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Peer review of ISC OP Malhotra Solutions Class 11 Chapter 19 Differentiation Ex 19(d) can encourage collaborative learning.

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(d)

Question 1.
sin 5x
Solution:
Let y – sin 5x ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin 5x = cos 5x\(\frac{d}{d x}\)(5x) = 5 cos 5x

Question 2.
cos 8x
Solution:
Let y = cos 8x ; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (cos 8x) = -sin8x\(\frac{d}{d x}\) (8x) = – 8 sin x

Question 3.
sin (5x + 9)
Solution:
Let y = sin (5x + 9); Diff. both sides w.r.t. x,
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin (5x + 9) = cos(5x + 9)\(\frac{d}{d x}\)(5x + 9) = cos (5x + 9) (5.1 + 0) = 5 cos (5x + 9)

Question 4.
cos (2x – 3)
Solution:
Let y = cos (2x – 3) ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)cos(2x – 3) = -sin(2x – 3)\(\frac{d}{d x}\)(2x – 3) = – 2 sin (2x – 3)

Question 5.
tan 7x
Solution:
Let y = tan 7x ; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) tan 7x = sec² 7x\(\frac{d}{d x}\) (7x) = 7 sec² 7x

Question 6.
cot nx
Solution:
Let y = cot nx ; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) cot nx = – cosec² nx\(\frac{d}{d x}\) (nx) = – n cosec² nx

Question 7.
tan (6x + 11)
Solution:
Let y = tan (6x + 11); Diff. both sides w.r.t. x
\(\frac{d y}{d x}\) = sec²(6x +11) \(\frac{d}{d x}\)(6x + 11) = sec² (6x + 11) (6.1 + 0) = 6 sec² (6x + 11)

Question 8.
sin \(\frac{x}{3}\)
Solution:
Let y = sin\(\frac{x}{3}\); diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos\(\frac{x}{3}\) \(\frac{d}{d x}\) \(\frac{x}{3}\) = \(\frac{1}{3}\) cos \(\frac{x}{3}\)

Question 9.
sec mx
Solution:
Let y = sec mx ; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sec mx) = sec mx tan mx \(\frac{d}{d x}\) (mx) = m sec mx tan mx

Question 10.
Solution:
Let y = sec \(\left(\frac{x}{2}-1\right)\); Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sec\(\left(\frac{x}{2}-1\right)\) = sec\(\left(\frac{x}{2}-1\right)\) tan \(\left(\frac{x}{2}-1\right)\) \(\frac{d}{d x}\) \(\left(\frac{x}{2}-1\right)\) = \(\frac{1}{2}\) sec \(\left(\frac{x}{2}-1\right)\) tan \(\left(\frac{x}{2}-1\right)\)

Question 11.
cosec \(\frac{2}{3}\)x
Solution:
Let y = cosec \(\frac{2}{3}\)x ; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = – cot\(\frac{2}{3}\) x cosec \(\frac{2}{3}\)x \(\frac{d}{dx}\) \(\left(\frac{2}{3} x\right)\) = –\(\frac{2}{3}\) cot \(\frac{2}{3}\) x cosec \(\frac{2x}{3}\)

Question 12.
x sin x
Solution:
Let y = x sin x; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x sin x) = x\(\frac{d}{d x}\) sin x + sin x \(\frac{d}{d x}\)(x)
[∵ \(\frac{d}{d x}\) (uv) = u\(\frac{dv}{d x}\) + v \(\frac{du}{d x}\)]
= x cos x + sin x . 1 = x cos x + sin x

Question 13.
x² cos 5x
Solution:
Let y = x² cos 5x; Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = x² \(\frac{d}{d x}\) cos 5x + cos 5x \(\frac{d}{d x}\)x²
[∵ \(\frac{d}{d x}\) (uv) = u\(\frac{dv}{d x}\) + v \(\frac{du}{d x}\)]
= x² (-sin 5x) \(\frac{d}{d x}\) (5x) + cos 5x . 2x = – 5x² sin 5x + 2x cos 5x

Question 14.
\(\sqrt{x}\) cosec (5x + 7)
Solution:
Let y = \(\sqrt{x}\) cosec (5x + 7); Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\sqrt{x}\) \(\frac{d}{d x}\)cosec (5x + 7) + cosec (5x + 7) \(\frac{d}{d x}\) = \(\sqrt{x}\)
= \(\sqrt{x}\) {- cot (5x + 7) cosec (5x + 7)} \(\frac{d}{d x}\) (5x + 7) + cosec (5x + 7) \(\frac{1}{2} x^{\frac{1}{2}-1}\)
= -5\(\sqrt{x}\) cot (5x + 7) cosec (5x + 7) + \(\frac{1}{2 \sqrt{x}}\) cosec (5x + 7)

Question 15.
\(\frac{\sin 3 x}{x-6}\)
Solution:
Let y = \(\frac{\sin 3 x}{x-6}\); Diff. both sides w.r.t. x, we have

Question 16.
\(\frac{\cos x}{5 x}\)
Solution:

Question 17.
\(\frac{\tan x}{2 x+3}\)
Solution:
Let y = \(\frac{\tan x}{2 x+3}\); Diff. both sides w.r.t. x, we have

Question 18.
\(\frac{\sec (a x-b)}{x^2-2}\)
Solution:
Let y = \(\frac{\sec (a x-b)}{x^2-2}\); diff. both sides w.r.t. x, we have

Question 19.
sin 2x
Solution:
Let y = sin 2x; Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) sin 2x = cos 2x \(\frac{d}{d x}\) (2x) = 2 cos 2x

Question 20.
cos 3x
Solution:
Let y = cos 3x; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (cos 3x) = – sin 3x\(\frac{d }{d x}\) (3x) = – 3 sin 3x

Question 21.
tan 2x
Solution:
Let y = tan 2x; diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (tan 2x) = sec² 2x\(\frac{d }{d x}\) (2x) = 2 sec² 2x

Question 22.
sin\(\frac{x}{2}\)
Solution:
Let y = sin\(\frac{x}{2}\); diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) sin \(\frac{x}{2}\) = cos\(\frac{x}{2}\) \(\frac{d}{dx}\) \(\frac{x}{2}\) = \(\frac{1}{2}\) cos \(\frac{x}{2}\)

Question 23.
sec ax
Solution:
Let y = sec ax ; diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) (sec ax) = sec ax tan ax \(\frac{d }{d x}\)(ax) = a sec ax tan ax

Question 24.
sec (px + q)
Solution:
Let y = sec (px + q); diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) = sec(px + q) = sec (px + q) tan (px + q) \(\frac{d }{d x}\) (px + q) = p sec (px + q) tan (px + q)

Question 25.
tan (4x – 7)
Solution:
Let y = tan (4x – 7) ; diff. both sides w.r.t. x,
\(\frac{d y}{d x}\) = \(\frac{d }{d x}\) tan(4x – 7) = sec²(4x – 7) \(\frac{d }{d x}\)(4x – 7)
= sec² (4x – 7) (4 × 1 – 0) = 4 sec² (4x – 7)

Regular engagement with ISC OP Malhotra Solutions Class 11 Chapter 19 Differentiation Ex 19(c) can boost students’ confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(c)

