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Practice AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Complex Numbers and Quadratic Equations MCQ

Question 1.
The value of i^-999 is
(1) 1
(2) -1
(3) i
(4) -i
Answer:
(3) i

Explanation:
i^-999 = \(\frac{1}{i^{999}}=\frac{1}{i^3 i^{996}}=\frac{1}{-i \cdot\left(i^4\right)^{249}}=\frac{-1}{i}\)
= i

Question 2.
The multiplicative inverse of 1 + i is
(1) \(\frac{1}{2}\)(1 – i)
(2) \(\frac{1}{2}\)(1 + i)
(3) 1 – i
(4) i
Answer:
(1) \(\frac{1}{2}\)(1 – i)

Explanation:
Multiplicative inverse of 1 + i = \(\frac{1}{1+i}=\frac{1}{1+i} \times \frac{1-i}{1-i}=\frac{1-i}{1+1}=\frac{1}{2}\)
= \(\frac{1}{2}\)(1 – i)

Question 3.
The modulus of 5 + 4i is
(1) 41
(2) -41
(3) \(\sqrt{41}\)
(4) \(\sqrt{-41}\)
Answer:
(3) \(\sqrt{41}\)

Explanation:
|5 + 4i| = \(\sqrt{5^2+4^2}=\sqrt{25+16}=\sqrt{41}\)

Question 4.
The value of \(\sqrt{-25}\) \(\sqrt{-9}\) is
(1) 15
(2) -15
(3) 15i
(4) None of these
Answer:
(2) -15

Explanation:
\(\frac{1}{2}\)
= 5i × 3i
= 15i²
= – 15

Question 5.
The Conjugate of \(\frac{2-i}{1-2 i}\) is
(1) \(\frac{4+3 i}{5}\)
(2) 4 – 3i
(3) \(\frac{4-3 i}{5}\)
(4) 1
Answer:
(3) \(\frac{4-3 i}{5}\)

Explanation:
Conjugate of z = z̄; z = \(\frac{2-i}{1-2 i}\)
z̄ = \(\frac{2-i}{1-2 i}=\frac{2+i}{1+2 i}=\frac{2+i}{1+2 i} \times \frac{1-2 i}{1-2 i}\)
= \(\frac{2-4 i+i-2 i^2}{1+4}=\frac{4-3 i}{5}\)

Question 6.
The value of (z + 3)(z̄ + 3) is equivalent to
(1) |z + 3|²
(2) |z – 3|
(3) z² + 3
(4) None of these
Answer:
(1) |z + 3|²

Explanation:
(z + 3)\((\overline{z+3})\) = |z + 3|² [∵ z.z̄ = |z|²]

Question 7.
Let x, y ∈ R, then x + iy is a non-real complex number if
(1) x = 0
(2) y = 0
(3) x ≠ 0
(4) None of these
Answer:
(4) None of these

Explanation:
y ≠ 0

Question 8.
If a + ib = c + id then
(1) a² + c² = 0
(2) b² + c² = 0
(3) b² + d² = 0
(4) a² + b² = c² + d²
Answer:
(4) a² + b² = c² + d²

Explanation:
a + ib = c + id
⇒ a = c, b = d
⇒ a² = c², b² = d²
⇒ a² + b² = c² + d²

Question 9.
The sum of the series i + i² + i³ + ……………. upto 1000 terms is
(1) 1
(2) 0
(3) -1
(4) -2
Answer:
(2) 0

Explanation:
i + i² + i³ + ……………. upto 1000 terms
= \(\frac{i\left(i^{1000}-1\right)}{i-1}=\frac{i\left(\left(i^4\right)^{250}-1\right)}{i-1}\)
= \(\frac{i(1-1)}{i-1}\) = 0

Question 10.
The complex number z which satisfies the condition \(\left|\frac{\mathrm{i}+\mathrm{z}}{\mathrm{i}-\mathrm{z}}\right|\) = 1 lies on
1) Circle x² + y² = 1
2) The X-axis
3) The Y-axis
4) The line x + y = 1
Answer:
2) The X-axis

Explanation:
Given \(\left|\frac{\mathrm{i}+\mathrm{z}}{\mathrm{i}-\mathrm{z}}\right|\) = 1
⇒ |i + z| = |i – z|
⇒ |x + iy + i| = |i – (x + iy)|
⇒ x² + (1 + y)² = x² + (1 – y)²
⇒ 1 + y² + 2y = 1 + y² – 2y
⇒ 4y = 0
⇒ y = 0
Which is the X-axis.
∴ Locus of z lies on the X-axis

Referring to the AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations Exercise 4c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4c

I. Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

Question 1.
z = – 1 – i√3
Solution:
Given that z = -1 – i√3
Let r cos θ = -1 and r sin θ = -√3
Squaring and adding, we obtain (rcos θ)² + (rsin θ)² = (-1)² + (-√3)²
⇒ r² (cos θ)² + (rsin θ)² = (-1)² + (√3)²
⇒ r²(cos²θ + sin²θ) = 1 + 2
⇒ r² = 4 [∵ cos²θ + sin²θ = 1]
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, Modulus = 2
Hence, 2cos θ = -1 and 2 sin θ = -√3
⇒ cos θ = \(\frac{1}{2}\) and sin θ = –\(\frac{\sqrt{3}}{2}\)
Since both the values of sin 0 and cos 0 are negative in III quadrant.
Argument = -(π – \(\frac{\pi}{3}\)) = \(\frac{-2 \pi}{3}\)
Thus, the modulus and argument of the complex number -1 – i√3 are 2 and \(\frac{-2 \pi}{3}\) respectively.

Question 2.
z = -√3 + i
Solution:
Given that z = -√3 + i
Let r cos θ = -√3 and r sin θ = 1
On squaring and adding, we obtain
r² cos²θ + r² sin²θ = (-√3)² + 1²
⇒ r² = 3 + 1 = 4 [∵ cos²θ + sin²θ = 1]
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, Modulus = 2, Hence, 2 cos θ = -√3 and 2 sin θ = 1
⇒ cos θ = \(-\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
Since, θ lies in the quadrant II, θ = π – \(\frac{\pi}{6}=\frac{5 \pi}{6}\)
Thus, the modulus and argument of the complex number – √3 + i are 2 and \(\frac{5 \pi}{3}\) respectively.

Convert each of the complex numbers given in exercises 3 to 8 in the polar form.

Question 3.
1 – i
Solution:
Given that z = 1 – i.
Let r cos θ = 1 and r sin θ = -1
On squaring and adding, we obtain
r² cos²θ + r² sin²θ = 1² + (-1)²
⇒ r² (∵ cos²θ + sin²θ) = 1 + 1
⇒ r² = 2
⇒ r = √2 [∵ Conventionally, r > 0]
Since, 0 lies in the IV quadrant θ = \(\left(\frac{-\pi}{4}\right)\)
Hence, 1 – i = r cos θ + ir sin θ
= √2cos \(\left(\frac{-\pi}{4}\right)\) + i sin\(\left(\frac{-\pi}{4}\right)\)
= √2[cos\(\left(\frac{-\pi}{4}\right)\) + isin\(\left(\frac{-\pi}{4}\right)\)]
Thus, this is the required polar form.

Question 4.
-1 + i
Solution:
Given that z = -1 + i.
Let r cos θ = -1 and r sin θ = 1
On squaring and adding, we obtain r² cos²θ + r² sin²θ = (- 1)² + 1²
⇒ r²(cos²θ + sin²θ) = 1 + 1
⇒ r² = 2
⇒ r = √2 [∵ Conventionally, r > 0]
Therefore,√2cos θ = -1 and √2 sin θ = 1
⇒ cos θ = –\(\frac{1}{\sqrt{2}}\) and sin θ = \(\frac{1}{\sqrt{2}}\)
Since, θ lies in the II quadrant, θ = π – \(\frac{\pi}{4}=\frac{3 \pi}{4}\)
⇒ r² (cos²θ + sin²θ) = 1 + 1
Hence, -1 – i = r cos θ + i r sin θ = √2 cos \(\frac{-3 \pi}{4}\) + i √2 sin \(\frac{-3 \pi}{4}\)
= √2(cos \(\frac{-3 \pi}{4}\) + i sin \(\frac{-3 \pi}{4}\))
Thus, this is the required polar form.

