 or \(\frac{14}{3}\).

Question 5.
Find the equation of the circum circle of the triangle formed by the straight lines given in each of the following:
i) 2x + y = 4; x + y = 6; x + 2y = 5
Solution:
AB: 2x + y = 4, AB : 2x + y = 4
BC : x + y = 6, AC : x + 2y = 5
B : (-2, 8), A : (1, 2)
AC: x + 2y = 5
BC : x + y = 6
C : (7, -1)

Equation of circle
x² + y² + 2gx + 2fy + c = 0
passes through A, B, C
∴ 4 + 64 – 4g + 16f + c = 0 ……… (i)
1 + 4 + 2g + 4f + c = 0 ………… (ii)
49 + 1 + 14g – 2f + c = 0 ……….. (iii)
(i) – (ii) we get
21 – 2g + 4f = 0 ……….. (iv)
(iii) – (ii) we get
15 + 4g – 2f = 0 ……….. (v)
Solving (iv) and (v) we get f = – \(\frac{19}{2}\)
g = –\(\frac{17}{2}\); c = 50
We get substituting g, f in equation (i)
Required equation of circle be
x² + y² – 17x – 19y + 50 = 0

ii) x + 3y – 1 = 0; x + y + 1 = 0; 2x + 3y + 4 = 0
Solution:
AB : x + 3y – 1 = 0 AB : x + 3y – 1 = 0
BC : x + y +1 = 0 AC: 2x + 3y‘ + 4 = 0
B : (-2, 1) A: (-5, 2)
AC : 2x + 3y + 4 = 0
BC : x + y + 1 = 0
C : (1, -2)

Equation of circle x² + y² + 2gx + 2fy + c = 0
A, B, C are points on circumference.
25 + 4 – 10g + 4f + c = 0 ……….. (i)
4 + 1 – 4g + 2f + c = 0 ……… (ii)
∴ 1 + 4 + 2g – 4f + c = 0 ……… (iii)
Subtracting (iii) – (ii) we get
6g – 6f = 0 (or) g = f ……… (iv)
Subtracting (iii) – (i) we get
24 – 12g + 8f = 0 ……… (v)
Solving (iv) and (v) we get
g = 6, f = 6, c = 7
Required equation of circle be
x² + y² + 12x + 12y + 7

iii) 5x – 3y + 4 = 0; 2x + 3y – 5 = 0; x + y = 0.
Solution:
AB : 5x – 3y + 4 = 0
AC : 2x + 3y – 5 = 0
BC : x + y = 0
A : (\(\frac{1}{7}\), \(\frac{11}{7}\)) C : (-5, 5)
B : (-\(\frac{1}{2}\), \(\frac{1}{2}\))
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
Points A, B, C are on circumference of circle
\(\frac{1}{49}\) + \(\frac{121}{49}\) + \(\frac{2}{7}\)g + \(\frac{22}{7}\)f + c = 0 ………. (i)
25 + 25 – 10g + 10f + c = 0 ………. (ii)
\(\frac{1}{4}\) + \(\frac{1}{4}\) – g + f + c ……….. (iii)
(Or)
1 – 2g + 2f + 2c = 0
(ii) — (iii) we get
(50 – \(\frac{1}{2}\) – 9g + 9f = 0
\(\frac{11}{2}\) – g + f = 0 ………… (iv)
(iii) – (i) we get

Solving (iv) and (v) we get
g = \(\frac{40}{14}\); f = \(\frac{-37}{14}\); h = \(\frac{70}{14}\)
x² + y² + \(\frac{80}{2}\)x + \(\frac{74}{14}\)y + \(\frac{70}{14}\) = 0
Required equation of circle be
7(x² + y²) + 40x – 37y + 35 = 0

iv) x – y – 2 = 0;
2x – 3y + 4 = 0;
3x – y + 6 = 0
Solution:
AB : x – y – 2 = 0
B : (10, 8) A: (-4, -6)
BC : 2x – 3y + 4 = 0
AC : 3x – y + 6 = 0
C : (- 2, 0)
Equation of circle is
x² + y² + 2gx + 2fy + c = 0
Points A, B, C are on circumference of circle
100 + 64 + 20g + 16f + c = 0 ……… (i)
16 + 36 – 8g – 12f + c = 0 ……… (ii)
4 – 4g + c = 0 ……….. (iii)
Solving above equations we get
g = – 12, f = 8, c = – 52
Required equation of circle is
x² + y² – 24x + 16y – 52 = 0

Question 6.
Show that the locus of the point of intersection of the lines x cos a + y sin a = a, x sin a – y cos a = b (a is a parameter) is a circle.
Solution:
Equations of the given lines are
x cos α + y sin α = a
x sin α -y cos α = b
Let P (x₁, y₁) be the point of intersection
x₁ cos α + y₁ sin α = a ………. (1)
x₁ sin α – y₁ cos α = b ……… (2)
Squaring and adding (1) and (2)
(x₁cos α + y₁sin α)² + (x₁sin α – y₁cos α)² = a² + b²
x²₁ cos²α + y²₁ sin²α + 2x₁y₁ cos α sin α + x²₁ sin² α + y²₁ cos² α – 2x₁y₁ cos α sin α = a² + b²
x²₁ (cos²α +sin²α) + y²₁ (sin² α + cos² α) = a² + b²
x²₁ + y²₁ = a² + b²
Locus of P(x₁, y₁) is the circle
x² + y² = a² + b²

Question 7.
Show that the locus of a point such that the ratio of distance of it from two given points is constant k (≠ ± 1) is a circle.
Solution:
Let P(x₁, y₁) be a point on the locus
Let A (a, 0), B(-a, 0) be two given points

By squaring and cross multiplying, we get
(x₁ – a)² + y₁² = k² [(x₁ + a)2 + y₁2 ]
⇒ (1 – k²) (x²₁ + y²₁ + a²) + (-1 – k²) (2ax₁) = 0
⇒ x²₁ + y²₁ – 2 \(\frac{(1+k^{2})}{1-k^{2}}\)ax + a² = 0
∴ Locus of P(x₁, y₁) is
x² + y² – 2\(\frac{1+k^{2}}{1-k^{2}}\)ax + a² = 0
which represents a circle. (Here k ≠± 1)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(b)

I.

Question 1.
In the experiment of tossing a coin n times, if the variable X denotes the number of heads and P(X = 4), P(X = 5), P(X = 6) are in arithmetic progression then find n.
Solution:
X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\) (∵ a coin is tossed)
Hint: a, b, c are in A.P.
⇒ 2b = a + c (or) b – a = c – a
Given, P(X = 4), P(X = 5), P(X = 6) are in A.P.

⇒ 2 × 30(n – 4) = 5[30 + n² – 9n + 20]
⇒ 12n – 48 = n² – 9n – 50
⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0
⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0
⇒ n = 7 or 14

Question 2.
Find the minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.8.
Solution:
Let n be the number of times a fair coin tossed
X denotes the number of heads getting
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Given P(X ≥ 1) ≥ 0.8
⇒ 1 – P(X = 0) ≥ 0.8
⇒ P(X = 0) ≤ 0.2
⇒ \({ }^n C_o\left(\frac{1}{2}\right)^n \leq 0.2\)
⇒ \(\left(\frac{1}{2}\right)^n \leq \frac{1}{5}\)
The Maximum value of n is 3

Question 3.
The probability of a bomb hitting a bridge is \(\frac{1}{2}\) and three direct hits (not necessarily consecutive) are needed to destroy it. Find the minimum number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
Solution:
Let n be the minimum number of bombs required and X be the number of bombs that hit the bridge, then
X follows binomial distribution with parameters n and p = \(\frac{1}{2}\)
Now P(X ≥ 3) > 0.9
⇒ 1 – P(X < 3) > 0.9
⇒ P(X < 3) < 0.1
⇒ P(X = 0) + P(X = 1) + P (X = 2) < 0.1

By trial and error, we get n ≥ 9
∴ The least value of n is 9
∴ n = 9

Question 4.
If the difference between the mean and the variance of a binomial variate is \(\frac{5}{9}\) then, find the probability for the event of 2 successes, when the experiment is conducted 5 times.
Solution:
Given n = 5
Let p be the parameters of the Binomial distribution
Mean – Variance = \(\frac{5}{9}\)

∴ Probability of the event of 2 success = \(\frac{80}{243}\)

Question 5.
One in 9 ships is likely to be wrecked when they are set on the sail, when 6 ships are on the sail, find the probability for
(i) Atleast one will arrive safely
(ii) Exactly, 3 will arrive safely
Solution:
p = probability of ship to be wrecked = \(\frac{1}{9}\)
q = 1 – p
= 1 – \(\frac{1}{9}\)
= \(\frac{8}{9}\)
Number of ships = n = 6

Question 6.
If the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X ≤ 4).
Solution:
Mean = np = 2.4 ………(1)
Variance = npq = 1.44 ……….(2)
Dividing (2) by (1)
\(\frac{n p q}{n p}=\frac{1.44}{2.4}\)
⇒ q = o.6 = \(\frac{3}{5}\)
p = 1 – q
= 1 – 0.6
= 0.4
= \(\frac{2}{5}\)
Substituting in (1)
n(0.4) = 2.4
⇒ n = 6
P(1 < X ≤ 4) = P(X = 2) + P(X = 3) + P(X = 4)
= \({ }^6 C_2 q^4 \cdot p^2+{ }^6 C_3 q^3 \cdot p^3+{ }^6 C_4 q^2 \cdot p^4\)

Question 7.
It is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.
Solution:
p = probability of defective bulb = \(\frac{1}{10}\)
q = 1 – p
= 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\)
n = Number of bulbs in the sample = 20
P(X > 2) = 1 – P(X ≤ 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

Question 8.
On average, rain falls on 12 days every 30 days, find the probability that, the rain will fall on just 3 days of a given week.
Solution:
Given p = \(\frac{12}{30}=\frac{2}{5}\)
q = 1 – p
= 1 – \(\frac{2}{5}\)
= \(\frac{3}{5}\)
n = 7, r = 3
P(X = 3) = ⁿC_r . q^n-r . p^r
= \({ }^7 C_3\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\text { 35. }\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^3\)
= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Let n, p be the parameters of a binomial distribution
Mean (np) = 6 ……..(1)
and variance (npq) = 2 ……..(2)
then \(\frac{n p q}{n p}=\frac{2}{6}\)
⇒ q = \(\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
From (1) np = 6
n(\(\frac{2}{3}\)) = 6
∴ n = 9
The first two terms of the distribution are
P(X = 0) = \({ }^9 C_0\left(\frac{1}{3}\right)^9=\frac{1}{3^9}\)
and P(X = 1) = \({ }^9 C_1\left(\frac{1}{3}\right)^8\left(\frac{2}{3}\right)=\frac{2}{3^7}\)

Question 10.
In a city, 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.
Solution:
Average number of accidents per day
λ = \(\frac{10}{50}=\frac{1}{5}\) = 0.2
The probability that there win be 3 or more accidents in a day
P(X ≥ 3) = \(\sum_{\mathrm{K}=3}^{\infty} \mathrm{e}^{-\lambda} \cdot \frac{\lambda^{\mathrm{K}}}{\mathrm{K} !}, \lambda=0.2\)

II.