Question 1.
(ax + b) (cx + d)
Solution:
Let y = (ax + b) (cx + d)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = (ax + b)\(\frac { d }{ dx }\)(cx + d) + (cx + d)\(\frac { d }{ dx }\)(ax + b)
[∵ \(\frac { d }{ dx }\) (uv) = u\(\frac { dv }{ dx }\) + v\(\frac { du }{ dx }\)]
= (ax + b) (c1 + 0) + (cx + d) (a1 + 0)
= c (ax + b) + a (ax + d)

Question 2.
(x¹⁰⁰ + 2x⁵⁰ – 3) (7x⁸ + 20x + 5)
Solution:
Let y = (x¹⁰⁰ + 2x⁵⁰ – 3) (7x⁸ + 20x + 5)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = (x¹⁰⁰ + 2x⁵⁰ – 3) \(\frac { d }{ dx }\) (7x⁸ + 20x + 5) + (7x⁸ + 20x + 5) \(\frac { d }{ dx }\) (x¹⁰⁰ + 2x⁵⁰ – 3)
= (x¹⁰⁰ + 2x⁵⁰ – 3) (56x⁷ + 20) + (7x⁸ + 20x + 5) (100x⁹⁹ + 100x⁴⁹)

Question 3.
x (2x – 1)(x + 2)
Solution:
Let y = x (2x – 1) (x + 2); Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = x(2x – 1)\(\frac { d }{ dx }\)(x + 2) + x (x + 2) \(\frac { d }{ dx }\)(2x – 1) (x + 2)\(\frac { d }{ dx }\)x
[∵ \(\frac { d }{ dx }\) (uvw) = uw\(\frac { dw }{ dx }\) + uw\(\frac { dv }{ dx }\) + uw\(\frac { du }{ dx }\)]
= x (2x – 1) + 2x (x + 2) + (2x – 1) (x + 2)

Question 4.
(x – 2) (x + 3) (2x + 5)
Solution:
Let y = (x – 2) (x + 3) (2x + 5)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = (x – 2) (x + 3) \(\frac { d }{ dx }\) (2x + 5) + (x – 2) (2x + 5) \(\frac { d }{ dx }\)(x + 3) + (x + 3) (2x + 5) \(\frac { d }{ dx }\)(x – 2)
= 2 (x – 2) (x + 3) + (x – 2) (2x + 5) + (x + 3) (2x + 5) [using \(\frac { d }{ dx }\) (uvw) = uv\(\frac { dw }{ dx }\) + uw\(\frac { du }{ dx }\) + uw\(\frac { du }{ dx }\)]

Question 5.
y = \(\frac{2 x+5}{3 x-2}\)
Solution:
Let y = \(\frac{2 x+5}{3 x-2}\); Diff. both sides w.r.t. x; we have

Question 6.
y = \(\frac{x^2-3}{x+4}\)
Solution:
Let y = \(\frac{x^2-3}{x+4}\); diff. both sides w.r.t. x; we have

Question 7.
y = \(\frac{2 x-3}{3 x+4}\)
Solution:
Let y = \(\frac{2 x-3}{3 x+4}\); diff. both sides w.r.t. x; we have

Question 8.
y = \(\frac{x^5-x+2}{x^3+7}\)
Solution:
Let y = \(\frac{x^5-x+2}{x^3+7}\); diff. both sides w.r.t. x; we have

Question 9.
s = t² (t + 1)^-1
Solution:
Let s = t² (t + 1)^-1 = \(\frac{t^2}{t+1}\); diff. both sides w.r.t. x; we have

Question 10.
z = \(\frac{u}{u^2+1}\)
Solution:
given z = \(\frac{u}{u^2+1}\); diff. both sides w.r.t. x; we have

Question 11.
y = \(\frac{x^2+2 x+5}{x^3+2 x+4}\)
Solution:
Given y = \(\frac{x^2+2 x+5}{x^3+2 x+4}\); diff. both sides w.r.t. x; we have

Question 12.
f(x) = \(\frac{x^3+2 x}{x^2+4}\)
Solution:
Given f(x) = \(\frac{x^3+2 x}{x^2+4}\); diff. both sides w.r.t. x, we have

Question 13.
If f(x) = \(\frac{x+2}{x-2}\) for all x ≠ 2, find f ‘ ( – 2).
Solution:
Given f(x) = \(\frac{x+2}{x-2}\), x ≠ 2; Diff. both sides w.r.t. x, we have

Question 14.
Differentiate \(\frac{x+2}{x^2-3}\) and find the value of the derivative at x = 0.
Solution:
Given f(x) = \(\frac{x+2}{x^2-3}\); Diff. both sides w.r.t. x, we have

Question 15.
If y = \(\frac{x}{x+a}\), prove that x\(\frac{dy}{dx}\) = y (1 – y).
Solution:
Given y = \(\frac{x}{x+a}\); Diff. both sides w.r.t. x, we have

Question 16.
If \(x \sqrt{1+y}+y \sqrt{1+x}=0\) = 0, prove that \(\frac{d y}{d x}\) = –\(\frac{1}{(1+x)^2}\).
Solution:
Given \(x \sqrt{1+y}+y \sqrt{1+x}=0\) = 0 ⇒ \(x \sqrt{1+y}=-y \sqrt{1+x}\) …(1)
On squaring both sides, we have
x² (1 + y) = y² (1 + x)
⇒ x² + x²y – y² – xy² = 0
⇒(x – y)(x + y) + xy(x – y) = 0
⇒ (x – y) (x + y + xy) = 0
⇒ x + y + xy = 0 [∵ x ≠ y if x = y then given eqn. is meaningless]
⇒ x + y (1 + x) = 0
⇒ y = \(\frac{-x}{1+x}\)
Diff. both sides w.r.t. x, we have

Question 17.
Given that y = \(\sqrt{\frac{1-x}{1+x}}\) show that (1 – x²)\(\frac{dy}{dx}\) + y = 0
Solution:
Given y = \(\sqrt{\frac{1-x}{1+x}}\)
Diff. both sides w.r.t. x, we have

Question 18.
Given that y = (3x – 1)² + (2x – 1)³, find \(\frac{dy}{dx}\) and the points on the curve for which \(\frac{dy}{dx}\) = 0.
Solution:
Given y = (3x – 1)² + (2x – 1)³
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{dy}{dx}\) = 2(3x -1)\(\frac{d}{dx}\) (3x -1) + 3 (2x -1)² \(\frac{d}{dx}\) (2x -1)
= 2 (3x – 1) (3.1 -0) + 3(2x – 1)² (2 × 1 – 0) = 6 (3x – 1) + 6 (2x – 1)²
= 6 [3x – 1 + 4x² – 4x + 1] = 6 [4x² – x] = 6x (4x- 1)
Now \(\frac{dy}{dx}\) = 0 ⇒ 6x (4x – 1) = 0 ⇒ x = 0, \(\frac{1}{4}\)
when x = 0 ∴ from (1); y = (0 – 1)² + (0 – 1)³ = 1 – 1 = 0
When x = \(\frac{1}{4}\) ∴ from (1); y = \(\left(\frac{3}{4}-1\right)^2\) + \(\left(\frac{1}{2}-1\right)^3\) = \(\frac{1}{16}\) – \(\frac{1}{8}\) = \(\frac{-1}{16}\)
Hence the required points on curve are (0, 0) and \(\left(\frac{1}{4},-\frac{1}{16}\right) .\)