Question 5.
-1 -i ,
Solution:
Given that z = -1 – i.
Let r cos θ = – 1 and r sin θ = -1
On squaring and adding, we obtain
r² cos² θ + r² sin²θ = (- 1)² + (- 1)²
Therefore, √2 cos θ = -1 and √2 sin θ = -1
⇒ cos θ = \(\\frac{-1}{\sqrt{2}}\) and sin θ = \(\frac{-1}{\sqrt{2}}\)
Hence, -1 – i = r cos θ + i r sin θ
= √2 cos \(\frac{-3 \pi}{4}\) + i 2 sin\(\frac{-3 \pi}{4}\)
= √2(cos \(\frac{-3 \pi}{4}\) + is sin \(\frac{-3 \pi}{4}\))
Thus, this is the required polar form.

Question 6.
-3
Solution:
Given that z = -3.
Let r cos θ = -3 and r sin θ = 0
On squaring and adding, we obtain
r² cos²θ + r² sin²θ = (- 3)² + (0)²
⇒ r² (cos²θ + sin²θ) = 9 + 0
⇒ r² = 9
⇒ r = 3 [∵ Conventionally, r > 0]
Therefore, 3 cos θ = – 3 and 3 sin θ = 0
⇒ cos θ = – 1 and sin θ = 0.
Since, θ lies in the II quadrant, θ = n
Hence, – 3 = r cos θ + ir sin θ = 3 cos π + i3 sin π = 3(cos π + i sin π)
Thus, this is the required polar form.

Question 7.
√3 + i
Solution:
Given that z = √3 + i
Let r cos θ = √3 and r sin θ = 1
On squaring and adding, we obtain
r² cos²θ + r² sin²θ = (^3 )² + 1²
⇒ r² (cos²0 + sin²0) = 3+l
⇒ r² = 4
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, 2 cos θ = √3 and 2 sin θ = 1
⇒ cos θ = \(\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
Since, θ lies in I quadrant, θ = \(\frac{\pi}{6}\)
Hence, √3+ 1 = r cos θ + ir sin θ = 2cos \(\frac{\pi}{6}\) + i² sin \(\frac{\pi}{6}\) = 2(cos \(\frac{\pi}{6}\) + isin \(\frac{\pi}{6}\))
Thus, this is the required polar form.

Question 8.
i
Solution:
Given that z = i
Let r cos θ = 0 and r sin θ = 1
On squaring and adding, we obtain
r² cos²θ + r² sin²θ = (√3 )² + 1²
⇒ r² (cos²θ + sin²θ) = 1
⇒ r² = 1
⇒ r = √1 =1 [∵ Conventionally, r > 0]
Therefore, cos θ = 0 and sin θ = 1
Since, θ lies in the I quadrant, θ = \(\frac{\pi}{2}\)
Hence, i = r cos θ + ir sin θ = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
Thus, this is the required polar form.

Referring to the AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations Exercise 4d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4d

I. Solve each of the following equations.

Question 1.
x² + 3 = 0
Solution:
The given quadratic equation is x² + 3 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 1, b = 0 and c = 3
Therefore, the discriminant of the given equation is
D = b² – 4ac = 0² – 4 × 1 × 3 = – 12
Therefore, the required solutions are

Question 2.
2x² + x + 1 = 0
Solution:
The given quadratic equation is 2x² + x + 1 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 2, b = 1 and c = 1
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 × 2 × 1 × = -7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times 2}=\frac{-1 \pm \sqrt{7} i}{4}\) [∵ √-1 = i]

Question 3.
x² + 3x + 9 = 0
Solution:
The given quadratic equation is x² + 3x + 9 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 1, b = 3 and c = 9
Therefore, the discriminant of the given equation is
D = b² – 4ac = 3² – 4 × 1 × 9 = -27
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{-27}}{2 \times 1}=\frac{-3 \pm 3 \sqrt{-3}}{2}=\frac{-3 \pm 3 \sqrt{3} i}{2}\) [∵ √-1 = i]

Question 4.
-x² + x – 2 = 0
Solution:
The given quadratic equation is -x² + x – 2 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = -1, b = 1 and c = -2
Therefore, the discriminant of the given equation is
D = b² – 4ac = 1² – 4 × (-1) × (-2) = -7
Hence, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times(-1)}=\frac{-1 \pm \sqrt{7} i}{-2}\) [∵ √-1 = i]

Question 5.
x² + 3x + 5 = 0
Solution:
The given quadratic equation is x² + 3x + 5 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 1, b = 3 and c = 5
Therefore, the discriminant of the given equation is D = b² – 4ac = 3² – 4 × 1 × 5 = -11
Hence, the required solutions are
\(\frac{-\mathrm{b} \pm \sqrt{\mathrm{D}}}{2 \mathrm{a}}=\frac{-3 \pm \sqrt{-11}}{2 \times 1}=\frac{-3 \pm \sqrt{11} \mathrm{i}}{2}\) [∵ √-1 = i]

Question 6.
x² – x + 2 = 0
Solution:
The given quadratic equation is x² – x + 2 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 1, b = -1 and c = 2
Therefore, the discriminant of the given equation is
D = b² – 4ac = (-1)² – 4 × 1 × 2 = -7
Hence, the requird solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1(-1) \pm \sqrt{-7}}{2 \times 1}=\frac{1 \pm \sqrt{7} i}{2}\) [∵ √-1 = i]

Question 7.
√2x² + x + √2 = 0
Solution:
The given quadratic equation is √2x² + x + √2 = 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = 72 , b = 1 and c = 72
Therefore, the discriminant of the given equation is
D = b² – 4ac = (-1)² – 4 × √2 × √2 = -7
Hence, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=-\frac{1 \pm \sqrt{7} i}{2 \sqrt{2}}\) [∵ √-1 = i]

Question 8.
√3x² – √2x + 3√3 = 0
Solution:
The given quadratic equation is √3x² – √2x + 3√3 = 0
comparing the given equation with ax² + bx + c = 0,
We obtain a = √3 , b = -√2 and c = 375
Therefore, the discriminant of the given equation is
D = b² – 4ac = (-√2)² – 4 × (√3) × (3√3) = -34
Hence, the required solutions are
\(\frac{-\mathrm{b} \pm \sqrt{\mathrm{D}}}{2 \mathrm{a}}=\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times \sqrt{3}}=\frac{\sqrt{2} \pm \sqrt{34} \mathrm{i}}{2 \sqrt{3}}\) [∵ √-1 = i]

Question 9.
x² + x + \(\frac{1}{\sqrt{2}}\) = 0
Solution:
This equation can also be written as x² + x + \(\frac{1}{\sqrt{2}}\)= 0
On comparing the given equation with ax² + bx + c = 0,
We obtain a = √2 , b = √2 and c = 1
Therefore, the discriminant of the given equation is
D = b² – 4ac = (√2)² – 4 × (√5) × 1 = 2 – 4√2
Hence, the required solutions are

Question 10.
x² + \(\frac{\mathbf{x}}{\sqrt{2}}\) + 1 = o
Solution:
The given quadratic equation is x² + \(\frac{\mathbf{x}}{\sqrt{2}}\) + 1 = 0
75
This equation can also be written as x² + \(\frac{\mathbf{x}}{\sqrt{2}}\) + 1 = o
On comparing the given equation with ax² + bx + c = 0,
We obtain a = √5, b = 1 and c = √2
Therefore, the discriminant of the given equation is
D = b² – 4ac = (1)² – 4 × (√2) × (√2) = 1 – 8 = -7
Hence, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations Exercise 4a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4a

I. Express each of the complex number given in the exercises 1 to 10 in the form a + ib.

Question 1.
(5i) (-\(\frac{3}{5}\)i)
Solution:
(5i) (-\(\frac{3}{5}\)i)
= -5i × \(\frac{3}{5}\) × i
= -3i² [∵ i² = -1]
= -3(-1) = 3
= 3 + i0

Question 2.
i⁹ + i¹⁹
Solution:
i⁹ + i¹⁹
= i⁹ + i¹⁰.i⁹
= i⁹[1 + (i²)⁵]
= i⁹(1 – 1)
= 0 = 0 + i0