Question 1.
Five coins are tossed 320 times. Find the frequencies of the distribution of a number of heads and tabulate the result.
Solution:
5 coins are tossed 320 times
Probability of getting a head on a coin
p = \(\frac{1}{2}\), n = 5
Probability of having x heads

Question 2.
Find the probability of guessing at least 6 out of 10 answers in (i) True or false type examination (ii) multiple choice with 4 possible answers.
Solution:
(i) Since the answers are in True or false type
probability of success p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10-6} \cdot\left(\frac{1}{2}\right)^6\)
= \({ }^{10} \mathrm{C}_6\left(\frac{1}{2}\right)^{10}\)

(ii) Since the answers are multiple-choice with 4 possible answers
Probability of success p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)
Probability of guessing at least 6 out of 10
P(X = 6) = \({ }^{10} C_6\left(\frac{3}{4}\right)^{10-6}\left(\frac{1}{4}\right)^6\)
= \({ }^{10} C_6 \cdot \frac{3^4}{4^{10}}\)

Question 3.
The number of persons joining a cinema ticket counter in a minute has Poisson distribution with parameter 6. Find the probability that
(i) no one joins the queue in a particular minute
(ii) two or more persons join the queue in a minute.
Solution:
Here λ = 6
(i) Probability that no one joins the queune in a particular minute
P(X = 0) = \(\frac{\mathrm{e}^{-\lambda} \lambda^0}{0 !}=\mathrm{e}^{-6}\)

(ii) Probability that two or more persons join the queue in a minute

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a)

I.

Question 1.
A p.d.f of a discrete random variable is zero except at the points x = 0, 1, 2. At these points it has the value P(0) = 3c³, P(1) = 4c – 10c², P(2) = 5c – 1 for some c > 0. Find the value of c.
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c³ + 4c – 10c² + 5c – 1 = 1
3c³ – 10c² + 9c – 2 = 0
Put c = 1, then 3 – 10 + 9 – 2 = 12 – 12 = 0
c = 1 satisfies the above equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c² – 7c + 2 = 0
(c – 2) (3c – 1) = 0
∴ c = 2 or c = \(\frac{1}{3}\)
c = 2
⇒ P(X = 0) = 3 . 2³ = 24 which is not possible
∴ c = \(\frac{1}{3}\)

Question 2.
Find the constant C, so that F(x) = \(C\left(\frac{2}{3}\right)^x\), x = 1, 2, 3,……… is the p.d.f of a discrete random variable X.
Solution:

Question 3.

is the probability distribution of a random variable X. Find the value of K and the variance of X.
Solution:
Sum of the probabilities = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 – 0.6 = 0.4
⇒ k = \(\frac{0.4}{4}\) = 0.1
Mean = (-2) (0.1) + (-1) (k) + 0 (0.2) + 1 (2k) + 2(0.3) + 3k
= – 0.2 – k + 0 + 2k + 0.6 + 3k
= 4k + 0.4
= 4(0.1) + 0.4
= 0.4 + 0.4
= 0.8
μ = 0.8
Variance (σ²) = \(\sum_{i=1}^n x_i^2 P\left(x=x_i\right)-\mu^2\)
∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1 (2k) + 4 (0.3) + 9k – μ²
= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ²
= 12k + 0.4 + 1.2 – (0.8)²
= 12(0.1) + 1.6 – 0.64
= 1.2 + 1.6 – 0.64
= 2.8 – 0.64
= 2.16
∴ σ² = 2.16

Question 4.

is the probability distribution of a random variable X. Find the variance of X.
Solution:

Question 5.
A random variable X has the following probability distribution.

Find (i) k (ii) the mean and (iii) P(0 < X < 5)
Solution:
Sum of the probabilities = 1
⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
⇒ 10k² + 9k = 1
⇒ 10k² + 9k – 1 = 0
⇒ 10k² + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\), -1 Since k > 0
∴ k = \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\)

(ii) Mean = \(\sum_{i=1}^n x_i P\left(x=x_i\right)\)
Mean (μ) = 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k2) + 6(2k²) + 7 (7k² + k)
= 0 + k + 4k + 6k + 12k + 5k² + 12k² + 49k² + 7k
= 66k² + 30k
= 66(\(\frac{1}{100}\)) + 30(\(\frac{1}{10}\))
= 0.66 + 3
= 3.66

(iii) P(0 < x < 5)
P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k
= 8k
= 8(\(\frac{1}{10}\))
= \(\frac{4}{5}\)

II.

Question 1.
The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c³, P(X = 1) = 4c – 10c², P(X = 2) = 5c – 1
(i) Find the value of c
(ii) P(X < 1), P(1 ≤ X < 2) and P(0 < X ≤ 3)
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c³ + 4c – 10c² + 5c – 1 = 1
3c³ – 10c² + 9c – 2 = 0
c = 1 satisfy this equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c² – 7c + 2 = 0
(c – 2) (3c – 1) = 0
c = 2 or c = \(\frac{1}{3}\)
c = 2 ⇒ P(X = 0) = 3 . 2³ = 24 which is not possible
∴ c = \(\frac{1}{3}\)

(i) P(X < 1) = P(X = 0)
= 3 . c³
= 3 . \(\left(\frac{1}{3}\right)^3\)
= 3 . \(\frac{1}{2}\)
= \(\frac{1}{9}\)

(ii) P(1 < X ≤ 2) = P(X = 2)
= 5c – 1
= \(\frac{5}{3}\) – 1
= \(\frac{2}{3}\)

(iii) P(0 < X ≤ 3) = P(X = 1) + P(X = 2)
= 4c – 10c² + 5c – 1
= 9c – 10c² – 1
= 9 . \(\frac{1}{3}\) – 10 . \(\frac{1}{9}\) – 1
= 3 – \(\frac{10}{9}\) – 1
= \(\frac{8}{9}\)

Question 2.
The range of a random variable X is {1, 2, 3, …..} and P(X = k) = \(\frac{C^K}{K !}\), (k = 1, 2, 3, ……), Find the value of C and P(0 < X < 3)
Solution:
Sum of the probabilities = 1

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(c)

I.

Question 1.
Three screws are drawn at random from a lot of 50 screws, 5 of which are defective. Find the probability of the event that all 3 screws are non-defective assuming that the drawing is (a) with replacement (b) without replacement.
Solution:
Let S be the sample space
∴ The total number of screws = 50
The number of defective screws is 5 and the remaining 45 screws are non-defective.
Let A be the event of getting a drawing of the 3 screws is non-defective.
(a) With replacement

(b) Without replacement

Question 2.
If A, B, C are three independent events of an experiment such that P(A ∩ B^C ∩ C^C) = \(\frac{1}{4}\), P(A^C ∩ B ∩ C^C) = \(\frac{1}{8}\), P(A^C ∩ B^C ∩ C^C) = \(\frac{1}{4}\), then find P(A), P(B) and P(C).
Solution:
Since A, B, C are independent events.

Question 3.
There are 3 black and 4 white balls in one bag. 4 black and 3 white balls in the second bag. A die is rolled and the first bag is selected if it is 1 or 3 and the second bag for the rest. Find the probability of drawing a black ball from the bag thus selected.
Solution:
Probability of selecting first bag = \(\frac{2}{6}=\frac{1}{3}\)
Probability of selecting second bag = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Probability of getting a black ball from first bag = \(\frac{3}{7}\)
Probability of getting a black ball from the second bag = \(\frac{4}{7}\)
Probability of drawing a black ball = \(\frac{1}{3} \cdot \frac{3}{7}+\frac{2}{3} \cdot \frac{4}{7}\) = \(\frac{11}{21}\)

Question 4.
A, B, C are aiming to shoot a balloon, A will succeed 4 times out of 5 attempts. The chance of B shooting the balloon is 3 out of 4 and that of C is 2 out of 3. If three aim at the balloon simultaneously, then find the probability that atleast two of them hit the balloon.
Solution:

Question 5.
If A, B are two events, then show that \(P\left(\frac{A}{B}\right) P(B)+P\left(\frac{A}{B^C}\right) P\left(B^C\right)=P(A)\)
Solution:

Question 6.
A pair of dice are rolled. What is the probability that they sum to 7, given that neither die shows a 2?
Solution:
Let A be the event that the sum of the two dice is 7, then
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Let B be the event that neither die shows a 2
B = {(1, 1), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 25
A ∩ B = {(1, 6), (3, 4), (4, 3), (6, 1)}
n(A ∩ B) = 4
Required probability
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
= \(\frac{n(A \cap B)}{n(B)}\)
= \(\frac{4}{25}\)

Question 7.
A pair of dice are rolled. What is the probability that neither die shows a 2, given that they sum to 7?
Solution:
Let A be the event that the sum on two dice is 7
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ n(A) = 6
Let B be the event that neither die shows a 2
B = {(1, 1), (1, 3), (1, 4), (1, 5), (1, 6)
(3, 1), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1),(4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 3), (6, 4), (6, 5), (6, 6)}
A ∩ B = {(1, 6), (3, 4), (4, 3), (6, 1)}
n(A ∩ B) = 4
Required probability
\(P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}\)
= \(\frac{n(A \cap B)}{n(A)}\)
= \(\frac{4}{6}\)
= \(\frac{2}{3}\)

Question 8.
If A, B are any two events, in an experiment, and P(B) ≠ 1 Show that \(P\left(\frac{A}{B^C}\right)=\frac{P(A)-P(A \cap B)}{1-P(B)}\)
Hint: P(A ∩ B^C) = P(A) – P(A ∩ B)
Solution:
By definition of condition probability
\(P\left(\frac{A}{B^C}\right)=\frac{P\left(A \cap B^C\right)}{P\left(B^C\right)}\)
= \(\frac{P(A)-P(A \cap B)}{1-P(B)}\)
∵ P(B^C) = \(P(\bar{B})\) = 1 – P(B)

Question 9.
An urn contains 12 red balls and 12 green balls. Suppose two balls are drawn one after another without replacement. Find the probability that the second ball drawn is green given that the first ball drawn is red.
Solution:
Total number of balls in an urn n(S) = 24
Let E₁ be the event of drawing a red ball in the first draw
P(E₁) = \(\frac{{ }^{12} C_1}{24}=\frac{1}{2}\)
Now the number of balls remaining is 23
Let \(\frac{E_2}{E_1}\) be the events of drawing a green ball in the second drawn
P(E₂/E₁) = \(\frac{12}{23}\)
∴ Required probability
P(E₁ ∩ E₂) = P(E₁) . P(E₂/E₁)
= \(\frac{1}{2} \times \frac{12}{23}\)
= \(\frac{6}{23}\)
∴ The probability of the second ball drawn is green given that the first ball drawn is red = \(\frac{6}{23}\)

Question 10.
A single die is rolled twice in succession. What is the probability that the number showing on the second toss is greater than that on the first rolling?
Solution:
A single die is rolled twice.
Let S be the sample space n(S) = 6² = 36
Let A be the event of getting the required event.
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}
n(A) = 15
∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

Question 11.
If one card is drawn at random from a pack of cards then show that events of getting an ace and getting a heart are independent events.
Solution:
Suppose A is the event of getting an ace and B is the event of getting a heart.
∴ P(A) = \(\frac{4}{52}=\frac{1}{13}\)
P(B) = \(\frac{13}{52}=\frac{1}{14}\)
A ∩ B is the event of getting a Heart’s ace
P(A ∩ B) = \(\frac{1}{52}=\frac{1}{13} \cdot \frac{1}{4}\) = P(A) . P(B)
∴ A and B are independent events.