Question 19.
(i) If y = \(\frac{x-1}{2 x^2-7 x+5}\), find \(\frac{dy}{dx}\) at x = 2.
(ii) If y = \(\frac{x^2+3}{x^3+2 x}\), find \(\frac{dy}{dx}\) at x = 1.
Solution:
(i) Given y = \(\frac{x-1}{2 x^2-7 x+5}\); Diff. both sides w.r.t. x, we have

(ii) Given y = \(\frac{x^2+3}{x^3+2 x}\)
Diff. both sides w.r.t. x, we have

Question 20.
Find the coordinates of the points on the curve y = \(\frac{x}{1-x^2}\) for which \(\frac{dy}{dx}\) = 1.
Solution:
Given eqn. of curve be y = \(\frac{x}{1-x^2}\)
Diff. both sides w.r.t. x, we have

Now \(\frac{dy}{dx}\) = 1
⇒ \(\frac{1+x^2}{\left(1-x^2\right)^2}\) = 1
⇒ 1 + x² = (1 – x²)²
⇒ 1 + x² = 1 + x⁴ – 2x²
⇒ x⁴ – 3x² = 0
⇒ x² (x² – 3) = 0 ⇒ x = 0, ± \(\sqrt{3}\)
When x = 0 ∴ from (1); y = 0
When x = \(\sqrt{3}\)
∴ from (1); y = \(\frac{\sqrt{3}}{1-3}\) = –\(\frac{\sqrt{3}}{2}\)
When x = –\(\sqrt{3}\)
∴ from (1); y = \(\frac{-\sqrt{3}}{1-3}\) = –\(\frac{\sqrt{3}}{2}\)
Hence the coordinates of required points on given curve are (0, 0);
\(\left(\sqrt{3}, \frac{-\sqrt{3}}{2}\right)\) and \(\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)\).

Peer review of ISC OP Malhotra Solutions Class 11 Chapter 19 Differentiation Ex 19(b) can encourage collaborative learning.

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(b)

Question 1.
(ax)^m + b^m
Solution:
Let y = (ax)^m + b^m = a^m x^m + b^m
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (a^m x^m) + \(\frac { d }{ dx }\) (b^m) = a^m \(\frac { d }{ dx }\) x^m + 0 = ma^m x^m-1

Question 2.
x³ + 4x² + 7x + 2
Solution:
Let y = x³ + 3x² + 7x + 2
Diff both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x^m + \(\frac { d }{ dx }\) 4x² + \(\frac { d }{ dx }\) (7x) + \(\frac { d }{ dx }\) (2) = 3x² + 4\(\frac { d }{ dx }\) x² + 7\(\frac { d }{ dx }\)(x) = 0 = 3x² + 8x + 7

Question 3.
7x⁶ + 8x⁵ – 3x⁴ + 11x² + 6x + 7
Solution:
Let y = 7x⁶ + 8x⁵ – 3x⁴ + 11x² + 6x + 7
diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 7 \(\frac { d }{ dx }\) x⁶ + 8\(\frac { d }{ dx }\)x⁵ – 3 \(\frac { d }{ dx }\) x⁴ + 11 \(\frac { d }{ dx }\) x² + 6\(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) (7)
= 7 × 6x⁵ + 8 × 5x⁴ – 3 × 4x³ + 11 × 2x + 6 + 0
= 42x⁵ + 40x⁴ – 12x³ + 22x + 6

Question 4.
3 + 4x – 7x² – √2x³ + πx⁴ – \(\frac { 2 }{ 5 }\)x⁵ + \(\frac { 4 }{ 3 }\)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) [3 + 4x – 7x² – √2x³ + πx⁴ – \(\frac { 2 }{ 5 }\)x⁵ + \(\frac { 4 }{ 3 }\)]
= \(\frac { d }{ dx }\) (3) + 4\(\frac { d }{ dx }\) (x) – 7\(\frac { d }{ dx }\) x² – √2 \(\frac { d }{ dx }\) x³ + π \(\frac { d }{ dx }\) x⁴ – \(\frac { 2 }{ 5 }\) \(\frac { d }{ dx }\) x⁵ + \(\frac { d }{ dx }\) \(\left(\frac{4}{3}\right)\)
= 0 + 4 – 14x – 3√2 x² + 4πx³ – 2x⁴ + 0
∴ \(\frac { dy }{ dx }\) = 4 – 14x – 3√2x² + 4πx³ – 2x⁴

Question 5.
\(\frac{3}{x^5}\)
Solution:
Let y = \(\frac{3}{x^5}\) = 3x^-5; diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (3x^-5) = \(\frac { d }{ dx }\) x^-5 = 3 (- 5) x^-5-1 = \(\frac{-15}{x^6}\) [∵ \(\frac { d }{ dx }\) xⁿ = nx^n-1]

Question 6.
\(x^{\frac{5}{3}}\)
Solution:
Let y = x^5/3; diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x^5/3 = \(\frac { 5 }{ 3 }\) = \(x^{\frac{5}{3}-1}\) = \(\frac{5}{3} x^{2 / 3}\)

Question 7.
\(\frac{7}{x^{\frac{2}{3}}}\)
Solution:
Let y = \(\frac{7}{x^{2 / 3}}\) = 7x^-2/3; diff. both sides w.r.t. x. we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (7x^-2/3) = 7 \(\frac { d }{ dx }\) x^-2/3 = 7\(\left(-\frac{2}{3}\right) x^{-\frac{2}{3}-1}\) = \(\frac{-14}{3} x^{-5 / 3}\)

Question 8.
\(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2, x \neq 0\)
Solution:
Let y = \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2\) = x + \(\frac { 1 }{ x }\) + 2√x . \(\frac{1}{\sqrt{x}}\)
⇒ y = x + \(\frac { 1 }{ x }\) + 2; Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (x) + \(\frac { d }{ dx }\) \(\left(\frac{1}{x}\right)\) + \(\frac { d }{ dx }\) (2) = 1 – \(\frac{1}{x^2}\)

Question 9.
\(\sqrt{x}-\frac{1}{\sqrt{x}}, x \neq 0\)
Solution:
Let y = \(\sqrt{x}-\frac{1}{\sqrt{x}}\); diff. both sides w.r.t. x, we have
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) \(\sqrt{x}\) – \(\frac{d}{d x} x^{-1 / 2}\) = \(\frac{1}{2} x^{\frac{1}{2}-1}\) – \(\left(\frac{-1}{2}\right) x^{-\frac{1}{2}-1}\)
[∵ \(\frac { d }{ dx }\) xⁿ =nx^n-1]
= \(\frac{1}{2 \sqrt{x}}\) + \(\frac{1}{2 x^{3 / 2}}\)

Question 10.
\(\frac { 1 }{ x }\) + \(\frac{3}{x^2}\) + \(\frac{2}{x^3}\)
Solution:
Let y = \(\frac { 1 }{ x }\) + \(\frac{3}{x^2}\) + \(\frac{2}{x^3}\); diff. both sides w.r.t. x, we have
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) \(\left(\frac{1}{x}\right)\) + 3 \(\frac { d }{ dx }\) x^-2 + 2\(\frac { d }{ dx }\) (x^-3) = -1 x^-1-1 + 3 (-2) x^-2-1 + 2 (- 3) x^-3-1 = – \(\frac{1}{x^2}\) – \(\frac{6}{x^3}\) – \(\frac{6}{x^4}\)