Question 3.
i^-39
Solution:
i^-39 = i^4×(-9)-3
= (i⁴)^-9 × i^-3
= (1)^-9 × i^-3 [∵ i⁴ = 1]
= \(\frac{1}{i^3}=\frac{1}{-i}\) [∵ i³ = -i]
= \(\frac{-1}{i} \times \frac{i}{i}=-\frac{i}{i^2}=\frac{-i}{-1}\) [∵ i² = -1]
= i
= 0 + i [∵ i² = -1]

Question 4.
3(7 + i7) + i(7 + i7)
Solution:
3(7 + i7) + i(7 + i7)
= 21 + 21i + 7i + 7i²
= 21 + 28i + 7 × (-1)
= 14 + 28i [∵ i² = -1]

Question 5.
(1 – i) – (-1 + i6)
Solution:
(1 – i) – (-1 + i6)
= 1 – i + 1 – 6i
= 2 – 7i

Question 6.
(\(\frac{1}{5}\) + i\(\frac{2}{5}\)) – (4 + i\(\frac{5}{2}\))
Solution:
(\(\frac{1}{5}\) + i\(\frac{2}{5}\)) – (4 + i\(\frac{5}{2}\))
= \(\frac{1}{5}\) + \(\frac{2}{5}\)i – 4 – \(\frac{5}{2}\)i
= (\(\frac{1}{5}\) – 4) + i(\(\frac{2}{5}\) – \(\frac{5}{2}\))
= \(\left(-\frac{19}{5}\right)\) + i\(\left(-\frac{21}{10}\right)\)
= \(\frac{19}{5}\) – i\(\frac{21}{10}\)

Question 7.
[(\(\frac{1}{3}\) + i\(\frac{7}{3}\)) + (4 + i\(\frac{1}{3}\))] – (-\(\frac{4}{3}\) + i)
Solution:
[(\(\frac{1}{3}\) + i\(\frac{7}{3}\)) + (4 + i\(\frac{1}{3}\))] – (-\(\frac{4}{3}\) + i)
= \(\frac{1}{2}\) + i + 4 + \(\frac{1}{3}\)i + \(\frac{4}{3}\) – i
= (\(\frac{1}{3}\) + 4 + \(\frac{4}{3}\)) + i(\(\frac{7}{3}\) + \(\frac{1}{3}\) – 1)
= \(\frac{17}{3}\) + i\(\frac{5}{3}\)

Question 8.
(1 – i)⁴
Solution:
(1 – i)⁴ = [(1 – i)²]²
= [i² + i² – 2i]²
= [1 – 1 – 2i]²
= [-2i]² = 4i²
= -4
= -4 + 0i [∵ i² = -1]

Question 9.
(\(\frac{1}{3}\) + 3i)³
Solution:
(\(\frac{1}{3}\) + 3i)³ = (\(\frac{1}{3}\))³ + (3i)³ + 3\(\frac{1}{2}\)(3i)(\(\frac{1}{2}\) + 3i)
= \(\frac{1}{27}\) + 27i³ + 3i(\(\frac{1}{3}\) + 3i)
= \(\frac{1}{27}\) + 27(-i) + i + 9i² (∵ i³ = -i)
= \(\frac{1}{27}\) – 27i + i – 9i (∵ i² = -1)
= (\(\frac{1}{27}\) – 9) – 26i
= –\(\frac{242}{27}\) – 26i

Question 10.
(-2 – \(\frac{1}{3}\)i)³
Solution:
(-2 – \(\frac{1}{3}\)i)³
= (-1)³ × (2 + \(\frac{1}{3}\)i)³
= -[2³ + (\(\frac{i}{3}\))³ + 3(2)(\(\frac{i}{3}\))(2 + \(\frac{i}{3}\))]
= -[8 + \(\frac{1}{2}\) +2i (2 + \(\frac{1}{3}\))]
= -[8 – \(\frac{i}{27}\) + 4i + \(\frac{2}{3}\)i²] [∵ i³ = -1]
= -[8 – \(\frac{1}{27}\) + 4i – \(\frac{2}{3}\)]
= –\(\left[\frac{22}{3}+\frac{107 \mathrm{i}}{27}\right]\)
= –\(\frac{22}{3}\) – i\(\frac{107}{27}\)

Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.

Question 11.
4 – 3i
Solution:
Let z = 4 – 3i .
Then, z̄ = 4 + 3i and |z|² = 4² + (-3)² = 16 + 9 = 25
Therefore, the multiplicative inverse of 4 – 3i is given by
z^-1 = \(\frac{\bar{z}}{|z|^2}=\frac{4+3 i}{25}\)
= \(\frac{4}{25}\) + i\(\frac{3}{25}\)

Question 12.
√5 + 3i
Solution:
Let z̄ = √5 + 3i
Then, z̄ = √5 – 3i and |z|² = (√5) + 3² = 5 + 9 = 14
Therefore, the multiplicative inverse of √5 + 3i is given by
z^-1 = \(\frac{\bar{z}}{|z|^2}=\frac{\sqrt{5}-3 i}{14}=\frac{\sqrt{5}}{14}-\frac{3}{14} i\)

Question 13.
-i
Solution:
Let z = -i
Let z = i and |z|² = 1² = 1.
Therefore the multiplicative inverse of -i is given by, z^-1 = \(\frac{\bar{z}}{|z|^2}=\frac{i}{1}\) = 0 + i(1)

II.

Question 1.
Express the following expression in the form of a + ib.
\(\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})}\)
Solution:

Practice AP Inter 1st Year Maths Study Material Chapter 13 Statistics MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Statistics MCQ

Question 1.
The range of the following data (marks scored by 10 students in mathematics) is 15, 20, 31, 62, 13, 6, 41, 86, 21, 74 is
1) 68
2) 56
3) 80
4) 71
Answer:
3) 80

Explanation:
Max = 86; Min = 6
Range = Max – Min = 86 – 6 = 80

Question 2.
In a test match two batsmen A and B scored 116, 2 and 76, 138 in two innings respectively. Whose score is more scattered ?
1) A
2) B
3) Can’t say
4) Equal
Answer:
1) A

Explanation:
A’s mean = \(\frac{(116+2)}{2}\) = 59;
B’s mean = \(\frac{(76+138)}{2}\) = 107

⇒ A : deviation from mean = |116 – 59| = 57, |2 – 59| = 57
⇒ B : deviation from mean = |76 – 107| = 31, |138 – 107| = 31
So, A has larger deviation ⇒ more scattered.

Question 3.
The formula \(\frac{1}{N} \sum_{i=1}^n\) f_i|x_i – x̄| is used to calculate which of the following dispersion?
1) Mean deviation
2) Standard deviation ( 1 )
3) Range
4) Q.D
Answer:
1) Mean deviation

Explanation:
The formula : This formula calculates the Mean Deviation from the mean.

Question 4.
Which of the following formula is used to measure mean deviation for grouped data about median?
(1) \(\sum_{i=1}^{\mathrm{n}}\)f_i|x_i – x̄|
(2) \(\frac{1}{N_i} \sum_{i=1}^n\)f_i|x_i – x̄|
(3) \(\frac{1}{N_i} \sum_{i=1}^n\)f_i|x_i – M|
(4) \(\frac{1}{N_i} \sum_{i=1}^n\)|x_i – M|
Answer:
(3) \(\frac{1}{N_i} \sum_{i=1}^n\)f_i|x_i – M|

Explanation:
Formula for mean deviation for grouped data about median is
\(\frac{1}{N_i} \sum_{i=1}^n\)f_i|x_i – M|

Question 5.
Standard deviation of a discrete frequency distribution is meansured by using
(1) \(\frac{1}{N} \sqrt{\sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2}\)
(2) \(\sqrt{\sum_{i=1}^n\left(x_i-\bar{x}\right)^2}\)
(3) \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)
(4) \(\sqrt{\frac{1}{\mathrm{~N}_{\mathrm{i}}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}_1\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}\)
Answer:
(3) \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)

Explanation:
Standard deviation formula: \(\sqrt{\frac{1}{N_1} \sum_{1=1}^n f_1\left(x_1-\bar{x}\right)^2}\)

Question 6.
If each observation is multiplied by a constant k (≠ 0), the variance of the resulting observations becomes times the original variance ( 4 )
1) k (remains unchanged)
2) √k
3) k³
4) k²
Answer:
4) k²

Explanation:
New variance = k² × original variance

Question 7.
The variance of 20 observations is 5. If each observation is added by 3 then the new variance of the resulting observations.
1) 5
2) 8
3) 15
4) 9
Answer:
1) 5

Explanation:
Adding (or) subtracting a constant doesn’t change variance.