Question 12.
The probability that boy A will get a scholarship is 0.9 and that another boy B will get one is 0.8. What is the probability that atleast one of them will get the scholarship?
Solution:
Suppose E₁ is the event of a boy ‘A’ getting a scholarship and E₂ is the event of another boy B getting the scholarship.
Given P(E₁) = 0.9, P(E₂) = 0.8
E₁ and E₂ are independent events.
P(E₁ ∩ E₂) = P(E₁) . P(E₂)
= (0.9) (0.8)
= 0.72
The probability that atleast one of them will get a scholarship = P(E₁ ∪ E₂)
= P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= 0.9 + 0.8 – 0.72
= 1.7 – 0.72
= 0.98

Question 13.
If A, B are two events with P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then find the value of P(A^C) + P(B^C).
Solution:
By addition theorem on probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A) + P(B) = P(A ∪ B) + P(A ∩ B)
= 0.65 + 0.15
= 0.8
P(A) + P(B) = 0.8 ……..(1)
P(A^C) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 0.8 [∵ by (1)]
= 1.2

Question 14.
If A, B, C are independent events, show that A ∪ B, and C are also independent events.
Solution:
∵ A, B, C are independent events.
⇒ A, B; B, C; C , A are also independent events
P(A ∩ B ∩ C) = P(A) P(B) P(C)
P(A ∩ C) = P(A) . P(C)
P(B ∩ C) = P(B) . P(C)
P(A ∩ B) = P(A) . P(B)
P[(A ∪ B) ∩ C] = P[(A ∩ C) ∪ (B ∩ C)]
= P(A ∩ C) + P(B ∩ C) – P[(A ∩ C) ∩ (B ∩ C)]
= P(A) . P(C) + P(B) . P(C) – P(A ∩ B ∩ C)
= P (A) . P(C) + P(B) . P(C) – P(A) . P(B) . P(C)
= [P(A) + P(B) – P(A) . P(B)] P(C)
= P(A ∪ B) . P(C)
∴ A ∪ B and C are independent events.

Question 15.
A and B are two independent events such that the probability of both the events occurring is \(\frac{1}{6}\) and the probability of both the events do not occur is \(\frac{1}{3}\). Find the probability of A.
Solution:
A and B are independent events.

Question 16.
A fair die is rolled. Consider the events. A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}. Find
(i) P(A ∩ B), P(A ∪ B)
(ii) P(\(\frac{A}{B}\)), P(\(\frac{B}{A}\))
(iii) P(\(\frac{A}{C}\)), P(\(\frac{C}{A}\))
(iv) P(\(\frac{B}{C}\)), P(\(\frac{C}{B}\))
Solution:
A fair die is rolled
P(A) = \(\frac{3}{6}=\frac{1}{2}\)
P(B) = \(\frac{2}{6}=\frac{1}{3}\)
P(C) = \(\frac{4}{6}=\frac{2}{3}\)
n(S) = 6¹ = 6
Given A = {1, 3, 5}, B = {2, 3}, C = {2, 3, 4, 5}
(i) A ∩ B = {3}
P(A ∩ B) = P{3} = \(\frac{1}{6}\)
∴ P(A ∩ B) = \(\frac{1}{6}\)
(A ∪ B) = {1, 2, 3, 5}
n(A ∪ B) = 4
n(S) = 6
P(A ∪ B) = \(\frac{4}{6}=\frac{2}{3}\)

Question 17.
If A, B, C are three events in a random experiment, prove the following:
(i) P(\(\frac{A}{A}\)) = 1
Solution:
\(P\left(\frac{A}{A}\right)=\frac{P(A \cap A)}{P(A)}=\frac{P(A)}{P(A)}=1\)

(ii) \(\mathbf{p}\left(\frac{\phi}{A}\right)=0\)
Solution:
\(P\left(\frac{\phi}{A}\right)=\frac{P(\phi \cap A)}{P(A)}=\frac{0}{P(A)}=0\)

(iii) A ⊂ B ⇒ P(\(\frac{A}{C}\)) ≤ P(\(\frac{B}{C}\))
Solution:

(iv) P(A – B) = P(A) – P(A ∩ B)
Solution:
A – B = {x / x ∈ A a x ∉ B}
A – B = A – (A ∩ B)

P(A – B) = P[A – (A ∩ B)] = P(A) – P(A ∩ B)

(v) If A, B are mutually exclusive and P(B) > 0, then P(\(\frac{A}{B}\)) = 0.
Solution:
We know \(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\)
= \(\frac{0}{\mathrm{P}(\mathrm{B})}\)
= 0 [∵ A, B are mutually exclusive events]
Hint: A, B are mutually exclusive then A ∩ B = φ ⇒ P(A ∩ B) = 0.

(vi) If A, B are mutually exclusive then P(A/B^C) = \(\frac{\mathrm{P}(\mathrm{A})}{1-\mathrm{P}(\mathrm{B})}\); when P(B) ≠ 1.
Solution:
Given P(A ∩ B) = 0 (∵ A and B are mutually exchanging)

(vii) If A, B are mutually exclusive and P(A ∪ B) ≠ 0, then \(P\left(\frac{A}{A \cup B}\right)=\frac{P(A)}{P(A)+P(B)}\)
Solution:
Hint: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Question 18.
Suppose that a coin is tossed three times. Let A be “getting three heads” and B be the event of “getting a head on the first toss”. Show that A and B are dependent events.
Solution:
Let event A be “getting their heads”, and B be the event of “getting a head on the first toss” when a coin is tossed three times.
∴ A = {HHH}
n(A) = 1
P(A) = \(\frac{1}{8}\)
B = {HTT, HTH, HHT, HHH}
n(B) = 4
P(B) = \(\frac{4}{8}\)
A ∩ B = {HHH}
n(A ∩ B) = 1
P(A ∩ B) = \(\frac{1}{8}\)
P(A) . P(B) = \(\frac{1}{8} \cdot \frac{4}{8}=\frac{1}{16}\) ≠ P(A ∩ B)
∴ P(A ∩ B) ≠ P(A) . P(B)
Hence A, B are dependent events.

Question 19.
Suppose that an unbiased pair of dice is rolled. Let A denote the event that the same number shows on each die. Let B denote the event that the sum is greater than 7. Find (i) P(\(\frac{A}{B}\)) (ii) P(\(\frac{B}{A}\))
Solution:
Given a denote the event that the same number on pair of dice is rolled.
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) (6, 6)}
n(A) = 6
P(A) = \(\frac{6}{36}=\frac{1}{6}\)
Given B denote the event that the sum is greater than 7 when pair of dice is rolled.
∴ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ n(B) = 15
P(B) = \(\frac{15}{36}\)
A ∩ B = {(4, 4), (5, 5), (6, 6)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)

Question 20.
Prove that A and B are independent events if and only if \(P\left(\frac{A}{B}\right)=P\left(\frac{A}{B^C}\right)\)
Solution:
Let A and B be independent

⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(A)-P(A \cap B)}{1-P(B)}\)
⇒ P(A ∩ B) – P(B) P(A ∩ B) = P(A) P(B) – P(B) . P(A ∩ B)
⇒ P(A ∩ B) = P(A) . P(B)
∴ A, B are independent.
Hence, A, B are independent iff
\(P\left(\frac{A}{B}\right)=P\left(\frac{A}{B^C}\right)\)

II.

Question 1.
Suppose A and B are independent events with P(A) = 0.6, P(B) = 0.7 then compute
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P(\(\frac{B}{A}\))
(iv) P(A^C ∩ B^C)
Solution:
Given A, B are independent events and
P(A) = 0.6, P(B) = 0.7
(i) P(A ∩ B) = P(A) . P(B)
= 0.6 × 0.7
= 0.42
(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.6 + 0.7 – 0.42
= 1.3 – 0.42
= 0.88
(iii) P(\(\frac{B}{A}\)) = P(B) = 0.7
(iv) P(A^C ∩ B^C) = P(A^C) . P(B^C) (∵ A^C & B^C are also independent events)
= [1 – P(A)] [1 – P(B)]
= (1 – 0.6) (1 – 0.7)
= 0.4 × 0.3
= 0.12

Question 2.
The probability that Australia wins a match against India in a cricket game is given to be \(\frac{1}{3}\). If India and Australia play 3 matches, what is the probability that,
(i) Australia will lose all three matches?
(ii) Australia will win atleast one match?
Solution:
Suppose A is the event of Australia winning the match.
Given P(A) = \(\frac{1}{3}\)
∴ P(\(\overline{\mathrm{A}}\)) = 1 – P(A)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
(i) Probability that Australia will loose all the three matches = P(\(\overline{\mathrm{A}}\))³
= \(\left(\frac{2}{3}\right)^3\)
= \(\frac{8}{27}\)
(ii) Probability that Australia will win atleast one match = 1 – P(\(\overline{\mathrm{A}}\))³
= 1 – \(\frac{8}{27}\)
= \(\frac{19}{27}\)

Question 3.
Three boxes numbered I, II, III contain balls as follows:

One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that it is from box II.
Solution:
Let B₁, B₂, B₃ be the events of selecting the Ist, IInd and IIIrd boxes respectively.
Then P(B₁) = P(B₂) = P(B₃) = \(\frac{1}{3}\)
Probability of selecting a red ball from the first box = \(\frac{3}{6}\) = P(R/B₁)
Probability of selecting a red ball from the second box = \(\frac{1}{4}\) = P(R/B₂)
Probability of selecting a red ball from the third box = \(\frac{3}{12}\) = P(R/B₃)
Assuming that the ball is red, the probability it is from box II,