Question 11.
\(2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}}\)
Solution:
Let y = \(2 x^{\frac{1}{2}}+6 x^{\frac{1}{3}}-2 x^{\frac{3}{2}}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 2\(\frac { d }{ dx }\) x^1/2 + 6\(\frac { d }{ dx }\) x^1/3 – 2\(\frac { d }{ dx }\) x^3/2
= 2 × \(\frac{1}{2} x^{\frac{1}{2}-1}\) + 6 × \(\frac{1}{3} x^{\frac{1}{3}-1}\) – 2 × \(\frac{3}{2} x^{\frac{3}{2}-1}\)
= \(\frac{1}{\sqrt{x}}\) + \(\frac{2}{x^{2 / 3}}\) – \(3 \sqrt{x}\)

Question 12.
8x³ – x² + 5 – \(\frac{2}{x}\) + \(\frac{4}{x^3}\)
Solution:
Let y = 8x³ – x² + 5 – \(\frac{2}{x}\) + \(\frac{4}{x^3}\)
Diff both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 8\(\frac { d }{ dx }\) x³ – \(\frac { d }{ dx }\) x² + \(\frac { d }{ dx }\) (5) – 2\(\frac { d }{ dx }\) x^-1 + 4\(\frac { d }{ dx }\) x^-3
= 8 × 3x² – 2x + 0 – 2 (- 1) x^-1-1 + 4 (- 3) x^-3-1
= 24x² – 2x + \(\frac{2}{x^2}\) – \(\frac{12}{x^4}\)

Question 13.
\(\frac{3 x^7+x^5-2 x^4+x-3}{x^4}\)
Solution:
Let y = \(\frac{3 x^7+x^5-2 x^4+x-3}{x^4}\) = 3x³ + x – 2 + \(\frac{1}{x^3}\) – \(\frac{3}{x^4}\)
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 3\(\frac { d }{ dx }\) x³ + \(\frac { d }{ dx }\) x – \(\frac { d }{ dx }\) (2) + \(\frac { d }{ dx }\) x^-3 – 3\(\frac { d }{ dx }\) x^-4
= 3 × 3x² + 1 – 0 – 3x^-3-1 – 3 ( – 4)x^-4-1 = 9x² + 1 – \(\frac{3}{x^4}\) + \(\frac{12}{x^5}\)

Question 14.
(i) (2x – 3)²
(ii) (2x – 3)¹⁰⁰
Solution:
(i) Let y = (2x – 3)² = 4x² – 12x + 9 ; diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 4\(\frac { d }{ dx }\) x² – 12 \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) (9) = 8x – 12

(ii) Let y = (2x – 3)¹⁰⁰ ; diff. both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (2x – 3)¹⁰⁰ = 100 (2x – 3)^{100 – 1} \(\frac { d }{ dx }\) (2x – 3) [∵ \(\frac { d }{ dx }\) xⁿ = nx^n-1]
= 100 (2x – 3)⁹⁹ [2 (1) – 0] = 200 (2x – 3)⁹⁹

Question 15.
\(\sqrt{3 x+2}\)
Solution:
Let y = \(\sqrt{3 x+2}\) ; diff. both sides w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)\((3 x+2)^{\frac{1}{2}}\) = \(\frac{1}{2}(3 x+2)^{\frac{1}{2}-1} \frac{d}{d x}(3 x+2)\)
= \(\frac{1}{2}(3 x+2)^{-\frac{1}{2}}(3+0)\) = \(\frac{3}{2} \frac{1}{\sqrt{3 x+2}}\)
[∵ \(\frac { d }{ dx }\) (ax + b)ⁿ = n (ax + b)^n-1 \(\frac { d }{ dx }\) (ax + b)]

Question 16.
Given f(x) = \(\frac { 7 }{ 4 }\)x², find f ‘ \(\left(\frac{1}{7}\right)\)
Solution:
Given f(x) = \(\frac { 7 }{ 4 }\)x²;diff. both sides w.r.t. x, we have
f ‘ (x) = \(\frac { 7 }{ 4 }\) × 2x = \(\frac { 7 }{ 2 }\)x
⇒ f ‘ \(\left(\frac{1}{7}\right)\) = \(\frac{7}{2}\) × \(\frac{1}{7}\) = \(\frac{1}{2}\)

Question 17.
Find the derivative with respect to x of the following:
(i) x – \(\frac { 1 }{ x }\)
(ii) √x + \(\frac{1}{\sqrt{x}}\)
(iii) 3x² + \(\frac{3}{x^2}\)
(iv) \(\frac{x^2+1}{x}\)
(v) \(\frac{2 x+x^4}{x^2}\)
(vi) \(\frac{1+x^2}{x^3}\)
Solution:
(i) Let y = x – \(\frac { 1 }{ x }\); diff. both sides, w.r.t. x, we get
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\)(x) – \(\frac { d }{ dx }\) x^-1 = 1 – (-1) x^-1-1 = 1 + \(\frac{1}{x^2}\)

(ii) Let y = \(\sqrt{x}+\frac{1}{\sqrt{x}}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x^1/2 + \(\frac { d }{ dx }\) x^-1/2 = \(\frac { 1 }{ 2 }\) \(\frac{1}{2} x^{\frac{1}{2}-1}\) + \(\left(\frac{-1}{2}\right) x^{\frac{-1}{2}-1}\) = \(\frac{1}{2 \sqrt{x}}\) – \(\frac{1}{2 x^{3 / 2}}\)

(iii) Let y = \(3 x^2+\frac{3}{x^2}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = 3\(\frac { d }{ dx }\) x^-2 = 3 × 2x + 3 (-2) x^-2-1 = 6x – \(\frac{6}{x^3}\)

(iv) Let y = \(\frac{x^2+1}{x}\); diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) x^-1 = 1 + (-1)x^-1-1 = 1 – \(\frac{1}{x^2}\)

(v) Let y = \(\frac{2 x+x^4}{x^2}\) = \(\frac{2}{x}\) + x²; diff. both sides w.r.t. x,
∴ \(\frac { dy }{ dx }\) = 2\(\frac { d }{ dx }\) x^-1 + \(\frac { d }{ dx }\) x^-2 = 2 (- 1)x^-1-1 + 2x = \(\frac{-2}{x^2}\) + 2x

(vi) Let = \(\frac{1+x^2}{x^3}\) = \(\frac{1}{x^3}\) + \(\frac{1}{x}\); diff. both sides w.r.t. x
∴ \(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) (x^-3) + \(\frac { d }{ dx }\) x^-1 = 3x^-3-1 + (-1)^-1-1 = –\(\frac{3}{x^4}\) – \(\frac{1}{x^2}\)

Question 18.
If y = x + \(\frac { 1 }{ x }\), prove that x²\(\frac { dy }{ dx }\) – xy + 2 = 0
Solution:
Given y = x + \(\frac { 1 }{ x }\); diff, both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dx }\) x + \(\frac { d }{ dx }\) x^-1 = 1 + (-1)x^-1-1 = 1 – \(\frac{1}{x^2}\)
⇒ x²\(\frac { dy }{ dx }\) = x²\(\frac { dy }{ dx }\) – xy + 2 = x² – 1 – x\(\left(x+\frac{1}{x}\right)\) + 2
= x² – 1 x² – 1 + 2 = 0 = R.H.x.