Question 8.
Mean of the squares of the deviations from mean is known as ( 4 )
1) Mean deviation
2) Standard deviation
3) Quartile deviation
4) Variance
Answer:
4) Variance

Explanation:
This is the variance by defintion.

Question 9.
Relation between the Mean deviation and the Standard deviation is ( 2 )
1) σ ≤ MD
2) MD ≤ σ
3) σ ≤ \(\sqrt{MD}\)
4) σ² ≤ \(\sqrt{MD}\)
Answer:
2) MD ≤ σ

Explanation:
The relation between mean deviation (MD) and Standard Deviation is always MD ≤ a

Question 10.
The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is
(a) \(\frac{4}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{1}{5}\)
Answer:
Mean x̄ = \(\frac{12}{10}\) = 1.2
Variance : \(\frac{1}{10}\) Σx²= \(\frac{18}{10}\) – (1.2)²
= 1.8 – 1.44
= 0.36.
Standard deviation = \(\sqrt{0.36}\) = 0.6 = \(\frac{3}{5}\),

Referring to the AP Inter 1st Year Maths Study Material Chapter 13 Statistics Exercise 13c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Statistics Solutions Exercise 13c

Question 1.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Solution:
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y
Mean x̄ = \(\frac{6+7+10+12+12+13+x+y}{8}\) = 9
⇒ 60 + x + y = 72
⇒ x + y = 12 ………(1)
Variance = 9.25 = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^8\)(x_i – x̄)²
9.25 = \(\frac{1}{8}\)[(-3)² + (-2)² + (1)² + (3)² + (3)² + (4)² + x² + y² – 2 × 9(x + y) + 2(9)²]
9.25 = \(\frac{1}{8}\)[9 + 4 + 1 + 9 + 9 + 16 + x² + y² – 18(12) + 162] ………..[Using (1)]
9.25= \(\frac{1}{8}\)[48 + x² + y² – 216 + 162]
9.25 = \(\frac{1}{8}\)[x² + y² – 6]
⇒ x² + y² = 80 …. (2)

From (1), we obtain (x + y)² = (12)²
⇒ x² + y² + 2xy = 144
From (2) and (3), we obtain 2xy = 64
Subtracting (4) from (2), we obtain
x² + y² – 2xy = 80 – 64 = 16
⇒ x – y = ±4
Therefore, from (1) and (5), we obtain x = 8 and y = 4, when x – y = 4 x = 4 and y = 8, when x – y = – 4 Thus, the remaining observations are 4 and 8.

Question 2.
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Solution:
Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.
i.e Mean, x̄ = \(\frac{2+4+10+12+14+x+y}{7}\)
⇒ 56 = 42 + x + y
⇒ x + y = 14 ………..(1)

Variance = 16 = \(\frac{1}{n} \sum_{i=1}^7\left(x_i-\bar{x}\right)^2\)
16 = \(\frac{1}{7}\)[36 + 16 + 4 +16 + 36 + x² + y² – 16(x + y) + 2(64)]
16 = \(\frac{1}{7}\) [36 + 16 + 4 + 16 + 36 + x² + y² – 16(14) + 2(64)]
16 = \(\frac{1}{7}\)[108 + x² + y² – 224 + 128]
16 = \(\frac{1}{7}\)[12 + x² + y²]
⇒ x² + y² = 112 – 12
⇒ x² + y² = 100
From (1), we obtain x² + y² + 2xy = 196
From (2) and (3), we obtain 2xy = 196 – 100
⇒ 2xy = 96 …. (4)
Subtracting (4) from (2), we obtain x² + y² – 2xy = 100 – 96
⇒ (x – y)2 = 4 ⇒ x – y = ±2 ….(5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2; x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8.

Question 3.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Solution:
Let the observations be x₁, x₂, x₃, x₄, x₅ and x₆
It is given that mean is 8 and standard deviation is 4.
Mean x̄ = \(\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}\) …. (1)
If each observation is multiplied by 3 and the resulting observation be y., then
y_i = 3x_i ⇒ x_i = \(\frac{1}{3}\)y_i, for i = 1 to 6
New Mean ȳ = \(\frac{y_1+y_2+y_3+y_4+y_5+y_6}{6}=\frac{3\left(x_1+x_2+x_3+x_4+x_5+x_6\right)}{6}\)
= 3 × 8 = 24 ……….. [using (1)]
Standard deviation σ = \(\sqrt{\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^6\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}\)
⇒ (4)² = \(\frac{1}{6} \sum_{i=1}^6\)(x_i – x̄)²

From (1) and (2), it can be observed that, ȳ = 3x̄ ⇒ x̄ = \(\frac{1}{3}\) ȳ
Substituting the values of x_i and x̄ in (2), we obtain
\(\sum_{i=1}^6\left(\frac{1}{3} y_i-\frac{1}{3} \bar{y}\right)^2\) = 96 ⇒ \(\sum_{\mathrm{l}=1}^6\)(y_i – ȳ)² = 864

Therefore, variance of new observations = (\(\frac{1}{6}\) × 864) = 144
Hence, the standard deviation of new observations is \(\sqrt{144}\) = 12.

Question 4.
Given that x̄ is the mean and σ² is the variance of n observations x₁, x₂,…. x_n. Prove that the mean and variance of the observations ax₁, ax₂, ax₃ ………….. ax_n are ax̄ and a²σ², respectively, (a * 0)
Solution:
The given n observations x₁, x₁ ….x_n.
Mean = x̄; Variance = σ².
∴ σ² = \(\frac{1}{n} \sum_{i=1}^n\) y_n …………(1)
If each observation is multiplied by a and the new observations be y_i, then
y_i = ax_i ⇒ x_i = \(\frac{1}{a}\)y_i
∴ ȳ = \(\frac{1}{n} \sum_{i=1}^n\)y₁
= \(\frac{1}{n} \sum_{i=1}^n\) ax_i
= \(\frac{a}{n} \sum_{i=1}^n\) x_i
= ax

(x̄ = \(\frac{1}{n} \sum_{i=1}^n\)x_i)
Therefore, mean of the observations ax₁, ax₂, ……… ax_n, is ax̄
Substituting the values of x_i and x̄ in (!), we obtain
σ² = \(\frac{1}{n} \sum_{i=1}^n\left(\frac{1}{a} y_i-\frac{1}{a} \bar{y}\right)^2\)
⇒ a²σ² = \(\frac{1}{n} \sum_{i=1}^n\)(y_i – ȳ)²
Thus, the variance of the observations, ax₁, ax₂ ……….. ax_n is a2a2.

Question 5.
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases.
i) If wrong item is omitted
ii) If it is replaced by 12.
Solution:
i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
x̄ = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{20}\)x_i
⇒ 10 = \(\frac{1}{20} \sum_{i=1}^{20}\)x_i
⇒ \(\sum_{i=1}^{20}\)x_i = 200
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean = \(\frac{\text { correct sum }}{19}=\frac{192}{19}\) = 10.105

Standard deviation σ = \(\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}\)

⇒ 2 = \(\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2}\)

⇒ 4 = \(\frac{1}{20}\) Incorrect \(\sum_{i=1}^{\mathrm{n}}\) x_i² – 100

⇒ Incorrect \(\sum_{i=1}^n\) x_i = 2080
∴ Correct = \(\sum_{i=1}^n\)x_i² = Incorrect \(\sum_{i=1}^n\)x – (8)² + (12)²
= 2080 – 64 + 144
= 2016

Corect standard deviation = \(\sqrt{\frac{\text { Correct } \sum x_1^2}{n}-(\text { Correct mean })^2}\)
= \(\sqrt{\frac{2016}{19}-(10.105)^2}\)
= \(=\sqrt{106.105-102.111}\)
= \(\sqrt{3.984}\) = 1.995

ii) When 8 is replaced by 12.
Incorrect sum of observations = 200
∴ Correct sum of observation = 200 – 8 + 12 = 204