Question 4.
A person secures a job in a construction company in which the probability that the workers go on strike is 0.65 and the probability that the construction job will be completed on time if there is no strike is 0.80. If the probability that the construction job will be completed on time even if there is a strike is 0.32, determine the probability that the constructed job will be completed on time.
Solution:
Let P(S) = Probability of the workers going on strike = 0.65
P(\(\bar{S}\)) = Probability on the workers go on strike
= 1 – P(S)
= 1 – 0.65
= 0.35
\(P\left(\frac{E}{S}\right)\) = Probability that the job completed if there is no strike = 0.32
\(\mathrm{P}\left(\frac{\mathrm{E}}{\overline{\mathrm{S}}}\right)\) = Probability that the job completed if there is a strike = 0.80
P(E) = Probability that the construction job will be completed on time
= P(S) \(P\left(\frac{E}{S}\right)\) + P(\(\bar{S}\)) \(\mathrm{P}\left(\frac{\mathrm{E}}{\overline{\mathrm{S}}}\right)\)
= (0.65) (0.32) + (0.35) (0.08)
= 0.2080 + 0.2800
= 0.4880

Question 5.
For any two events A, B show that
P(A ∩ B) – P(A) P(B) = P(A^C) P(B) – P(A^C ∩ B) = P(A) P(B^C) – P(A ∩ B^C)
Solution:
R.H.S. I = P(A^C) P(B) – P(A^C ∩ B)
= [(1 – P(A)] P(B) – [P(B) – P(A ∩ B)]
= P(B) – P(A) P(B) – P(B) + P(A ∩ B)
= P(A ∩ B) – P(A) P(B)
= L.H.S
R.H.S. II = P(A) P(B^C) – P(A ∩ B^C)
= P(A) [1 – P[P(B)] – [P(A) – P(A ∩ B)]
= P(A) – P(A) P(B) – P(A) + P(A ∩ B)
= P(A ∩ B) – P(A) P(B)
= L.H.S
∴ L.H.S = R.H.S I = R.H.S II
Hence P(A ∩ B) – P(A) . P(B) = P(A^C) P(B) – P(A^C ∩ B) = P(A) P(B^C) – P(A ∩ B^C)

III.

Question 1.
Three Urns have the following composition of balls.
Urn I: 1 White, 2 black
Urn II: 2 White, 1 black
Urn III: 2 White, 2 balck
One of the Urn is selected at random and a ball is drawn. It turns out to be white. Find the probability that it comes from Urn III.
Solution:
Let E_i be the event of Choosing the Urn i = 1, 2, 3, and P(E_i) be the probability of choosing the Urn i = 1, 2, 3
Then P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)
Having choosen the Urn i, the probability of drawing a white ball, P(W/E_i), is given by
P(W/E₁) = \(\frac{1}{3}\)
P(W/E₂) = \(\frac{2}{3}\)
P(W/E₃) = \(\frac{2}{4}\)
We have to find the probability P(E₃/W) by Baye’s theorem.

Question 2.
In a shooting test the probability of A, B, C hitting the targets are \(\frac{1}{2}\), \(\frac{2}{3}\) and \(\frac{3}{4}\) respectively. If all of their fire is at the same target. Find the probability that
(i) Only one of them hits the target
(ii) At atleast one of them hits the target
Solution:
The probabilities that A, B, C hit the targets are denoted by

Question 3.
In a certain college, 25% of the boys and 10% of the girls are studying mathematics. The girls constitute 60% of the student’s strength. If a student selected at random is found studying mathematics. Find the probability that the student is a girl.
Solution:
The probability that a student selected to be a girl
P(G) = \(\frac{60}{100}=\frac{6}{10}\)
The probability that a student selected to be a boy
P(B) = \(\frac{40}{100}=\frac{4}{10}\)
The probability that a boy studying mathematics
P(M/B) = \(\frac{25}{100}=\frac{1}{4}\)
Similarly probability that a girl studying mathematics
P(M/G) = \(\frac{10}{100}=\frac{1}{10}\)
We have to find P(G/M) By Baye’s theorem

Question 4.
A person is known to speak the truth 2 out of 3 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.
Solution:
P(T) = Probability that a person speaks truth 2 out of 3 times = \(\frac{2}{3}\)
P(F) = 1 – P(T)
= 1 – \(\frac{2}{3}\)
= \(\frac{1}{3}\)
After he reports that it is 1, it is true if it actually shows 1 otherwise false if does not show 1.
P(1) = \(\frac{1}{6}\) and P(T) = \(\frac{5}{6}\)
P(T/1) = P(reports true if it is 1) = \(\frac{2}{3}\)
P(F/T) = P(report False if it is T) = \(\frac{1}{3}\)
By Baye’s theorem

∴ The probability that reports that it is 1 and actually it is 1 is \(\frac{2}{7}\).

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(b)

I.

Question 1.
If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear.
Solution:
4 coins are tossed simultaneously.
Total number of ways = 2⁴ = 16
n(S) = 16
From 4 heads we must get 2 heads.
Number of ways of getting 2 heads = ⁴C₂
= \(\frac{4.3}{1.2}\)
= 6
∴ n(E) = 6
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)
∴ Probability of getting 2 heads and 2 tails = \(\frac{3}{8}\)

Question 2.
Find the probability that a non-leap year contains
(i) 53 Sundays
(ii) 52 Sundays only
Solution:
A non-leap year contains 365 days 52 weeks and 1 day more.
(i) We get 53 Sundays when the remaining day is Sunday.
Number of days in the week = 7
∴ n(S) = 7
The number of ways getting 53 Sundays.
n(E) = 1
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{1}{7}\)
∴ Probability of getting 53 Sundays = \(\frac{1}{7}\)
(ii) Probability of getting 52 Sundays
P(\((\overline{\mathrm{E}})\)) = 1 – P(E)
= 1 – \(\frac{1}{7}\)
= \(\frac{6}{7}\)

Question 3.
Two dice are rolled, what is the probability that none of the dice shows the number 2?
Solution:
The random experiment is rolling 2 dice.
n(S) = 6² = 36
Let E be the event of not getting 2
n(E) = 5 × 5 = 25
∴ P(E) = \(\frac{\mathrm{n}(E)}{\mathrm{n}(s)}=\frac{25}{36}\)

Question 4.
In an experiment of drawing a card at random from a pack, the event of getting a spade is denoted by A, and getting a pictured card (King, Queen, or Jack) is denoted by B. Find the probabilities of A, B, A ∩ B, and A ∪ B.
Solution:
A is the event of getting a spade from the pack
∴ P(A) = \(\frac{13}{52}=\frac{1}{4}\)
B is the event of getting a picture card
P(B) = \(\frac{4 \times 3}{52}=\frac{3}{13}\)
A ∩ B is the event of getting a picture card in spades.
n(A ∩ B) = 3, n(s) = 52
P(A ∩ B) = \(\frac{3}{52}\)
A ∪ B is the event of getting a spade or a picture card.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{4}+\frac{3}{13}-\frac{3}{52}\)
= \(\frac{13+12-3}{52}\)
= \(\frac{11}{26}\)

Question 5.
In a class of 60 boys and 20 girls, half of the boys and half of the girls know cricket. Find the probability of the event that a person selected from the class is either a boy or a girl who knows cricket.
Solution:
Let A be the event that the selected person is a boy and B be the event that the selected person knows a cricket when a person is selected from the class and S be the sample space.
Now, n(S) = ⁸⁰C₁ = 80
n(A) = ⁶⁰C₁ = 60
n(B) = ⁴⁰C₁ = 40
n(A ∩ B) = ³⁰C₁ = 30
∴ P(A) = \(\frac{60}{80}\), P(B) = \(\frac{40}{80}\), P(A ∩ B) = \(\frac{30}{80}\)
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{60}{80}+\frac{40}{80}-\frac{30}{80}\)
= \(\frac{70}{80}\)
= \(\frac{7}{8}\)

Question 6.
For any two events A and B, show that P(A^C ∩ B^C) = 1 + P(A ∩ B) – P(A) – P(B).
Solution:
A^C ∩ B^C = \(\overline{A \cup B}\)
P(A^C ∩ B^C) = P(\(\overline{A \cup B}\))
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 + P(A ∩ B) – P(A) – P(B)

Question 7.
Two persons A and B are rolling die on the condition that the person who gets 3 will win the game. If A starts the game, then find the probabilities of A and B respectively to win the game.
Solution:
p = Probability of getting 3 = \(\frac{1}{6}\)
q = 1 – p
= 1 – \(\frac{1}{6}\)
= \(\frac{5}{6}\)
Probability success (p) = \(\frac{1}{6}\)
Probability of failure (q) = \(\frac{5}{6}\)
A may win the game either in I trial or in trial or in V trial etc.
Probability of A win = p + q . q . p + q . q . q . q . p + ……..

Question 8.
A, B, C are 3 newspapers from a city. 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C, and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = \(\frac{20}{100}\) = 0.2
P(B) = \(\frac{16}{100}\) = 0.16
P(C) = \(\frac{14}{100}\) = 0.14
P(A ∩ B) = \(\frac{8}{100}\) = 0.08
P(B ∩ C) = \(\frac{4}{100}\) = 0.04
P(A ∩ C) = \(\frac{5}{100}\) = 0.05
P(A ∩ B ∩ C) = \(\frac{2}{100}\) = 0.02

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= 0.2 + 0.16 + 0.14 – 0.08 – 0.04 – 0.05 + 0.02
= 0.52 – 0.17
= 0.35
Percentage of population who read atleast one newspaper = 0.35 × 100 = 35%

Question 9.
If one ticket is randomly selected from tickets numbered 1 to 30. Then find the probability that the number on the ticket is
(i) a multiple of 5 or 7
(ii) a multiple of 3 or 5
Solution:
(i) Number of ways drawing one ticket = n(S) = ³⁰C₁ = 30
Suppose A is the event of getting a multiple of 5 and B is the event of getting a multiple of 7.
A = {5, 10, 15, 20, 25, 30}
B = {7, 14, 21, 28}
A ∩ B = φ ⇒ A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
= \(\frac{6}{30}+\frac{4}{30}\)
= \(\frac{1}{3}\)
Probability of getting a multiple of 5 or 7 = \(\frac{1}{3}\)
(ii) Suppose A is the event of getting a multiple of 3 and B is the event of getting a multiple of 5.
A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
B = {5, 10, 15, 20, 25, 30}
A ∩ B = {15, 30}
P(A) = \(\frac{10}{30}\)
P(B) = \(\frac{6}{30}\)
P(A ∩ B) = \(\frac{2}{30}\)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{10}{30}+\frac{6}{30}-\frac{2}{30}\)
= \(\frac{14}{30}\)
= \(\frac{7}{15}\)
Probability of getting a multiple of 3 or 5 = \(\frac{7}{15}\)