Question 19.
If y= \(\sqrt{x}\) – \(\frac{1}{\sqrt{x}}\), prove that 2x\(\frac{d y}{d x}\) + y = \(2 \sqrt{x}\).
Solution:

Question 20.
If y = \(\frac{1}{a-z}\), show that \(\frac{d z}{d y}\) = (z – a)²
Solution:
Given y = \(\frac{1}{a-z}\)
diff. both sides w.r.t. z, we have
\(\frac { dy }{ dz }\) = \(\frac { d }{ dz }\)\(\left(\frac{1}{a-z}\right)\) = \(\frac { d }{ dz }\)(a – z)^-1 = (-1)(a – z)^-1-1 \(\frac { d }{ dz }\) (a – z)
= 1 (a – z)^-2 (0 – 1) = \(\frac{1}{(a-z)^2}\)
∴ \(\frac { dz }{ dy }\) = \(\frac{1}{\frac{d y}{d z}}\) = (a – z)²

Question 21.
If y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …… to ∞, show that \(\frac { dy }{ dx }\) = y.
Solution:
Given y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ …(1)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 0 + 1 + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ …(1)
= 1 + x + \(\frac{x^2}{2}\) + \(\frac{x^3}{6}\) …. ∞
= 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + ….. ∞ = y
Thus \(\frac { dy }{ dx }\) = y
Aliter = Given y = 1 + x + \(\frac{x^2}{2 !}\) + \(\frac{x^3}{3 !}\) + \(\frac{x^4}{4 !}\) + …. ∞ = e^x
Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = \(\frac { d }{ dy }\) e^x = e^x = y

The availability of ISC OP Malhotra Solutions Class 11 Chapter 19 Differentiation Ex 19(a) encourages students to tackle difficult exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(a)

Question 1.
2x
Solution:
Let y = 2x …(1)
Let δx be the increment in x and corresponding increment in y be δy
∴ y + δy = 2 (x + δx) …(2)
subtracting eqn. (1) from eqn. (2); we get
δy = 2δx ; On dividing both sides by δx
∴ \(\frac{\delta y}{\delta x}\) = 2, Taking limits as δx → 0
Thus, \(\underset{\delta x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\delta y}{\delta x}\) = \(\frac{d y}{d x}\) = \(\underset{\delta x \rightarrow 0}{\mathrm{Lt}}\) = 2 = 2
∴ \(\frac{d }{d x}\)(2x) = 2

Question 2.
(x – 1)²
Solution:
Let y = f(x) = (x – 1)²
∴ f(x + δx) = (x + δx – 1)²
Thus by first principle, we have

Question 3.
x³
Solution:
Let y = f(x) = x³
∴ f (x + δx) = (x + δx)³
Then by first principle, we have

Question 4.
\(\frac{1}{\sqrt{x}}\)
Solution:

Question 5.
\(\sqrt{x+1}\); x > – 1
Solution:
Let y = f(x) = \(\sqrt{x+1}\) ∴ f(x + δx) = \(\sqrt{x+\delta x+1}\)
Then by first principle, we have

Question 6.
\(\frac{2 x+3}{3 x+2}\)
Solution:
Let y = f(x) = \(\frac{2 x+3}{3 x+2}\)
∴ f(x + δx) = \(\frac{2(x+\delta x)+3}{3(x+\delta x)+2}\)
Then by first principle, we have

Question 7.
\(\frac{1}{\sqrt{x+a}}\)
Solution:
Given y = f(x) = \(\frac{1}{\sqrt{x+a}}\)
∴f(x + δx) = \(\frac{1}{\sqrt{x+\delta x+a}}\)
Then by first principle, we have

Question 8.
x + \(\frac{1}{x}\)
Solution:
Let y = f(x) = x + \(\frac{1}{x}\)
∴f(x + δx) = (x + δx) + \(\frac{1}{x+δx}\)
Then by first principle, we have

Question 9.
\(\frac{1}{\sqrt{2 x+3}}\)
Solution:
Let y = f(x) = \(\frac{1}{\sqrt{2 x+3}}\)
∴ f (x + δx) = \(\frac{1}{\sqrt{2(x+\delta x)+3}}\)
Then by first principle, we have

Question 10.
\(\frac{1}{x^{\frac{3}{2}}}\)
Solution:
Let y = f(x) = \(\frac{1}{x^{\frac{3}{2}}}\)
∴ f(x + δx) = \(\frac{1}{(x+\delta x)^{3 / 2}}\)
Then by first principle, we have

Question 11.
(x + 1) (2x – 3)
Solution:
Let y = f(x) = (x + 1) (2x – 3)
∴f(x + δx) = (x + δx + 1) (2x – 3 + 2 δx)
Then by first principle, we have

Question 12.
\(\frac{x^2+1}{x}\)
Solution:
Let y = f(x) = \(\frac{x^2+1}{x}\) = x + \(\frac { 1 }{ x }\)
∴f(x + δx) = (x + δx) + \(\frac{1}{x+\delta x}\)
Then by first principle, we have

Parents can use S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Ex 21(b) to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(b)

Question 1.
(i) Five students secured marks as ; 8, 10, 15, 30, 22. Find the standard deviation.
(ii) For a set of ungrouped values the following sums are found :
n = 15, Σx = 480, Σx² = 15735.
Find the standard deviation.
(iii) The standard deviation of the numbers 2, 3, 11, x is 3 1/2. Calculate the values of x.
Solution:
(i)

(iii)

Question 2.
Calculate the standard deviation and variance for the integers 11,12,13,…, 20.
Solution:
Given observations are 11, 12, 13, …., 20

Question 3.
Find the standard deviation of the following set of numbers :
25, 50, 45, 30, 70, 42, 36, 48, 34, 60
Solution:

Question 4.
Calculate the possible values of x, if the standard deviation of the numbers 2, 3, 2x and 11 is 3.5.
Solution:

Question 5.
Calculate the standard deviation for the following distribution :

Solution:
The table of values is given as under:

Then by direct method, we have

Question 6.
Calculate the standard deviation of the following data :

Solution:

Then by short cut method, we have

Question 7.
Calculate the standard deviation of the following data:

Solution:

By short cut method, we have

Question 8.
Calculate the standard deviation for the following data giving the age distribution of persons.

Solution:
Same age be continuous variable

Then by step deviation method, we have

Question 9.
The heights, to the nearest cm, of 30 men are given below :

Using class intervals 155-160, 160-165,… draw up a grouped frequency distribution and use this to estimate the Arithmetic mean and standard deviation.
Solution:
Table of values for given data is given as under :

Question 10.
Find the mean and the standard deviation from the following :

Solution:

Question 11.
The following table shows the I.Q. of 480 school children. Find
(i) the mean.
(ii) the standard deviation using the step deviation method. Use Chrlier’s check to verify the computation of the standard deviation.