Correct mean = \(\frac{\text { Correct sum }}{20}=\frac{204}{20}\) = 10.2
Standard deviation σ = \(\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-\frac{1}{n^2}\left(\sum_{i=1}^n x_i\right)^2}=\sqrt{\frac{1}{n} \sum_{i=1}^n x_i^2-(\bar{x})^2}\)
2 = \(\sqrt{\frac{1}{20} \text { Incorrect } \sum_{i=1}^n x_i^2-(10)^2}\)
⇒ 4 = \(\frac{1}{20}\) Incorrect \(\sum_{l=1}^{\mathrm{n}}\) x_i² – 100

Incorrect \(\sum_{l=1}^{\mathrm{n}}\)x_i = 2080
Correct Σx_i² = Incorrect \(\sum_{i=1}^{\mathrm{n}}\) _i – (8)² + (12)²
= 2080 – 64 + 144
= 2160

Correct standard deviation
= \(\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct mean })^2}=\sqrt{\frac{2160}{20}-(10.2)^2}\)
= \(\sqrt{108-104.04}=\sqrt{3.96}\) = 1.98

Question 6.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Solution:
Number of observations (n) = 100
Incorrect mean (x) = 20 ; Incorrect standard deviation (σ) = 3
20 = \(\frac{1}{100} \sum_{i=1}^{100}\)x_i ⇒ \(\sum_{i=1}^{100}\)x_i = 20 × 100 = 2,000
Incorrect sum of observations = 2,000

Correct sum of observations = 2000 – 21 – 21 – 18 = 1940
Now \(\frac{1}{2}\)x_i² – (x̄)²
⇒ \(\frac{1}{100}\) Σx_i² – (20)² = 9
⇒ Σx_i² = 40, 900

Correct mean = \(\frac{1940}{97}\) = 20
Also Σx_i² = incorrect Σx_i² – (21)² – (21)² – (18)²
= 40,900 – 441 – 441 – 324
= 39,694

Correct variance = \(\frac{1}{97}\) correct Σx_i² – (correct mean)²
= \(\frac{1}{97}\) × 39,694 – (20)²
= 409.22 – 400 = 9.22
⇒ Correct SD = \(\sqrt{9.22}\) = 3.036

Referring to the AP Inter 1st Year Maths Study Material Chapter 13 Statistics Exercise 13b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Statistics Solutions Exercise 13b

I.

Question 1.
Find the mean and variance for each of the following data.
a) 6, 7, 10, 12, 13, 4, 8, 12.
b) First ‘n’ natural numbers,
c) First 10 multiples of 3.
Solution:
a) Given observations : 6, 7, 10, 12, 13, 4, 8, 12.

Mean, x̄ = \(\frac{\sum_{i=1}^8 x_i}{n}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}\) = 9

The following table is obtained for the given data.

Variance (σ²) = \(\frac{1}{n} \sum_{i=1}^n\) (x_i – x̄)²
= \(\frac{1}{8}\) × 74
= 9.25

b) The mean of first ‘n’ natural numbers is calculated as follows.
Mean = \(\frac{\text { Sum of all observations }}{\text { No. of observations }}=\frac{n\left(\frac{n+1}{2}\right)}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}\)

c) The first 10 multiples of ‘3’ are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
Here, number of observations, n = 10
Mean, x̄ = \(\frac{\sum_{i=1}^{10} x_i}{10}=\frac{165}{10}\) = 16.5

The following table is obtained for the given data.

Variance (σ²) = \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{10}\)(x_i – x̄)²
= \(\frac{1}{10}\) × 742.5
= 74.25

II.

Question 1.
Find the mean and variance for each of the following data.

Solution:

Here, N = 40, \(\sum_{i=1}^7\)f_ix_i = 760
x̄ = \(\frac{\sum_{i=1}^7 f_i x_i}{N}=\frac{760}{40}\) = 19
Variance (σ²) = \(\frac{1}{N} \sum_{i=1}^7\) f_i(x_i – x̄)² = \(\frac{1}{40}\) × 1736
= 43.4

Solution:
The data id obtained in tabular form as follows.

Here, N = 22, \(\frac{1}{2}\)f_ix_i = 2200
∴ x̄ = \(=\frac{1}{N} \sum_{i=1}^7\)f_ix_i

Variance (σ²) = \(\frac{1}{N} \sum_{i=1}^7\) f_i(x_i – x̄)² = \(\frac{1}{22}\) × 640 = 29.09

Question 2.
Find the mean and standard deviation using short-cut method.

Solution:
The data is obtained in tabular form as follows:

Mean x̄ = A + \(\frac{\sum_{i=1}^9 f_i y_i}{N}\) × h
= 64 + \(\frac{0}{100}\) × 1
= 64 + 0
= 64

Variance (σ²) = \(\frac{h^2}{N^2}\left[N \sum_{i=1}^9 f_i y_i^2-\left(\sum_{i=1}^9 f_i y_i\right)^2\right]\)
= \(\frac{1}{100^2}\)[100 × 286 – 0]
= 2.86

∴ Standard deviation (σ) = \(\sqrt{2.86}\) = 1.69

III.

Question 1.
Find the mean and variance for the following frequency distribution.

Solution:
The data is obtained in tabular form as follows:

Mean, x̄ = A + \(\frac{\sum_{i=1}^9 f_i y_i}{N}\) × h
= 105 + \(\frac{2}{30}\) × 30
= 105 + 2
= 107

Variance (σ²) = \(\frac{h^2}{N^2}\left[N \sum_{i=1}^9 f_i y_i^2-\left(\sum_{i=1}^9 f_i y_i\right)^2\right]\)
= \(\frac{(30)^2}{(30)^2}\)[30 × 76 – (2)²]
= 2280 – 4
= 2276

Solution:
The data is obtained in tabular form as follows:

Mean, x̄ = A + \(\frac{\sum_{i=1}^9 f_i y_i}{N}\) × h
= 25 + \(\frac{10}{50}\) × 10
= 25 + 2 = 27

Variance (σ²) = \(\frac{h^2}{N^2}\left[N \sum_{i=1}^9 f_i y_i^2-\left(\sum_{i=1}^9 f_i y_i\right)^2\right]\)
= \(\frac{1}{2}\)[50 × 68 – (10)²]
= \(\frac{1}{25}\)[3400 – 100]
= \(\frac{3300}{25}\) = 132

Question 2.
Find the mean, variance and standard deviation using short-cut method.

Solution:
The data is obtained in tabular form as follows:

Mean, x̄ = A + \(\frac{\sum_{i=1}^9 f_i y_i}{N}\) × h
= 92.5 + \(\frac{6}{60}\) × 5
= 92.5 + 0.5
= 93

Variance (σ²) = \(\frac{h^2}{N^2}\left[N \sum_{i=1}^9 f_i y_i^2-\left(\sum_{i=1}^9 f_i y_i\right)^2\right]\)
= \(\frac{(5)^2}{(60)^2}\)[60 × 254 – (6)²]
= \(\frac{25}{3600}\)[15240 – 36]
= \(\frac{25}{3600}\)(15204)
= 105.58

∴ Standard deviation (σ) = \(\sqrt{105.58}\) = 10.27

Question 3.
The diameters of circles (in mm) drawn in a design are given below.

Calculate the standard deviation and mean diameter of the circles.
Solution:

Here, N = 100, h = 4. Let the assumed mean A, be 42.5
Mean, x̄ = A + \(\frac{\sum_{i=1}^9 f_i y_i}{N}\) × h
= 42.5 + \(\frac{25}{100}\) × 4
= 43.5

Variance (σ²) = \(\frac{h^2}{N^2}\left[N \sum_{i=1}^9 f_i y_i^2-\left(\sum_{i=1}^9 f_i y_i\right)^2\right]\)
= \(\frac{16}{10000}\) [100 × 199 – (25)²]
= \(\frac{16}{10000}\) [19900 – 625]
= \(\frac{16}{10000}\) × 19275
= 30.84

∴ Standard deviation (σ) = 5.55

Referring to the AP Inter 1st Year Maths Study Material Chapter 13 Statistics Exercise 13a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Statistics Solutions Exercise 13a

I.