Question 10.
If two numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is (i) an even number (ii) an odd number.
Solution:
(i) Let A be the event that the sum of the numbers is even when two numbers are selected out of 20 consecutive natural numbers.
In 20 consecutive natural numbers, we have 10 odd and 10 even natural numbers.
∵ The sum of two odd natural numbers is an even number and the sum of two even natural numbers is also an even number

(ii) Probability that the sum of two numbers is an odd number
P(\(\overline{\mathrm{A}}\)) = 1 – P(A)
= 1 – \(\frac{9}{19}\)
= \(\frac{10}{19}\)

Question 11.
A game consists of tossing a coin 3 times and noting its outcome. A boy wins if all tosses give the same outcome and lose otherwise. Find the probability that the boy loses the game.
Solution:
Let A be the event that the boy wins the game getting the same outcome when a coin is tossed 3 times and S be the sample space.
∴ n(S) = 2³ = 8
A = {HHH, TTT}
n(A) = 2
P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{8}=\frac{1}{4}\)
∴ The probability that the boy loses the game = P(\(\overline{\mathrm{A}}\))
= 1 – P(A)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

Question 12.
If E₁, E₂ are two events with E₁ ∩ E₂ = φ then show that \(P\left(E_1^c \cap E_2^c\right)=P\left(E_1^c\right)-P\left(E_2\right)\)
Solution:
Given E₁ ∩ E₂ = φ
∴ P(E₁ ∩ E₂) = 0
\(P\left(E_1^c \cap E_2^c\right)=P\left(\overline{E_1 \cup E_2}\right)\)
= 1 – P(E₁ ∪ E₂)
= 1 – [P(E₁) + P(E₂) – P(E₁ ∩ E₂)]
= 1 – P(E₁) – P(E₂) + P(E₁ ∩ E₂)
= \(P\left(E_1^c\right)\) – P(E₂) + 0
∴ \(P\left(E_1^c \cap E_2^c\right)=P\left(E_1^c\right)-P\left(E_2\right)\)

II.

Question 1.
A pair of dice rolled 24 times. A person wins by not getting a pair of 6’s on any of the 24 rolls. What is the probability of his winning?
Solution:
Random experiment is tossing two dice 24 times = 36 × 36 × ……. 36 = (36)²⁴
∴ n(S) = (36)²⁴
Let A be the event of not getting a pair of 6’s on any of the 24 rolls.
∴ number of ways favourable to an event A = 35 × 35 × ……. × 35 = (35)²⁴
n(A) = (35)²⁴
∴ P(A) = \(\frac{(35)^{24}}{(36)^{24}}=\left(\frac{35}{36}\right)^{24}\)

Question 2.
If P is a probability function, then show that for any two events A and B.
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)
Solution:
For any sets A, B we have

A ∩ B ≤ A ≤ A ∪ B
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B)
By Addition theorem of probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ≤ P(A) + P(B) ……….(2)
From (1), (2) we get
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)

Question 3.
In a box containing 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, find the probability of the event, that
(i) None of them is defective
(ii) Only one of them is defective
(iii) Atleast one of them is defective
Solution:
Out of 15 bulbs, 5 are defective
probability of selecting a defective bulb = P
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
We are selecting 5 bulbs
n(S) = ¹⁵C₅
(i) None of them is defective.
All 5 bulbs must be selected from 10 good bulbs. This can be done in ¹⁰C₅ ways.
P(A) = \(\frac{{ }^{10} C_5}{{ }^{15} C_5}=\frac{10.9 .8 .7 .6}{15.14 .13 .12 .11}=\frac{12}{143}\)
(ii) Only one of them is defective in 4 good and 1 defective ball.
This can be done in (¹⁰C₄) (⁵C₁) = \(\frac{10.9 .8 .7}{1.2 \cdot 3.4} \cdot 5\)
= 210 × 5
= 1050
Probability of selecting one defective = \(\frac{1050}{{ }^{15} C_5}\)
= \((1050) \frac{1.2 .3 .4 .5}{15.14 .13 .12 .11}\)
= \(\frac{50}{143}\)
(iii) Probability atleast one of them is defective = P(\(\overline{A}\))
= 1 – P(A)
= 1 – \(\frac{12}{143}\)
= \(\frac{131}{143}\)

Question 4.
A and B are seeking admission into I.I.T. If the probability for A to be selected is 0.5 and that both to be selected is 0.3. Is it possible that the probability of B being selected is 0.9?
Solution:
Given P(A) = 0.5; P(A ∩ B) = 0.3
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.5 + P(B) – 0.3
= 0.2 + P(B)
P(A ∪ B) ≤ 1
0.2 + P(B) ≤ 1
P(B) ≤ 0.8
∴ It is not possible to have P(B) = 0.9

Question 5.
The probability for a contractor to get a road contract is \(\frac{2}{3}\) and to got a building contract is \(\frac{5}{9}\). The probability to get atleast on contract is \(\frac{4}{5}\). Find the probability to get both contracts.
Solution:
Suppose A is the event of getting a road contract
B is the event of getting a building contract
Given P(A) = \(\frac{2}{3}\); P(B) = \(\frac{5}{9}\); P(A ∪ B) = \(\frac{4}{5}\)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{2}{3}+\frac{5}{9}-\frac{4}{5}\)
= \(\frac{30+25-36}{45}\)
= \(\frac{19}{45}\)
∴ Probability to get both contracts = \(\frac{19}{45}\)

Question 6.
In a committee of 25 members, each member is proficient either in Mathematics or Statistics or in both. If 19 of these are proficient in Mathematics, and 16 in Statistics, find the probability that a person selected from the committee is proficient in both.
Solution:
When a person is chosen at random from the academy consisting of 25 members
Let A be the event that the person is proficient in Mathematics
B be the event that the person is proficient in Statistics and
S is the sample space.
Since 19 members are proficient in Mathematics and 16 members are proficient in Statistics.
P(A) = \(\frac{19}{25}\), P(B) = \(\frac{16}{25}\)
Since everyone is either proficient in Mathematics or Statistics or in both
A ∪ B = S
⇒ P(A ∪ B) = P(S)
⇒ P(A) + P(B) – P(A ∩ B) = P(S)
⇒ \(\frac{19}{25}+\frac{16}{25}\) – P(A ∩ B) = 1
⇒ P(A ∩ B) = \(\frac{19}{25}+\frac{16}{25}-1\) = \(\frac{10}{25}\)
∴ P(A ∩ B) = \(\frac{2}{5}\)

Question 7.
A, B, C are three horses in a race. The probability of A winning the race is twice that of B and the probability of B is twice that of C. What are the probabilities of A, B, and C winning the race?
Solution:
Let A, B, C be the events that horses A, B, and C win the race respectively.
Given P(A) = 2P(B), P(B) = 2P(C)
∴ P(A) = 2P(B) = 2[2P(C)] = 4P(C)
Since the horses A, B and C run the race,
A ∪ B ∪ C = S and A, B, C are mutually disjoint.
P(A ∪ B ∪ C) = P(A) + P(B) + P(C)
⇒ P(S) = 4P(C) + 2P(C) + P(C)
⇒ 1 = 7P(C)
⇒ P(C) = \(\frac{1}{7}\)
P(A) = 4P(C) = 4 × \(\frac{1}{7}\) = \(\frac{4}{7}\)
P(B) = 2P(C) = 2 × \(\frac{1}{7}\) = \(\frac{2}{7}\)
∴ P(A) = \(\frac{4}{7}\), P(B) = \(\frac{2}{7}\), P(C) = \(\frac{1}{7}\)

Question 8.
A bag contains 12 two rupee coins, 7 one rupee coins, and 4 half rupee coins. If 3 coins are selected at random find the probability that
(i) The sum of the 3 coins is maximum
(ii) The sum of the 3 coins is minimum
(iii) Each coin is of a different value
Solution:
In the bag, there are 12 two rupees, 7 one rupees, and 4 half rupee coins.
Total number of coins = 12 + 7 + 4 = 23
Number of ways drawing 3 coins ²³C₃
n(S) = ²³C₃
(i) We get the maximum amount if the coins are 2 rupee coins.
Number of drawing 3 two rupee coins = ¹²C₃
n(E₁) = ¹²C₃
P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{{ }^{12} C_3}{{ }^{23} C_3}\)
(ii) We get a minimum amount if 3 coins are taken from 4 half rupee coins.
Number of ways of drawing 3 half rupee coins = ⁴C₃
n(E₂) = ⁴C₃
P(E₂) = \(\frac{n\left(E_2\right)}{n(S)}=\frac{{ }^4 C_3}{{ }^{23} C_3}\)
(iii) Each coin is of different value we must draw one coin each.
This can be done in ¹²C₁, ⁷C₁, ⁴C₁ ways
n(E₃) = ¹²C₁ × ⁷C₁ × ⁴C₁ = 12 × 7 × 4
P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{12 \times 7 \times 4}{{ }^{23} C_3}\)

Question 9.
The probabilities of three events A, B, C are such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09 and P(A ∪ B ∪ C) ≥ 0.75. Show that P(B ∩ C) lies in the interval [0.23, 0.48]
Solution:
P(A ∪ B ∪ C) ≥ 0.75
0.75 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0.75 ≤ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) ≤ 1
⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 – 0.28 – P(B ∩ C) + 0.09 ≤ 1
⇒ 0.75 ≤ 1.23 – P(B ∩ C) ≤ 1
⇒ -0.75 ≥ P(B ∩ C) – 1.23 ≥ -1
⇒ 0.48 ≥ P(B ∩ C) ≥ 0.23
⇒ 0.23 ≤ P(B ∩ C) ≤ 0.48
∴ P(B ∩ C) lies in the interval [0.23, 0.48]

Question 10.
The probabilities of three mutually exclusive events are respectively given as \(\frac{1+3 P}{3}, \frac{1-P}{4}, \frac{1-2 P}{2}\). Prove that \(\frac{-1}{3} \leq P \leq \frac{1}{2}\)
Solution:
Three mutually exclusive events probabilities are given.
∴ 0 ≤ \(\frac{1+3 P}{3}\) ≤ 1
⇒ 0 ≤ 1 + 3P ≤ 3
⇒ -1 ≤ P ≤ 2
⇒ \(\frac{-1}{3} \leq P \leq \frac{2}{3}\) …….(1)
Also 0 ≤ \(\frac{1-\mathrm{P}}{4}\) ≤ 1
⇒ 0 ≤ 1 – P ≤ 4
⇒ -1 ≤ -P ≤ 3
⇒ 1 ≥ P ≥ -3
⇒ -3 ≤ P ≤ 1 …….(2)
Also 0 ≤ \(\frac{1-2 P}{2}\) ≤ 1
⇒ 0 ≤ 1 – 2P ≤ 2
⇒ -1 ≤ -2P ≤ 1
⇒ 1 ≥ 2P ≥ -1
⇒ -1 ≤ 2P ≤ 1
⇒ \(\frac{-1}{2} \leq P \leq \frac{1}{2}\) ………(3)
From (1), (2), (3) we have \(\frac{-1}{3} \leq P \leq \frac{1}{2}\)