Solution:
The table of values is given as under :

Question 12.
In a certain test, the 30 scores were grouped as follows :

Calculate the mean and the standard deviation:
Solution:
The table of values is given as under:

Then by step deviation method, we have

Question 13.
The number of faults on the surface of each of 1000 tiles were distributed as follows :

Calculate the mean and the standard deviation.
Solution:
The table of values is given as under :

Question 14.
The mean and the standard deviation of 25 observations and 60 and 3. Later on it was decided to omit an observation which was incorrectly recorded as 50. Calculate the mean and the standard deviation of the remaining 24 observations.
Solution:

Question 15.
The scores of two golfers for 10 rounds each are:

Which may be regarded as the more consistent player ?
Solution:
For golfer A ; no. of observations = 10
and Sum of all observations = 58 + 59 + 60 + 54 + 65 + 66 + 52 + 75 + 69 + 52 = 610
∴Mean \(\overline{\mathrm{x}}\) = \(\frac { Sum of all observations }{ 10 }\) = \(\frac { 610 }{ 10 }\) = 61
For golfer B : Sum of all observations = 84 + 56 + 92 + 65 + 86 + 78 + 44 + 54 + 78 + 68 = 705
∴Mean \(\overline{\mathrm{Y}}\) = \(\frac { Sum of all observations }{ 10 }\) = \(\frac { 705 }{ 10 }\) = 70.5
We construct the table of values is given as under :

Question 16.
Goals scored by two teams A and B in a football season were as follows :

By calculating the coefficient of variation in each case find which team may be considered more consistent.
Solution:
The table of values is given as under :

Question 17.
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b ?
(a) a = 0, b = 7
(b) a = 5, b = 2
(c) a = l, b = 6
(d) a = 3, b = 4
Solution:
Given observations are a, b, 8, 5 and 10 ∴ no. of observations = n = 5, \(\overline{\mathrm{x}}\) = 6, \(\sigma^2\) = 6.80
∴ Mean = \(\frac { Sum of all given observations }{ n }\) ⇒ 6 = \(\frac { a+b+8+5+10 }{ 5 }\)
⇒ 30 = a + b + 23
⇒ a + b = 7

From (1) and (2); we have
a² + (7 – a)² = 25
⇒ 2a² – 10a + 24 = 0
⇒ a² – 7a + 12 = 0
⇒ (a – 3)(a – 4) = 0
⇒ a = 3, 4
When a = 3 ∴ from (1); b = 4
When a = 4 ∴ from (1); b = 3
∴ Ans. (d)

Students can track their progress and improvement through regular use of S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Chapter Test.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Chapter Test

Question 1.
Find the mean deviation from the mean for the following data :
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Mean \(\bar{x}\) = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\) = \(\frac{500}{10}\) = 50

∴ M.D about mean = \(\frac{\Sigma\left|x_i-\bar{x}\right|}{n}\) = \(\frac{84}{10}\) = 8.4

Question 2.
Find the mean deviation from the mean for the following data:

Solution:

By direct method, Mean \(\bar{x}\) = \( \frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{528}{66}\) = 8
Thus M.D from mean = \(\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}\) = \(\frac{138}{66}\) = 2.09

Question 3.
Find the mean deviation for the mean for the following data:

Solution:

Thus by direct method, Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{1350}{50}\) = 27
∴ M.D about mean = \(\frac{\Sigma f_i\left|x_i-27\right|}{\Sigma f_i}\) = \(\frac{512}{50}\) = 10.24

Question 4.
Find the mean deviation about the median for the following data :
11, 3, 8, 7, 5, 14, 10, 2, 9
Solution:
Arranging the given data in ascending order ; we have
2, 3, 5, 7, 8, 9, 10, 11, 14
Here no. of observations = n = 9 (odd)
∴ M_d = \(\left(\frac{n+1}{2}\right)\)th observation = \(\left(\frac{9+1}{2}\right)\)th obs = 5th obs = 8

∴ M.D about Median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{27}{9}\) = 3

Question 5.
Find the variance and standard deviation of the following data :

Solution:
The table of values is given as under:

∴ Variance = \(\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2\) = \(\frac{728}{22}\) – \(\left(\frac{44}{22}\right)^2\) = 33.090909 – 4 = 29.09
and S.D = \(\sqrt{\text { Variance }}\) = \(\sqrt{29.09}\) = 5.3935

Question 6.
Calculate the mean and variance after the following data :

Solution:
The table of values is given as under:

Well-structured S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Ex 21(a) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(a)

Question 1.
15, 17, 19, 25, 30, 35, 48
Solution:
The table of values is given as under :

Question 2.
21, 23, 25, 28,30, 32, 38, 39, 46, 48
Solution:
Here Mean \(\vec{x}\) = \(\frac{21+23+25+28+30+32+38+39+46+48}{10}\) = \(\frac{330}{10}\) = 3
The table of values is given as under:

∴ M.D. about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{78}{10}\) = 7.8 and coeff. of M.D. = \(\frac{\text { M.D }}{\bar{x}}\) = \(\frac{7.8}{33}\) = 0.236

Question 3.
10, 70, 50, 53, 20, 95, 55, 42, 60, 48, 80
Calculate the mean deviation from the mean for the following frequency distributions.
Solution:
Mean \(\bar{x}\) = \(\frac{10+70+50+53+20+95+55+42+60+48+80}{11}\) = \(\frac { 583 }{ 11 }\) = 53

∴ required M.D about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{190}{11}\) = 17.27 and coeff. of M.D = \(\frac{\text { M.D }}{\text { Mean }}\) = \(\frac{190}{11 \times 53}\) = 0.326

Question 4.

Solution:

∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{660}{44}\) = 15
Thus M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{312}{44}\) = 7.09

Question 5.

Solution:
The table of values is given as under :

∴ \(\bar{x}\) = \(\frac { 576 }{ 48 }\) = 12
∴ M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{48}\) = \(\frac{36}{48}\) = \(\frac{3}{4}\) = 0.75

Question 6.

Solution:
We construct the table of values is given as under :

∴ by direct method, mean \(\bar{x}\) = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac{1350}{50}\) = 27
Thus, M.D from mean = \(\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}\) = \(\frac{472}{50}\) = 9.44

Question 7.

Solution:
The table of values is given as under:

∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{8560}{50}\) = 171.2
Thus M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{528}{50}\) = 10.56

Question 8.

Solution:

∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{9200}{182}\) = 50.55
∴ M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{2390.6}{182}\) = 13.135

Question 9.
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Solution:
Arranging the data in ascending order, we get 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Here no. of observations = n = 11 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 6th observation = 9

∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{58}{11}\) = 5.273
and Coefficient of M.D = \(\frac{\text { M.D }}{\text { Median }}\) = \(\frac{58}{11 \times 9}\) = 0.586

Question 10.
100, 150, 200, 250,360, 490, 500, 600, 671
Solution:
Data given is already in ascending order we have, no. of observations = n = 9 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 5th observation = 360

∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{1561}{9}\) = 173.44
and Coeff. of M.D = \(\frac{\text { M.D }}{\text { median }}\) = \(\frac{1561}{9 \times 360}\) = 0.48

Question 11.

Solution:
The table of values is given as under :

Question 12.