Question 1.
Find the mean deviation about the mean for the following data.
a) 4, 7, 8, 9, 10, 12, 13, 17
b) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Solution:
a) The given data is 4, 7, 8, 9, 10, 12, 13, 17.
Mean of the data, \(\frac{1}{n} \sum_{i=1}^n x_i\) = x̄ = \(\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}\) = 10
The deviations of the respective observations from the mean x̄ i.e., x_i – x̄ are -6, -3, -2, -1, 0, 2, 3, 7.
The required mean deviation about the mean is 8
M.D. (x̄) = \(\frac{\sum_{i=1}^8\left|x_i-\bar{x}\right|}{8}=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}\) = 3

b) The given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Mean of the given data, x̄ = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}\) = 50
The deviations of the respective observations from the mean Xs i.e.,
x_i – x̄ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.
\(\sum_{i=1}^{10}\)|x_i – x̄| = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84
The required mean deviation about the mean is
M.D. (x̄) = \(\frac{\sum_{j=1}^{10}\left|x_i-\bar{x}\right|}{10}=\frac{84}{10}\) = 8.4

Question 2.
Find the mean deviation about the median for the following data.
a) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
b) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Solution:
a) Here, the number of observations are 12, which is even.
Arranging the data in ascending order, we have,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

\(\sum_{i=1}^{12}\)|x_i – M| = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5 = 28
∴ The required mean deviation about the median is
M.D. (M) = \(\frac{\sum_{i=1}^{12}\left|x_i-M\right|}{12}=\frac{28}{12}\) = 2.33

b) The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Here, the number of observations is 10, which is even.
Arranging the data in an ascending order, we have
36, 42, 45, 46, 46, 49, 51, 53, 60, 72

\(\sum_{i=1}^{10}\)|x_i – M| = 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 = 70
Thus, the required mean deviation about the median is
M.D (M) = \(\frac{\sum_{i=1}^{10}\left|x_i-M\right|}{10}=\frac{70}{10}\) = 7

II.

Question 1.
Find the mean deviation about the mean for the following data.

Solution:
We make following table from the given data.

N = \(\sum_{i=1}^5\) f_i = 25 and \(\sum_{i=1}^5\)f_ix_i = 350;
∴ x̄ = \(\frac{1}{N_i} \sum_{i=1}^5\) f_ix_i = \(\frac{1}{25}\) × 350 = 14
∴ M.D.(x̄) = \(\frac{1}{2}\) f_i |x_i – x̄| = \(\frac{1}{25}\) x 158 = 6.32

Solution:
We make following table from the given data.

Question 2.
Find the mean deviation about the median for the following data.

Solution:
The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we have the following table.

Here, N = 26, which is even. Median is the Mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
∴ Median = \(\frac{\left(13^{\text {th }} \text { observation }+14^{\text {th }} \text { observation }\right)}{2}=\frac{7+7}{2}=\frac{14}{2}\)
Median = 7
\(\sum_{i=1}^6\)f_i = 26 and \(\sum_{i=1}^6\)f_i|x_i – M| = 84;
M.D (M) = \(\frac{1}{N_i} \sum_{i=1}^6\)f_i|x_i – M| = \(\frac{1}{26\) × 84 = 3.23

Solution:
The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we have the following table.

Here, N = 29, which is odd.
Median = \(\left(\frac{29+1}{2}\right)^{\text {th }}\) observation = 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
M = 30
\(\sum_{i=1}^5\) f_i = 29, \(\sum_{i=1}^5\) f_i|x_i – M| = 148
∴ M.D. (M) = \(\frac{1}{N} \sum_{i=1}^5\)f_i|x_i – M| = \(\frac{1}{29}\) × 148 = 5.1

III.

Question 1.
Find the mean deviation about the mean for the following data.

Solution:
The following table is formed for the given data.

Here, N = \(\frac{1}{2}\) f_i = 50 and f_ix_i = 17900
x = \(\frac{1}{2}\)f_ix_i = \(\frac{1}{50}\) × 17900 = 358
and M.D. (x) = \(\frac{1}{N} \sum_{i=1}^8\) f_i|x_i – x̄| = \(\frac{7896}{50}\) = 157.92

Solution:
The following table is formed for the given data.

Here, N = \(\sum_{i=1}^6\) f_i = 100 and \(\sum_{i=1}^6\) f_ix_i = 12530
x = \(\frac{1}{N_i} \sum_{i=1}^6\)f_ix_i = \(\frac{1}{100}\) × 12530 = 125.3
and M.D (x̄) = \(\frac{1}{N_i} \sum_{i=1}^6\) f_i|x_i – x̄| = \(\frac{1}{100}\) × 1128.8 = 11.28

Question 2.
Find the mean deviation about median for the folloeing data:

Solution:
The following table is formed for the given data.

The class interval containing the \(\left[\frac{\mathrm{N}}{2}\right]^{\mathrm{th}}\) or 25th item is 20 – 30.
Therefore, 20 – 30 is the median class. It is known that
Median = l + \(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\) × h
Here, 1= 20, C = 14, f = 14, h = 10 and N = 50
Median = 20 + \(\frac{25-14}{14}\) × 10 = 20 + \(\frac{110}{14}\) = 20 + 7.85 = 27.85
Thus, mean deviation about the median is given by,
M.D. (M) = \(\frac{1}{N_i} \sum_{i=1}^6\)f_i|x_i – x̄| = \(\frac{1}{50}\) × 517.1 = 10.34

Question 3.
Calculate the mean deviation about median age for the age distribution of 100 persons given below.

Solution:
The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

The class interval containing the \(\left[\frac{\mathrm{N}}{2}\right]^{\text {th }}\) or 50th item is 35.5 – 50.5
Therefore, 35.5 – 40.5 is the median class. It is known that
Median (M) = l + \(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\) × h
Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100
∴ Median = 35.5 + \(\frac{50-37}{26}\) × 5 = 35.5 + \(\frac{13 \times 5}{26}\) = 35.5 + 2.5 = 38
Thus, mean deviation about the median is given by,
M.D. (M) = \(\frac{1}{N_i} \sum_{i=1}^8\)f_i|x_i – x̄| = \(\frac{1}{100}\) × 735 = 7.35

Referring to the AP Inter 1st Year Maths Study Material Chapter Trigonometric Functions Exercise 3e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Trigonometric Functions Exercise 3e

I. Find the principal and general solutions of the following equations.

Question 1.
tan x = √3
Solution:
We know that tan \(\frac{1}{2}\) = √3 and tan\(\left(\frac{4 \pi}{3}\right)\) = tan\(\left(\pi+\frac{\pi}{3}\right)\)
= tan\(\frac{\pi}{3}\) = √3
The principal solutions are x = \(\frac{\pi}{3}\) and \(\frac{4 \pi}{3}\)
tan x = tan\(\frac{\pi}{3}\) ⇒ x = nπ + \(\frac{\pi}{3}\) where n ∈ Z
The general solution is x = nπ + \(\frac{\pi}{3}\) where n ∈ Z

Question 2.
sec x = 2
Solution:
We know that sec\(\frac{\pi}{3}\) = 2 and sec \(\frac{5 \pi}{3}\)
= sec (2π – \(\frac{\pi}{3}\)) = sec \(\frac{\pi}{3}\) = 2
The principal solutions are x = \(\frac{\pi}{3}\) and \(\frac{5 \pi}{3}\)
sec x = sec \(\frac{\pi}{3}\) ⇒ cos x = cos \(\frac{\pi}{3}\) [∵ sec x = \(\frac{1}{\cos x}\)]
⇒ x – 2nπ ± \(\frac{\pi}{3}\) where n ∈ Z
∴ The general solution is x = 2nπ ± \(\frac{\pi}{3}\); where n ∈ Z

Question 3.
cot x = √3
Solution:
We know that, cot \(\frac{\pi}{3}\) = √3
cot (π – \(\frac{\pi}{6}\)) = -cot\(\frac{\pi}{6}\) = -√3 and cot(2π – \(\frac{\pi}{6}\))
= -cot\(\frac{\pi}{6}\) = -√3
⇒ cot \(\frac{5 \pi}{6}\) = -√3 and cot \(\frac{11 \pi}{6}\) = -√3