Question 11.
On a Festival day, a man plans to visit 4 holy temples A, B, C, D in random order. Find the probability that he visits (i) A before B (ii) A before B and B before C.
Solution:
(i) It is nothing but arranging A, B, C, D in 4 chains so that A seats before B.
For this, first, we arrange C, D is the 4 chains it can be done in ⁴P₂ = 12 ways and the remaining 2 seats A & B can be sit in only one way (A before B).
Also n(S) = 24
∴ The probability that he visit 4 temples, A before B = \(\frac{12 \times 1}{24}\) = \(\frac{1}{2}\)
(ii) Similarly to the above problem
First, we arrange D in any one of 4 chains it can be done in 4 ways. Then the remaining 3 seats A, B, C can sit in only one way (A before 3 and B before C)
Also n(A) = 4! = 24
The probability that he visit 4 temples, A before B and B before C = \(\frac{4 \times 1}{24}\) = \(\frac{1}{6}\)

Question 12.
From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. The particulars of 5 persons are as follows.

A person is selected at random from this group to act as a spokesperson. Find the probability that the spokesperson will be either male or above 35 years.
Solution:
Let A be the event that the selected person is male and B be the event that the selected person is above 35 years when a person is selected at random from the group of 5 persons to act as a spokesperson and S be the sample space.
∴ n(S) = ⁵C₁ = 5
n(A) = ³C₁ = 3
P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{3}{5}\)
n(B) = ²C₁ = 2
P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{2}{5}\)
n(A ∩ B) = ¹C₁ = 1
P(A ∩ B) = \(\frac{n(A \cap B)}{n(S)}\) = \(\frac{1}{5}\)
By Addition theorem on probability
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\)
= \(\frac{4}{5}\)

Question 13.
Out of 100 students, two sections of 40 are 60 and formed. If you and your friend are among the 100 students, find the probability that
(i) You both enter the same section.
(ii) You both enter the different sections.
Solution:
Let S be the sample space
n(S) = no.of ways of doing 100 students into 2 sections of 40 and 60 = \(\frac{100 !}{40 ! 60 !}\)
(i) You both enter the same section:
First, you and your friend both enter the section the remaining 98 students and I can be divided into two sections (First section 38 and second section 60) is \(\frac{98 !}{38 ! 60 !}\)
You and your friend both enter section II and the remaining 98 students can be divided into two sections (First section 40 and second section 58) is \(\frac{98 !}{40 ! 58 !}\)

= \(\frac{26}{165}+\frac{59}{165}\)
= \(\frac{17}{33}\)
(ii) You both enter the different sections:
The probability that the both in different sections = 1 – P(E)
= 1 – \(\frac{17}{33}\)
= \(\frac{16}{33}\)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Probability Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Probability Solutions Exercise 9(a)

I. In the experiment of throwing a die, consider the following events:

Question 1.
A = {1, 3, 5}, B = {2, 4, 6}, C = {1, 2, 3} Are these events equally likely?
Solution:
Since events A, B, C has an equal chance to occur, hence they are equally likely events.

Question 2.
In the experiment of throwing a die, consider the following events:
A = {1, 3, 5}, B = {2, 4}, C = {6}
Are these events mutually exclusive?
Solution:
Since the happening of one of the given events A, B, C prevents the happening of the other two, hence the given events are mutually exclusive.
Otherwise A ∩ B = φ, B ∩ C = φ, C ∩ A = φ
Hence they are mutually exclusive events.

Question 3.
In the experiment of throwing a die, consider the events.
A = (2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Are these events exhaustive?
Solution:
A = {2, 4, 6}, B = {3, 6}, C = {1, 5, 6}
Let S be the sample space for the random experiment of throwing a die
Then S = {1, 2, 3, 4, 5, 6}
∵ A ⊂ S, B ⊂ S and C ⊂ S, and A ∪ B ∪ C = S
Hence events A, B, C are exhaustive events.

II.

Question 1.
Give two examples of mutually exclusive and exhaustive events.
Solution:
Examples of mutually exclusive events:
(i) The events {1, 2}, {3, 5} are disjoint in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) When two dice are thrown, the probability of getting the sums of 10 or 11.
Examples of exhaustive events:
(i) The events {1, 2, 3, 5}, (2, 4, 6} are exhaustive in the sample space S = {1, 2, 3, 4, 5, 6}
(ii) The events {HH, HT}, {TH, TT} are exhaustive in the sample space S = {HH, HT, TH, TT} [∵ tossing two coins]

Question 2.
Give examples of two events that are neither mutually exclusive nor exhaustive.
Solution:
(i) Let A be the event of getting an even prime number when tossing a die and let B be the event of getting even number.
∴ A, B are neither mutually exclusive nor exhaustive.
(ii) Let A be the event of getting one head tossing two coins.
Let B be the event of getting atleast one head tossing two coins.
∴ A, B are neither mutually exclusive nor exhaustive.

Question 3.
Give two examples of events that are neither equally likely nor exhaustive.
Solution:
(i) Two coins are tossed
Let A be the event of getting an one tail and
Let B be the event of getting atleast one tail.
∴ A, B are neither equally likely nor exhaustive.
(ii) When a die is tossed
Let A be the event of getting an odd prime number and
Let B be the event of getting odd number.
∴ B are are neither equally likely nor exhaustive.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Measures of Dispersion Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Measures of Dispersion Solutions Exercise 8(a)

I.

Question 1.
Find the mean deviation about the mean for the following data.
(i) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
The arithmetic mean of the given data
\(\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
= \(\frac{500}{10}\)
= 50
The absolute values of the deviations are
\(\left|x_i-\bar{x}\right|\), |50 – 38|, |50 – 70|, |50 – 48|, |50 – 40|, |50 – 42|, |50 – 55|, |50 – 63|, |50 – 46|, |50 – 54|, |50 – 44|
= 12, +20, 2, 10, 8, +5, +13, 4, +4, 6
∴ The mean deviation from the mean

(ii) 3, 6, 10, 4, 9, 10
Solution:
The arithmetic mean of the given data
\(\bar{x}=\frac{3+6+10+4+9+10}{6}\)
= \(\frac{42}{6}\)
= 7
The absolute values of the deviations are
\(\left|x_i-\bar{x}\right|\), |3 – 7|, |16 – 7|, |10 – 7|, |4 – 7|, |9 – 7|, |10 – 7|
= 4, 1, 3, 3, 2, 3
∴ The mean deviation from the mean

Question 2.
Find the mean deviation about the median for the following data.
(i) 13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
Given ungrouped data are
13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17
Expressing the data in the ascending order of magnitude, we have
10, 11, 11, 12, 13, 13, 16, 16, 17, 17, 18
∴ Median = 13 = b(say)
The absolute values are
|13 – 10|, |13 – 11|, |13 – 11|, |13 – 12|, |13 – 13|, |13 – 13|, |13 – 16|, |13 – 16|, |13 – 17|, |13 – 17|, |13 – 18|
= 3, 2, 2, 1, 0, 0, 3, 3, 4, 4, 5
∴ Mean deviation from the median

(ii) 4, 6, 9, 3, 10, 13, 2
Solution:
Given ungrouped data are 4, 6, 9, 3, 10, 13, 2
Expressing the data in the ascending order of magnitude, we have
2, 3, 4, 6, 9, 10, 13
∴ Median = 6 = b (say)
The absolute values are
|6 – 2|, |6 – 3|, |6 – 4|, |6 – 6|, |6 – 9|, |6 – 10|, |6 – 13|
= 4, 3, 2, 0, 3, 4, 7
∴ Mean deviation from the median = \(\frac{\sum_{i=1}^7\left|x_i-b\right|}{7}\)
= \(\frac{4+3+2+0+3+4+7}{7}\)
= 3.285

Question 3.
Find the mean deviation about the mean for the following distribution.
(i)

Solution:

(ii)

Solution:

Question 4.
Find the mean deviation about the median for the following frequency distribution.

Solution:

\(\frac{N}{2}\) = 13
∴ Median = 7
Hence mean deviation from the median = \(\frac{1}{N} \sum f_i \mid x_i-\text { Median } \mid\)
= \(\frac{1}{26}\) (84)
= 3.23

II.

Question 1.
(i) Find the mean deviation about the median for the following continuous distribution.

Solution:
Construct the table

\(\frac{N}{2}\) = 25
Observation lies in the class-interval 20-30.
This is the median class.

(ii)

Solution:
Construct the table
\(\frac{N}{2}=\frac{100}{2}\) = 50
Observation lies in the interval 40-50
This is the median class
∴ Median = \(L+\left\{\frac{\frac{N}{2}-\text { P.C.f }}{f}\right\} \times i\)
= 40 + \(\left\{\frac{50-32}{28}\right\} \times 10\)
= 40 + 6.43
= 46.43

∴ Median deviation about median = \(\frac{1}{N} \sum_{i=1}^8 f_i \mid x_i-\text { median } \mid\)
= \(\frac{1}{100}\) (1428.6)
= 14.286

Question 2.
Find the mean deviation about the mean for the following continuous distribution.

Solution:
Taking the assumed mean a = 130 and h = 10 construct the table

\(\bar{x}=a+\left(\frac{\sum f_i d_i}{N}\right) h\)
= 130 + \(\left(\frac{-47}{100}\right) 10\)
= 130 – 4.7
= 125.3
Mean deviation about the mean = \(\frac{1}{N} \sum_{i=1}^6 f_i\left|x_i-\bar{x}\right|\)
= \(\frac{1}{100}\) (1428.8)
= 14.288

Question 3.
Find the variance for the discrete data given below.
(i) 6, 7, 10, 12, 13, 4, 8, 12
Solution:
Given data are 6, 7, 10, 12, 13, 4, 8, 12
Mean \(\bar{x}=\frac{6+7+10+12+13+4+8+12}{8}\)
= \(\frac{72}{8}\)
= 9
Construct the table

Variance σ² = \(\frac{1}{n} \Sigma\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{8}\) (74)
= 9.25

(ii) 350, 361, 370, 373, 376, 379, 385, 387, 394, 395
Solution:
Given data are 350, 361, 370, 373, 376, 379, 385, 387, 394, 395
Mean \(\bar{x}=\frac{350+361+370+373+376+379+385+387+394+395}{10}\) = 377
Construct the table

Variance = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)
= \(\frac{1}{10}\) (1832)
= 183.2

Question 4.
Find the variance and standard deviation of the following frequency distribution.