Solution:
The table of values is given as under :

Practicing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Chapter Test is the ultimate need for students who intend to score good marks in examinations.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Chapter Test

Question 1.
In rolling two fair dice, what is the probability of obtaining a sum greater than 3 but not exceeding 6 ?
Solution:
In rolling two dice, total no. of outcomes = n (S) = 6² = 36
Let A : event of obtaining a sum greater than 3 but not exceeding 6 = {(1,3), (2, 2), (3, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1,5), (2, 4), (4, 2), (3,3), (5,1)}
∴ n( A) =12
Thus reqd/ probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\) = \(\frac{1}{3}\)

Question 2.
Find the probability of obtaining a sum 8 in a single throw of two dice.
Solution:
In a single throw of two dice
Then total no. of outcomes = n (S) = 6² = 36
Let A : obtaining a sum 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
∴ n (A) = 5
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{36}\)

Question 3.
A bag contains 4 red, 6 white and 5 black balls. 2 balls are drawn at random. Find the probability of getting one red and one Whitehall.
Solution:
Given no. of red balls = 4, no. of white balls = 6 and no. of black balls = 5
Total no. of balls = 4 + 6 + 5 = 15
∴ n (S) = Total no. of ways of drawing 2 balls out of 15 = ¹⁵C₂
Let A : getting one red and one white ball
∴ n (A) = Total no. of ways of drawing one red ball out of 4 and 1 white ball out of
6 = ⁴C₁ × ⁶C₁
Thus required probability
= P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{{ }^4 C_1 \times{ }^6 C_1}{{ }^{15} C_2}\) = \(\frac{\frac{4 \times 6}{15 \times 14}}{2}\) = \(\frac{24}{105}\) = \(\frac{8}{35}\)

Question 4.
Out of 26 cards numbered from 1 to 26, one card is chosen. Find the probability that it is not divisible by 4.
Solution:
Total no. of outcomes = n (S) = 26
Let A : number on card is divisible by 4 = {4, 8, 12, 16, 20, 24}
∴ n (A) = 6
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{6}{26}\) = \(\frac{3}{13}\)
∴ required probability that chosen card bear a number which is not divisible by 4
= \(P(\bar{A})\) = 1 – P (A) = 1 – \(\frac{3}{13}\) = \(\frac{10}{13}\)

Question 5.
If A and B are mutually exclusive events with \(\frac{1}{2}\) = P (B) and A ∪ B = S, the sample space. Find P (A).
Solution:
Given A and B are mutually exclusive events
∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
Given \(\frac{1}{2}\) = P (B)
Also S = A ∪ B ⇒ P(S) = P (A ∪ B) ⇒ P (A ∪ B) = 1 = P (A) + P (B) – P (A ∩ B)
⇒ 1 = P (A) + \(\frac{1}{2}\) – 0 ⇒ P (A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)

Question 6.
A and B are two events, such that
P (A) = 0.42, P (B) = 0.48, P (A and B) = 0.16 Determine (i) P (not A), (ii) P (not B), (iii) P (A or B)
Solution:
Given P (A) = 0.42 ; P (B) = 0.48 ; P (A and B) = P (A ∩ B) = 0.16
(i) P(not A) = \(P(\bar{A})\) = 1 – P (A) = 1 – 0.42 = 0.58
(ii) P (not B) = \(P(\bar{B})\) = l – P (B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P (A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.9 – 0.16 = 0.74

Accessing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(f) can be a valuable tool for students seeking extra practice.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(f)

Question 1.
(i) In a single throw of two coins, find the probability of getting both heads or both tails.
(ii) A dice is thrown twice. Find the probability that the sum of the two numbers obtain is 5 or 7 ?
(iii) Two dice are tossed once. Find the probability of getting an even number on first die or a total of 8.
Solution:
(i) In a single throw of two coins
Then sample space S = {HH, HT, TH, TT}
∴ Total no. of exhaustive cases = 4
P (both heads) = \(\frac { 1 }{ 4 }\) {HH}
P (both tails) = \(\frac { 1 }{ 4 }\) {TT}
Thus required probability = P (both heads) + P (tails) = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(ii) Here total no. of outcomes = 6² = 36 = n (S)
Let A : sum of two numbers be 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} ∴ n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{4}{36}\)
B : event that sum of numbers be 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ∴ n (B) = 6
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{6}{36}\)
∴ required probability of getting either sum 5 or 7 = P (A or B) = P (A) + P (B) = \(\frac { 4 }{ 36 }\) + \(\frac { 6 }{ 36 }\) = \(\frac { 10 }{ 36 }\) = \(\frac { 5 }{ 18 }\)
[since A and B are mutually exclusive events as A ∩ B = Φ) P (A ∩ B) = 0]

(iii) Here, total no. of exhaustive cases = n (S) = 6² = 36
Let A : event of getting an even number on first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 18

Question 2.
If the probability of a horse A winning a race is \(\frac { 1 }{ 5 }\) and the probability of horse B winning the same race is \(\frac { 1 }{ 4 }\), what is the probability that one of the horses will win ?
Solution:
Given P (horse A winning a race) = \(\frac { 1 }{ 5 }\); P (horse B winning a race) = \(\frac { 1 }{ 4 }\)
∴ required probability = P (A) + P(B) = \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 9 }{ 20 }\)

Question 3.
In a single throw of two dice, what is the probability of obtaining a total of 9 or 11 ?
Solution:
In a single throw of two dice
Then total no. of exhaustive cases = 6² = 36 = n (S)
Let A : event of obtaining a total of 9 = {(3, 6), (4, 5), (5,4), (6, 3)} ∴n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 26 }\)
Let B : obtaining a total of 11 = {(5, 6), (6, 5)} ∴ n(B) = 2
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 36 }\)
Here A ∩ B = Φ ⇒ P(A ∩ B) = 0
Thus required probability of obtaining a total of 9 or 11 = P (A or B)
= P (A) + P (B) – P (A ∩ B) = \(\frac { 4 }{ 36 }\) + \(\frac { 2 }{ 36 }\) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 4.
In a group there are 2 men and 3 women. 3 persons are selected at random from the group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected.
Solution:
Total no. of persons = 2 + 3 = 5
∴ Total no. of ways of selecting 3 persons out of 5 = ⁵C₃
Let A : event that 1 man and 2 women are selected ∴ n(A) = ²C₁ × ³C₂

Question 5.
In a class of 25 students with roll numbers 1 to 25, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of 5 or 7 ?
Solution:
Total exhaustive cases = 25 = n (S)
Let A : event that roll no. of selected student be a multiple of 5 = {5, 10, 15, 20, 25}
B : event that roll no. of selected student be a multiple of 7 = {7, 14,21}
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{25}\); P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{3}{25}\)
Thus required probability = P(A or B) = P(A) + P(B) = \(\frac { 5 }{ 25 }\) + \(\frac { 3 }{ 25 }\) = \(\frac { 8 }{ 25 }\)

Question 6.
If chance of A, winning a certain race be \(\frac { 1 }{ 6 }\) and the chance of B winning it is \(\frac { 1 }{ 3 }\), what is the chance that neither should win ?
Solution:
Given P(A’s Winning) = \(\frac { 1 }{ 6 }\), P(B’s Winning) = \(\frac { 1 }{ 3 }\)
∴ P (neither A nor B wins)

Question 7.
Discuss and criticise the following : P(A) = \(\frac { 2 }{ 3 }\), P(B) = \(\frac { 1 }{ 4 }\), P (C) = \(\frac { 1 }{ 3 }\) are the probabilities of three mutually exclusive events A, B and C.
Solution:
Since A, B and C are mutually exclusive events.
Now P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 5 }{ 4 }\) > 1
But P(A ∪ B ∪ C) ≤ 1 Thus, given statement is false.