The principal solutions are x = \(\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\)
cot x = cot \(\frac{5 \pi}{6}\) ⇒ tanx = tan\(\frac{5 \pi}{6}\)
⇒ x = nx + \(\frac{5 \pi}{6}\) , where n ∈ Z
∴ The general solution is x = nx + \(\frac{5 \pi}{6}\), where n ∈ Z

Question 4.
cosec x = -2
Solution:
We know that, cosec \(\frac{\pi}{6}\) = 2
cosec (π + \(\frac{\pi}{6}\)) = – cosec \(\frac{\pi}{6}\) = -2 and cosec (2π – \(\frac{\pi}{6}\)) = -cosec \(\frac{\pi}{6}\) = -2
⇒ cosec \(\frac{7 \pi}{6}\) = -2 and cosec \(\frac{11 \pi}{6}\) = -2
The principal solutions are x = \(\frac{7 \pi}{6}\) and \(\frac{11 \pi}{6}\)
cosec x = cosec \(\frac{7 \pi}{6}\) ⇒ sinx = sin\(\frac{7 \pi}{6}\)
x = nπ + (-1)ⁿ\(\frac{7 \pi}{6}\)
∴ The general solution is x = nπ + (-1)ⁿ\(\frac{7 \pi}{6}\) where n ∈ Z

II. Find the general solution for each of the following equations.

Question 1.
cos 4x = cos 2x
Solution:
Given that, cos 4x = cos 2x ⇒ cos 4x – cos 2x = 0
⇒ -2 sin\(\left(\frac{4 x+2 x}{2}\right)\)sin\(\left(\frac{4 x-2 x}{2}\right)\) = 0
⇒ sin 3x sin x = 0
⇒ sin 3x = 0 or sin x = 0
3x = nπ or x = nπ, where n ∈ Z ⇒ x = \(\frac{\mathrm{n} \pi}{3}\) or x = nx, where n ∈ Z

Question 2.
cos 3x + cos x – cos 2x = 0.
Solution:
Given that cos 3x + cos x – cos 2x = 0
⇒ 2cos 2x cos x -cos 2x = 0
⇒ cos 2x (2 cos x – 1) = 0
⇒ cos 2x = 0 or 2cos x – 1 = 0
⇒ cos 2x = 0 or cos x = \(\frac{1}{2}\)
⇒ cos\(\frac{\pi}{3}\) ⇒ x = (2n + 1)\(\frac{\pi}{4}\) or x = 2nx ± \(\frac{\pi}{3}\), where n ∈ Z

Question 3.
sin 2x + cos x = 0
Solution:
Given that sin 2x + cos x = 0
⇒ 2 sin x cos x + cos x = 0
⇒ cos x (2 sin x + 1) = 0
⇒ cos x = 0 or 2 sin x + 1 = 0
cos x = 0 ⇒ x = (2n + 1)\(\frac{\pi}{2}\), where n ∈ Z
2 sin x + 1 = 0
⇒ sin x = \(\frac{-1}{2}\) = -sin\(\frac{\pi}{2}\) = sin(π + \(\frac{\pi}{2}\)) = sin \(\frac{7 \pi}{2}\)
⇒ x = nπ + (-1)ⁿ\(\frac{7 \pi}{6}\), where n ∈ Z
∴ The general solution is (2n + 1)\(\frac{\pi}{2}\) or nπ + (-1)ⁿ\(\frac{7 \pi}{2}\), n ∈ Z

Question 4.
sec² 2x = 1 – tan 2x.
Solution:
Given that sec² 2x = 1 – tan 2x
⇒ 1 + tan² 2x = 1 – tan 2x
⇒ tan² 2x – tan 2x = 0
⇒ tan 2x (tan 2x – 1) = 0
⇒ tan 2x = 0 or tan 2x – 1 = 0
⇒ tan 2x = 0 or -1
⇒ 2x = nπ, where n ∈ Z
⇒ x = \(\frac{\mathrm{n} \pi}{2}\), where n ∈ Z

tan 2x = -1 = -tan \(\frac{\pi}{4}\) = tan (π – \(\frac{\pi}{4}\)) = tan \(\frac{3 \pi}{2}\)
⇒ 2x = nπ + \(\frac{3 \pi}{4}\), where n ∈ Z
⇒ x = \(\frac{\mathrm{n} \pi}{2}+\frac{3 \pi}{8}\), where n ∈ Z
∴ The general solution is \(\frac{\mathrm{n} \pi}{2}\) or \(\frac{\mathrm{n} \pi}{2}+\frac{3 \pi}{8}\), n ∈ Z

Question 5.
sin x + sin 3x + sin 5x = 0
Solution:
We know that, sin A + sin B = 2sin\(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) cos \(\left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\)
⇒ sin x + sin 5x = 2 sin \(\left(\frac{6 \mathrm{x}}{2}\right)\) cos \(\left(\frac{-4 x}{2}\right)\) = 2 sin 3x cos 2x
⇒ 2sin 3x cos 2x + sin 3x = 0
⇒ sin 3x(2 cos 2x + 1) = 0
⇒ sin 3x = 0 or 2 cos 2x + 1 = 0
⇒ sin 3x = 0
⇒ 3x = nπ
⇒ x = \(\frac{\mathrm{n} \pi}{3}\)

2 cos 2x + 1 = 0
⇒ cos 2x = –\(\frac{1}{2}\)
⇒ 2x = 2nπ ± \(\frac{2 \pi}{3}\)
⇒ x = nπ ± \(\frac{\pi}{3}\)
∴ The general solution is \(\frac{\mathrm{n} \pi}{3}\) or nπ ± \(\frac{\pi}{3}\), where n ∈ Z .

Referring to the AP Inter 1st Year Maths Study Material Chapter 14 Probability Exercise 14d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Probability Solutions Exercise 14d

In each of the following Exercises 1 to 7, describe the sample space for the indicated experiment.

Question 1.
A coin is tossed four times.
Solution:
Let S’ be a Sample space. Each toss has two outcomes: H(Head), T(Tail).
Total outcomes = 2⁴ = 16.
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}.

Question 2.
A coin is tossed and a dice is thrown.
Solution:
Let ‘S’ be a Sample space, Coin → {H, 1, Dice → {1, 2, 3, 4, 5, 6)
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Total outcomes = 12.

Question 3.
A coin is tossed and then a dice is rolled only in case a head is shown on the coin.
Solution:
Let S’ be a Sample space. If Head : roll dice → {1, 2, 3, 4, 5, 6), If Tail : no dice roll → T
S = {T, (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
Total outcomes = 7

Question 4.
2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for the experiment in which a room is selected and then a person.
Solution:
Let ‘S’ be a Sample Space. Room X : B1, B2, G1, G2 Room Y : B3, G4, G3, G4
S = {(X, B1), (X, B2), (X, G1), (X, G2), (Y, B3), (Y, B4), (Y, G3), (Y, G4)}
Total outcomes = 8

Question 5.
One dice of red colour, one of white colour and one of blue colour are placed in a bag. One dice is selected at random and rolled, its colour and the number on its uppermost face is noted. Describe the sample space.
Solution:
Dice colour → {R, W, B}, Each with outcomes 1 to 6.
S = {(R, 1), (R, 2), …, (R, 6), (W, 1), … , (W, 6), (B, 1) … , (B, 6)}
Total outcomes = 3 × 6 = 18

Question 6.
An experiment consists of recording boy-girl composition of families with 2 children.
i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
ii) What is the sample space if we are interested in the number of girls in the family?
Solution:
i) Order matters [first born, second bom) possible combinations
S = {BB, BG, GB, GG} (B = Boy, G = Girl)

ii) Only number of girls matters
S = {0, 1, 2} (0 girls → BB, 1 Girl → BG (or) GB, 2 girls → GG).

Question 7.
A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Solution:
Let R = Red, W = White, Balls : {R, W1, W2, W3} {W’s are identical, so indistinct)
Unique Combinations: RW, WR, WW
Hence, S = {RW, WR, WW}
Note : W1, W2, W3 are identical, so (W1, W2) – (W2, W3) = WW.