Solution:

III.

Question 1.
Find the mean and variance using the step deviation method, of the following tabular data, giving the age distribution of 542 members.

Solution:
If we take assumed mean A = 55 then y_i = \(\frac{x_i-55}{10}\)
Here h = 10
Construct the table

Question 2.
The coefficient of variation of the two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively. Find their arithmetic means.
Solution:
Given coefficient of variance C.V = 60
Standard deviation s = 21
We know C.V = \(\frac{\sigma}{x} \times 100\)
⇒ \(\bar{x}=\frac{\sigma}{C . V} \times 100\)
= \(\frac{21}{60}\) × 100
= 35
Arithmetic mean = 35
Given co-efficient of variance C.V = 70
Standard deviation σ = 16
We know C.V = \(\frac{\sigma}{x} \times 100\)
⇒ \(\bar{x}=\frac{\sigma}{C . V} \times 100\)
= \(\frac{16}{70}\) × 100
= 22.85

Question 3.
From the prices of Shares X and Y given below, for 10 days of trading, find out which share is more stable.

Solution:
For share x
\(\bar{x}=\frac{35+54+52+53+56+58+52+50+51+49}{10}\)
= \(\frac{510}{10}\)
= 51
For share y
\(\bar{y}=\frac{108+107+105+105+106+107+104+103+104+101}{10}\)
= \(\frac{1050}{10}\)
= 105

C.V of share x = \(\frac{\sigma_{\mathrm{x}}}{\mathrm{x}} \times 100\)
= \(\frac{5.92}{51}\) × 100
= 11.61
C.V of share y = \(\frac{\sigma_{\mathrm{y}}}{\mathrm{y}} \times 100\)
= \(\frac{2}{105}\) × 100
= 1.91
since \(\bar{x}<\bar{y}\) and C.V of share x > C.V of shares y
∴ Y share is more stable.

Question 4.
The mean of 5 observations is 4.4. Their variance is 8.24. If three of the observations are 1, 2, and 6. Find the other two observations.
Solution:
Let x and y be the required two observations.
Given, that the mean of 5 observations = 4.4
⇒ \(\frac{1+2+6+x+y}{5}\) = 4.4
⇒ 9 + x + y = 22
⇒ x + y = 13 ……..(1)
Also thin variance = 8.24
⇒ \(\frac{1}{5}\) [(1 – 4.4)² + (2 – 4.4)² + (6 – 4.4)² + (x – 4.4)² + (y – 4.4)²] = 8.24
⇒ (-3.4)² + (-2.4)² + (1.6)² + x² – 8.8x + 19.36 + y² – 8.8y + 19.36 = 41.2
⇒ x² + y² – (8.8) (x + y) = 41.2 – 11.56 – 5.76 – 2.56 – 19.36 – 19.36
⇒ x² + y² – (8.8) (13) = -17.4
⇒ x² + y² = -17.4 + 114.4
⇒ x² + y² = 97
We know (x + y)² = x² + y² + 2xy
⇒ 169 = 97 + 2xy
⇒ 2xy = 72
⇒ xy = 36
(x – y)² = (x + y)² – 4xy
⇒ (x – y)² = 169 – 144 = 25
⇒ x – y = 5 ……..(2)
(1) + (2) ⇒ 2X = 18
⇒ X = 9
(1) – (2) ⇒ 2Y = 8
⇒ Y = 4
∴ The other two observations are 4, 9.

Question 5.
The arithmetic mean and standard deviation of a set of 9 items are 43 and 5 respectively. If an item of value 63 is added to that set, find the new mean and standard deviation of 10 item set given.
Solution:
Given the arithmetic mean of 9 items is 43
∴ \(\bar{x}=\frac{1}{9} \sum_{i=1}^9 x_i=43\)
⇒ Σx_i = 387
If an item of value 63 is added to that set then
Σy_i = Σx_i + 63
= 387 + 63
= 450
∴ \(\bar{y}=\frac{1}{10} \sum y_i\)
= \(\frac{1}{10}\) (450)
= 45
∴ New mean = 45
Given the standard deviation of 9 items is 5
⇒ σ = 5
⇒ σ² = 25
⇒ \(\frac{1}{9} \sum_{i=1}^9\left(x_i-\bar{x}\right)^2\)
⇒ \(\sum_{i=1}^9\left(x_i-43\right)^2\) = 225 ……(1)
New variance σ² = \(\frac{1}{10} \sum_{i=1}^{10}\left(y_i-45\right)^2\)
= \(\frac{1}{10}\)[(x₁ – 45)² + (x₂ – 45)² + …….. + (x₉ – 45)² + (63 – 45)²]
Since first 9 items are same i.e, x_i = y_i, i = 1, 2, 3, …….., 9
= \(\frac{1}{10}\) [(x_i – 43 – 2)² + (x₂ – 43 – 2)² + …… + (x₉ – 43 – 2)² + 324]
= \(\frac{1}{10}\) [(x_i – 43)² + 4 – 2 . 2 . (x₁ – 43) + (x₂ – 43)² + 4 – 2 . 2 . (x₂ – 43) + …. (x₉ – 43)² + 4 – 2 . 2 . (x₉ – 43) + 324]
= \(\frac{1}{10}\) [(xi – 43)² + (x₂ – 43)² + ……. + (x₉ – 43)² + 36 – 4{x₁ – 43 + x₂ – 43 + …….. + x₉ – 43} + 324]
= \(\frac{1}{10}\) [225 + 36 – 4(x₁ + x₂ + …. + x₉ – 387) + 324]
= \(\frac{1}{10}\) [225 + 36 – 4(387 – 387) + 324]
= \(\frac{1}{10}\) [585]
= 58.5
New standard deviation σ = √58.5 = 7.65

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(d)

Question 1.
Find the coefficient of x³ in the power series expansion of \(\frac{5 x+6}{(x+2)(1-x)}\) specifying the region in which the expansion is valid.
Solution:

Question 2.
Find is the coefficient of x⁴ in the power series expansion of \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}\) specifying the interval in which the expansion is valid.
Solution:
Let \(\frac{3 x^2+2 x}{\left(x^2+2\right)(x-3)}=\frac{A}{x-3}+\frac{B x+C}{x^2+2}\)
Multiplying with (x² + 2) (x – 3)
3x² + 2x = A(x² + 2) + (Bx + C) (x – 3)
x = 3
⇒ 27 + 6 = A(9 + 2)
⇒ 33 = 11A
⇒ A = 3
Equating the coefficients of x²
3 = A + B
⇒ B = 3 – A = 3 – 3 = 0
Equating the constants,
2A – 3C = 0
⇒ 3C = 2A = 6
⇒ C = 2

Question 3.
Find the coefficient of xⁿ in the power series expansion of \(\frac{x-4}{x^2-5 x+6}\) specifying the region in which the expansion is valid.
Solution:
Let \(\frac{x-4}{x^2-5 x+6}=\frac{A}{x-2}+\frac{B}{x-3}\)
Multiplying with (x – 2) (x – 3)
x – 4 = A(x – 3) + B(x – 2)
x = 2
⇒ -2 = A(2 – 3) = -A
⇒ A = 2
x = 3
⇒ -1 = B(3 – 2) = B
⇒ B = -1

Question 4.
Find the coefficient of xⁿ in the power series expansion of \(\frac{3 x}{(x-1)(x-2)^2}\)
Solution:

∴ 3x = A(x – 2)² + B(x – 1) (x – 2) + C(x – 1) ……..(1)
putting x = 1,
3 = A(1 – 2)²
⇒ A = 3
putting x = 2,
6 = C(2 – 1)
⇒ C = 6
Now equating the co-efficient of x² terms in (1)
0 = A + B
⇒ B = -A
⇒ B = -3

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(c)

Resolve the following into partial fractions.

Question 1.
\(\frac{x^2}{(x-1)(x-2)}\)
Solution:
Let \(\frac{x^2}{(x-1)(x-2)}=1+\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with (x – 1) (x – 2)
x² = (x – 1) (x – 2) + A(x – 2) + B(x – 1)
Put x = 1, 1 = A(-1) ⇒ A = -1
Put x = 2, 4 = B(1) ⇒ B = 4
∴ \(\frac{x^2}{(x-1)(x-2)}=1-\frac{1}{x-1}+\frac{4}{x-2}\)

Question 2.
\(\frac{x^3}{(x-1)(x+2)}\)
Solution:

Question 3.
\(\frac{x^3}{(2 x-1)(x-1)^2}\)
Solution:
Let \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{A}{2 x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with 2(2x – 1) (x – 1)²
2x³ = (2x – 1) (x – 1)² + 2A(x – 1)² + 2B(2x – 1) (x – 1) + 2C(2x – 1)
Put x = \(\frac{1}{2}\),
⇒ 2(\(\frac{1}{8}\)) = 2A(\(\frac{1}{4}\))
⇒ A = \(\frac{1}{2}\)
Put x = 1,
⇒ 2(1) = 2C(1)
⇒ C = 1
Put x = 0,
0 = (-1) (1) + 2A(1) + 2B(-1) (-1) + 2C(-1)
⇒ 2A + 2B – 2C = 1
⇒ 2B = 1 + 2C – 2A
⇒ 2B = 1 + 2 – 1 = 2
⇒ B = 1
∴ \(\frac{x^3}{(2 x-1)(x-1)^2}\) = \(\frac{1}{2}+\frac{1}{2(2 x-1)}+\frac{1}{(x-1)}+\frac{1}{(x-1)^2}\)

Question 4.
\(\frac{x^3}{(x-a)(x-b)(x-c)}\)
Solution:
Let \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\)
Multiplying with (x – a)(x – b) (x – c),
x³ = (x – a)(x – b) (x – c) + A(x – b) (x – c) + B(x – a) (x – c) + C(x – a) (x – b)
Put x = a,
a³ = A(a – b) (a – c)
⇒ A = \(\frac{a^3}{(a-b)(a-c)}\)
Put x = b,
b³ = B(b – a) (b – c)
⇒ B = \(\frac{b^3}{(b-a)(b-c)}\)
Put x = c, c³ = C(c – a) (c – b)
⇒ C = \(\frac{c^3}{(c-a)(c-b)}\)
∴ \(\frac{x^3}{(x-a)(x-b)(x-c)}\) = \(1+\frac{a^3}{(a-b)(a-c)(x-a)}+\frac{b^3}{(b-a)(b-c)(x-b)}\) + \(\frac{c^3}{(c-a)(c-b)(x-c)}\)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(b)

Resolve the following into partial fractions.

Question 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{A}{x-1}+\frac{B x+C}{x^2+2}\)
Multiplying with (x – 1) (x² + 2)
2x² + 3x + 4 = A(x² + 2) + (Bx + C) (x – 1)
x = 1
⇒ 2 + 3 + 4 = A(1 + 2)
⇒ 9 = 3A
⇒ A = 3
Equating the coefficients of x²
2 = A + B
⇒ B = 2 – A = 2 – 3 = -1
Equating constants
4 = 2A – C
⇒ C = 2A – 4 = 6 – 4 = 2
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\)

Question 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}=\frac{A}{2+x}+\frac{B x+C}{1-x+x^2}\)
Multiplying with (2 + x) (1 – x + x²)
3x – 1 = A(1 – x + x²) (Bx + C) (2 + x)
x = -2
⇒ -7 = A(1 + 2 + 4) = 7A
⇒ A = -1
Equating the coefficients of x²
0 = A + B ⇒ B = -A = 1
Equating the constants
-1 = A + 2C
⇒ 2C = -1 – A = -1 + 1 = 0
⇒ C = 0
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(2+x)}=-\frac{1}{2+x}+\frac{x}{1-x+x^2}\)

Question 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
Solution:
Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
Multiplying with (x + 2) (x² + 1)
x² – 3 = A(x² + 1) + (Bx + C) (x + 2)
x = -2
⇒ 4 – 3 = A(4 + 1)
⇒ 1 = 5A
⇒ A = \(\frac{1}{5}\)
Equating the coefficients of x²
1 = A + B
⇒ B = 1 – A = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)
Equating the constants
-3 = A + 2C
⇒ 2C = -3 – A
⇒ 2C = -3 – \(\frac{1}{5}\)
⇒ 2C = \(-\frac{16}{5}\)
⇒ C = \(-\frac{8}{5}\)
∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}=\frac{1}{5(x+2)}+\frac{4 x-8}{5\left(x^2+1\right)}\)

Question 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
Solution:
Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{A x+B}{x^2+x+1}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)
Multiplying with (x² + x + 1)²
x² + 1 = (Ax + B) (x² + x + 1) + (Cx + D)
Equating the coefficients of x³,
A = 0
Equating the coefficients of x²,
A + B = 1 ⇒ B = 1
Equating the coefficients of x,
A + B + C = 0
⇒ 1 + C = 0
⇒ C = -1
Equating the constants,
B + D = 1
⇒ D = 1 – B = 1 – 1 = 0
∴ Ax + B = 1, Cx + D = -x
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}=\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\)

Question 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
Solution:

∴ x³ + x² + 1 = A(x – 1) (x² + x + 1) + B(x² + x + 1) + (Cx + D) (x – 1)² …….(2)
Put x = 1 in (2)
1 + 1 + 1 = A(0) + B(1 + 1 + 1) + (C(1) + D) (0)
⇒ 3B = 3
⇒ B = 1
Equating the coefficients of x³ in (2)
1 = A + C ………(3)
Equating the coefficients of x² in (2)
1 = A(1 – 1) + B(1) + C(-2) + D(1)
⇒ 1 = B – 2C + D
∵ B = 1,
⇒ 1 = 1 – 2C + D
⇒ 2C = D ………(4)
Put x = 0 in (2)
1 = A(-1)(1) + B(1) + D(-1)²
⇒ -A + B + D = 1
⇒ -A + 1 + D = 1
⇒ A = D ………(5)
From (3), (4) and (5)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Partial Fractions Solutions Exercise 7(a)

Resolve the following into partial fractions.

I.

Question 1.
\(\frac{2 x+3}{(x+1)(x-3)}\)
Solution:
Let \(\frac{2 x+3}{(x+1)(x-3)}=\frac{A}{x+1}+\frac{B}{x-3}\)
Multiplying with (x + 1) (x – 3)
2x + 3 = A(x – 3) + B(x + 1)
x = -1 ⇒ 1 = A(-4) ⇒ A = \(-\frac{1}{4}\)
x = 3 ⇒ 9 = B(4) ⇒ B = \(\frac{9}{4}\)
\(\frac{2 x+3}{(x+1)(x-3)}=\frac{-1}{4(x+1)}+\frac{9}{4(x-3)}\)

Question 2.
\(\frac{5 x+6}{(2+x)(1-x)}\)
Solution:
Let \(\frac{5 x+6}{(2+x)(1-x)}=\frac{A}{2+x}+\frac{B}{1-x}\)
Multiplying with (2 + x) (1 – x)
5x + 6 = A(1 – x) + B(2 + x)
Put x = -2,
-10 + 6 = A(1 + 2)
⇒ A = \(-\frac{4}{3}\)
Put x = 1,
5 + 6 = B(2 + 1)
⇒ B = \(\frac{11}{3}\)
∴ \(\frac{5 x+6}{(2+x)(1-x)}=-\frac{4}{3(2+x)}+\frac{11}{3(1-x)}\)

II.

Question 1.
\(\frac{3 x+7}{x^2-3 x+2}\)
Solution:
\(\frac{3 x+7}{x^2-3 x+2}=\frac{3 x+7}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}\)
Multiplying with x² – 3x + 2
3x + 7 = A(x – 2) + B(x – 1)
x = 1 ⇒ 10 = -A ⇒ A = -10
x = 2 ⇒ 13 = B ⇒ B = 13
∴ \(\frac{3 x+7}{x^2-3 x+2}=\frac{-10}{x-1}+\frac{13}{x-2}\)

Question 2.
\(\frac{x+4}{\left(x^2-4\right)(x+1)}\)
Solution:
\(\frac{x+4}{\left(x^2-4\right)(x+1)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x-2}\)
Multiplying with (x² – 4) (x + 1)
x + 4 = A(x² – 4) + B(x + 1) (x – 2) + C(x + 1) (x + 2)
x = -1
⇒ 3 = A(1 – 4)
⇒ 3 = -3A
⇒ A = -1
x = -2
⇒ 2 = B(-2 + 1) (-2 – 2)
⇒ 2 = 4B
⇒ B = \(\frac{1}{2}\)
x = 2
⇒ 6 = C(2 + 1)(2 + 2)
⇒ 6 = 12C
⇒ C = \(\frac{1}{2}\)
∴ \(\frac{x+4}{\left(x^2-4\right)(x+1)}=-\frac{1}{x+1}+\frac{1}{2(x+2)}\) + \(\frac{1}{2(x-2)}\)

Question 3.
\(\frac{2 x^2+2 x+1}{x^3+x^2}\)
Solution:
Let \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{2 x^2+2 x+1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\)
Multiplying with x²(x + 1)
2x² + 2x + 1 = Ax(x + 1) + B(x + 1) + Cx²
Put x = 0, 1 = B
Put x = -1, 2 – 2 + 1 = C(1) ⇒ C = 1
Equating the coefficients of x²,
2 = A + C
⇒ A = 2 – C = 2 – 1 = 1
∴ \(\frac{2 x^2+2 x+1}{x^3+x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}\)

Question 4.
\(\frac{2 x+3}{(x-1)^3}\)
Solution:
\(\frac{2 x+3}{(x-1)^3}\)
Put x – 1 = y ⇒ x = y + 1
⇒ \(\frac{2 x+3}{(x-1)^3}=\frac{2(y+1)+3}{y^3}=\frac{2 y+5}{y^3}\)
⇒ \(\frac{2 x+3}{(x-1)^3}\) = \(\frac{2}{y^2}+\frac{5}{y^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)
∴ \(\frac{2 x+3}{(x-1)^3}=\frac{2}{(x-1)^2}+\frac{5}{(x-1)^3}\)

Question 5.
\(\frac{x^2-2 x+6}{(x-2)^3}\)
Solution:
Let x – 2 = y then x = y + 2

III.

Question 1.
\(\frac{x^2-x+1}{(x+1)(x-1)^2}\)
Solution:
Let \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\)
Multiplying with (x + 1) (x – 1)²
x² – x + 1 = A(x – 1)² + B(x + 1) (x – 1) + C(x + 1)
Put x = -1,
1 + 1 + 1 = A(4)
⇒ A = \(\frac{3}{4}\)
Put x = 1,
1 – 1 + 1 = C(2)
⇒ C = +\(\frac{1}{2}\)
Equating the coefficients of x²,
A + B = 1
⇒ B = 1 – A
⇒ B = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
∴ \(\frac{x^2-x+1}{(x+1)(x-1)^2}=\frac{3}{4(x+1)}+\frac{1}{4(x-1)}\) + \(\frac{1}{2(x-1)^2}\)

Question 2.
\(\frac{9}{(x-1)(x+2)^2}\)
Solution:
Let \(\frac{9}{(x-1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
Multiplying with (x – 1) (x + 2)²
9 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1)
x = 1
⇒ 9 = 9A
⇒ A = 1
x = -2
⇒ 9 = -3C
⇒ C = -3
Equating the coefficients of x²
A + B = 0 ⇒ B = -A = -1
∴ \(\frac{9}{(x-1)(x+2)^2}=\frac{1}{x-1}-\frac{1}{x+2}-\frac{3}{(x+2)^2}\)

Question 3.
\(\frac{1}{(1-2 x)^2(1-3 x)}\)
Solution:
Let \(\frac{1}{(1-2 x)^2(1-3 x)}=\frac{A}{1-3 x}+\frac{B}{1-2 x}+\frac{C}{(1-2 x)^2}\)
Multiplying with (1 – 2x)² (1 – 3x)
1 = A(1 – 2x)² + B(1 – 3x) (1 – 2x) + C(1 – 3x)

Question 4.
\(\frac{1}{x^3(x+a)}\)
Solution:
Let \(\frac{1}{x^3(x+a)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+a}\) = \(\frac{A \cdot x^2(x+a)+B(x)(x+a)+C(x+a)+D x^3}{x^3(x+a)}\)
∴ 1 = A (x²) (x + a) + Bx (x + a) + C(x + a) + Dx³ ……..(1)
Put x = 0 in (1)
1 = A(0) + B(0) + C(0 + a) + D(0)
⇒ 1 = C(a)
⇒ C = \(\frac{1}{a}\)

Question 5.
\(\frac{x^2+5 x+7}{(x-3)^3}\)
Solution:
Let x – 3 = y ⇒ x = y + 3
\(\frac{x^2+5 x+7}{(x-3)^3}=\frac{(y+3)^2+5(y+3)+7}{y^3}\)

Question 6.
\(\frac{3 x^3-8 x^2+10}{(x-1)^4}\)
Solution:
Put x – 1 = y ⇒ x = y + 1