Question 8.
An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a tail occurs on the first toss, then a dice is rolled once. Find the sample space.
Solution:
Toss a coin. If head, toss again. If tail, roll a dice.
Sample Space, S :
First toss = H → toss again : (H, H), (H, T).
First toss = T → roll dice : (T, 1), (T, 2), … (T, 6).
S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Total outcomes = 2(From HH, HT) + 6 (From T1 – T6) = 8

Question 9.
Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as defective (D) or non-defective (N). Write the sample space of this experiment.
Solution:
3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-defective (N). The sample space of this experiment is given by
S = {DDD, DDN, DND, NDN, NND, NNN)}.

Question 10.
A coin is tossed. If the outcome is a head, a dice is thrown. If the dice shows up an even number, the dice is thrown again. What is the sample space for the experiment?
Solution:
When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5 or 6 S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}.

Question 11.
The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Solution:
Let ‘S’ be a Sample space.
Permutations of 2 slips out of 4 :
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
Total outcomes = ⁴P₂ = 12

Question 12.
An experiment consists of rolling a dice and then tossing a coin once if the number on the dice is even. If the number on the dice is odd, the coin is tossed twice. Write the sample space for this experiment.
Solution:
Experiment : Roll a dice, if it shows an even number, toss a coin once. If odd, toss the coin twice.
The possible outcomes when rolling dice are {1, 2, 3, 4, 5, 6}.
Outcomes for even numbers on dice : (2, H), (2, T), (4, H), (4, T), (6, H), (6, T).
Outcomes for odd numbers on dice : (1, HH), (1, TT), (1, TH), (1, HT), (3, HH), (3, TT), (3, TH), (3, HT), (5, HH), (5, TT), (5, TH), (5, HT).
So, S = {(2, H), (2, T), (4, H), (4, T), (6, H), (6, T), (1, HH), (1, TT), (1, TH), (1, HT), (3, HH), (3, TT), {3, TH), (3, HT), (5, HH), (5, TT), (5, TH), (5, HT)}
Total outcomes = 6 (even) + 12 (odd with coin twice) = 18
S = {(6), (x₁, 6), (x₁, x₂, 6), …} where each x_i ∈ {1, 2, 3, 4, 5}.
This is an infinite sample space of sequences ending with 6 and prior values being from (1, 2, 3, 4, 5}.

Question 13.
A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black balls. If it shows head, we throw a dice. Find the sample space for this experiment.
Solution:
Experiment : A coin is tossed if tail, draw a ball from box with 2 red and 3 black balls. If head, throw a dice.
Sample space, S :
If coin is Head → roll dice : H1, H2 … , H6
If coin is tail draw ball : TR1, TR2 (red), TB1, TB2, TB3 (Black).
5 = (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, Red 1), (T, Red 2), (T, Black 1), (T, Black 2), (T, Black 3).
Total outcomes = 6 + 5 = 11

Question 14.
A dice is thrown repeatedly untill a six comes up. What is the sample space for this experiment ?
Solution:
Experiment: A dice is thrown repeatedly untill a six comes up.
Sample space, S :
Stop only when 6 comes possible outcomes :
6
1, 6
1, 1, 6
2, 4, 3, 6
……………..
S = {6, (1, 6) ………….. (1, 1, 6) ………… (2,4, 3, 6) …………}
All sequences ending with 6 and before that only number (or) numbers from 1 to 5.

Practice AP Inter 1st Year Maths Study Material Chapter 14 Probability MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Probability MCQ

Question 1.
Let A and B be two events such that P(A) = 0.8, P(B) = 0.7. Then P (A ∩ B).
1) ≤ 0.6
2) ≥ 0.5
3) ≥ 0.7
4) ≤ 0.4
Answer:
2) ≥ 0.5

Explanation:
Given : P(A) = 0.8, P(B) = 0.7
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
So, P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
Maximum P(A ∪ B) = 1 ⇒ P(A ∩ B) = 0.8 + 0.7 – 1 = 0.5

Question 2.
The probability that a leap year selected at random will contain 53 Sundays is
(1) \(\frac{1}{7}\)
(2) \(\frac{2}{7}\)
(3) \(\frac{6}{7}\)
(4) \(\frac{5}{7}\)
Answer:
(2) \(\frac{2}{7}\)

Explanation:
Leap year has 366 days = 52 weeks + 2 days
So, it will have 53 Sundays if :
The two extra days are (Sunday and Monday) or (Saturday and Sunday) out of 7 possible day pairs; 2 pairs contain Sunday ⇒ Probability = \(\frac{2}{7}\).

Question 3.
If two dice are thrown, then the probability of getting the same number on both the faces is
(1) \(\frac{2}{6}\)
(2) \(\frac{3}{6}\)
(3) \(\frac{5}{6}\)
(4) \(\frac{1}{6}\)
Answer:
(4) \(\frac{1}{6}\)

Explanation:
Getting same number on both dice : (1, 1), (2, 2) ……….. (6, 6) outcomes.
Total outcomes = 36.
So, Probability = \(\frac{6}{36}=\frac{1}{6}\)

Question 4.
If two dice are thrown, then the probability of getting a total score of seven is
(1) \(\frac{1}{6}\)
(2) \(\frac{2}{6}\)
(3) \(\frac{3}{6}\)
(4) \(\frac{5}{6}\)
Answer:
(1) \(\frac{1}{6}\)

Explanation:
Total score of 7 means : (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 outcomes
Total outcomes = 36
So, Probability = \(\frac{6}{36}=\frac{1}{6}\)

Question 5.
If A and B are two events of a sample space with P(AuB) = P(A) + P(B), then they are
1) Mutually exclusive
2) Exhaustive
3) 1 and 2
4) None of these
Answer:
1) Mutually exclusive

Explanation:
P(A ∪ B) = P(A) + P(B) means P(A ∩ B) = 0 ⇒ A and B are mutually exclusive.

Question 6.
An integer is picked from 1 to 20 (both inclusive). Then the probability that it is a prime
(1) \(\frac{3}{20}\)
(2) \(\frac{3}{5}\)
(3) \(\frac{2}{5}\)
(4) \(\frac{1}{5}\)
Answer:
(3) \(\frac{2}{5}\)

Explanation:
Total numbers from 1 to 20 = 20.
Prime numbers ≤ 20 = 2, 3, 5, 7, 11, 13, 17, 19 → 8 primes
Probability = \(\frac{8}{20}=\frac{2}{5}\)

Question 7.
If P(A) = 0.5, P(B) = 0.3. (given that A and B are mutually exclusive) Then P(A’ ∩ B’) =
1) 0.6
2) 0.5
3) 0.7
4) 0.2
Answer:
4) 0.2

Explanation:
A and B are mutually exclusive ⇒ P(A ∩ B) = 0
P(A ∪ B) = P(A) + P(B) = 0.5 + 0.3 = 0.8
Then P(A’ ∩ B ) = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 8.
One card is drawn at random from a well shuffled deck of 52 cards. Then the probability that the card be not a red card.
(1) \(\frac{2}{5}\)
(2) \(\frac{1}{5}\)
(3) \(\frac{1}{2}\)
(4) \(\frac{1}{3}\)
Answer:
(3) \(\frac{1}{2}\)

Explanation:
In a deck of 52 cards, 26 are red (13 hearts + 13 diamonds). so, not red = 26
Probability = \(\frac{26}{52}=\frac{1}{2}\)

Question 9.
Which of the following is false ?
1) P(E) = 0 ⇔ E is an impossible event.
2) 0 ≤ P(E) < 1.
3) P(E) = 1 ⇔ E is a certain event.
4) P(E) + P(Ē) = 1.
Answer:
2) 0 ≤ P(E) < 1.

Explanation:
Option 2 is False, because a probability can be exactly 1; it’s not strictly less than one.

Question 10.
Which of the following statement is not correct ? P is a probability function of the sample space S.
1) P(E) > 0 ∀ E ∈ P(S).
2) P(S) = 1.
3) P(Φ) = 0.
4) P(E₁ ∩ E₂) = P(E₁) + P(E₂) where E₁, E₂ are mutually exclusive.
Answer:
4) P(E₁ ∩ E₂) = P(E₁) + P(E₂) where E₁, E₂ are mutually exclusive.

Explanation:
Option 4 is incorrect. In general, P(E₁ ∩ E₂) ≠ P(E₁) + P(E₂)
This equality is invalid unless both probabilities are 0 (which isn’t mentioned)
So, this is the false statement.