Question 8.
E and F are two events associated with a random experiment for which P (F) = 0.35, P (E or F) = 0.85, P (E and F) = 0.15. Find P (E).
Solution;
Given P (F) = 0.35 ; P (E or F) = 0.85 and P (E and F) = 0.15
We know that P (E ∪ F) = P (E) + P (F) – P (E ∩ F)
⇒ 0.85 = P(E) + 0.35 – 0.15
⇒ P (E) = 0.85 – 0.20 = 0.65

Question 9.
(i) Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Find the probability that neither A nor B occurs.
(ii) The probability of an event A occurring is 0.5 and of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
(i) Given P (A) = 0.25 ; P (B) = 0.50 and P (A ∩ B) = 0.14
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – [0.25 + 0.50 – 0.14] = 1 – 0.61 = 0.39

(ii) Given P (A) = 0.5 ; P (B) = 0.3
Since A and B are mutually exclusive events. ∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – 0.5 – 0.3 – 0 = 0.2

Question 10.
(i) A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is event ?
(ii) A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Solution:
(i) Total no. of ways of drawing 2 tickets from 25 tickets = ²⁵C₂
The product of numbers is even when either both are even or one is even and the other is odd
Let A : event that both are even and there are 12 even numbers from 1 to 25
∴ n (A) = Total no. of ways of drawing 2 tickets out of 12 = ¹²C₂
Let B : event that one is even and other is odd since there are 12 even and 13 odd numbers from 1 to 25.

(ii) Total no. of balls = 7 + 5 + 4 = 16
n (S) = Total no. of ways of drawing four balls out of 16 without replacement = ¹⁶C₄
Drawing atleast three black balls means drawing 3 black balls out of 5 and one ball from remaining 11 balls or 4 black balls out of 5.
Thus required probability = \(\frac{{ }^5 C_3 \times{ }^{11} C_1+{ }^5 C_4}{{ }^{16} C_4}\) = \(\frac{\frac{5 \times 4}{2} \times 11+5}{\frac{16 \times 15 \times 14 \times 13}{24}}\) = \(\frac { 23 }{ 364 }\)

Question 11.
(i) A and B are two mutually exclusive events of an experiment.
If P(not A) = 0.75, P(A ∪ B) = 0.65 and P(B) = p, find the value of p.
(ii) A and B are two mutually exclusive events. If P (A) = 0.5 and P \((\overline{\mathbf{B}})\) = 0.6, find P (A or B).
Solution:
(i) Given A and B are mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (not A) = 0.65 ⇒ 1 – P (A) = 0.65 ⇒ P (A) = 0.35
P(A ∪ B) = 0.65 and P (B) =p
We know that P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.65 = 0.35 + p – 0 ⇒ p = 0.30

(ii) Given A and B are two mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (A) = 0.5 ; P\((\overline{\mathbf{B}})\) = 0.6 ⇒ 1 – P (B) = 0.6 ⇒ P (B) = 0.4
∴ P (A or B) = P (A) + P (B) – P (A ∩ B) = 0.5 + 0.4 – 0 = 0.9

Question 12.
(i) A, B and C are three mutually exclusive and exhaustive events associated with a random experiment. Find P (A) given that P (B) = \(\frac { 3 }{ 2 }\) P (A) and P (C) = \(\frac { 1 }{ 2 }\) P (B).
Solution:
(i) Given A, B and C are mutually exclusive, exhaustive events
∴ P(A) + P(B) + P(C) = 1 …(1)
Given P(B) = \(\frac { 3 }{ 2 }\) P(A) and P (C) =\(\frac { 1 }{ 2 }\) P(B) = \(\frac { 1 }{ 2 }\) × \(\frac { 3 }{ 2 }\)P(A) = \(\frac { 3 }{ 4 }\) P (A)
∴ from (1); P(A) + \(\frac { 3 }{ 2 }\)P (A) + \(\frac { 3 }{ 4 }\) P(A) = 1
\(\left[1+\frac{3}{2}+\frac{3}{4}\right] \mathrm{P}(\mathrm{A})=1\)
\(\Rightarrow\left(\frac{4+6+3}{4}\right) P(A)=1\)
⇒ P (A) = \(\frac { 4 }{ 13 }\)

(ii) Since A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
Given P (A) = 2 P (B) = 3 P (C) …(1)

Question 13.
(i) In a single throw of two dice, find the probability that neither a doublet nor a total of a 10 will appear.
(ii) Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.
(iii) Two dice are thrown together ; what is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 5.
Solution:
(i) In a single throw of two dice, total exhaustive cases = 6² = 36
Let A: getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ∴ n (A) = 6
B : getting a total of 10 = {(4, 6), (5, 5), (6, 4)} ∴n (B) = 3
Thus P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 6 }{ 36 }\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 36 }\)
and A ∩ B = {(5, 5)} ∴ n (A ∩ B) = 1
Thus P (A ∩ B) = \(\frac{n(\mathrm{~A} \cup \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac { 6 }{ 36 }\) + \(\frac { 3 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\)
Thus, required probability = P (neither A nor B) = \(\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\) = \(P(\overline{A \cup B})\)

(ii) Total exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers obtained on two dice is multiple of 3
= {(U 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A)= 12
Let B = event that sum of the numbers on two dice is multiple of 4
= {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)} ∴ n (B) = 9
∴ A ∩ B = {(6, 6)} ⇒ n (A ∩ B) = 1
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{9}{36}\) and P (A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P(A ∪ B) = P (A) + P (B) = P (A ∩ B) = \(\frac { 12 }{ 36 }\) + \(\frac { 9 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)
∴ required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A u B) = 1 – \(\frac { 5 }{ 9 }\) = \(\frac { 4 }{ 9 }\)

(iii) When two dice are thrown together
Then total no. of exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers on the two faces is divisible by 3
= {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A) = 12
Let B : event that the sum of numbers on two faces is divisible by 5
= {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)} ∴ n (B) = 7
and A ∩ B = Φ ⇒ P (A ∩ B) = 0
Thus, P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – 0 = \(\frac { 12 }{ 36 }\) + \(\frac { 7 }{ 36 }\) = \(\frac { 19 }{ 36 }\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac { 19 }{ 36 }\) = \(\frac { 17 }{ 36 }\)

Question 14.
In a given race, the odds in favour of horses A, B, C and D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.
Solution:
Given odds in favour of horse A be 1 : 3
∴ P (horse A’s winning) = \(\frac{1}{1+3}\) = \(\frac{1}{4}\)
odds in favour of horse B be 1 : 4
∴ P (horse B’s winning) = \(\frac{1}{1+4}\) = \(\frac{1}{5}\)
odds in favour of horse C be 1 : 5
∴ P (horse C’s winning) = \(\frac{1}{1+5}\) = \(\frac{1}{6}\)
and odds in favour of horse D be 1 : 6
∴ D (horse D’s winning) = \(\frac{1}{1+6}\) = \(\frac{1}{7}\)
∴ Required probability that one of them will win the race = P (A) + P (B) + P (C) + P (D)
[since A, B, C and D are mutually exclusive and exhaustive events]
= \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + \(\frac{1}{7}\) = \(\frac{105+84+70+60}{420}\) = \(\frac{319}{420}\)

Question 15.
100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both examinations.
Solution:
Let A : event that students pass the fist examination
B : event that students pass the second exam.
Then P(A) = \(\frac{60}{100}\); P (B) = \(\frac{5}{100}\); P(A ∩ B) = \(\frac{60}{100}\) + \(\frac{50}{100}\) – \(\frac{30}{100}\) = \(\frac{80}{100}\)
Thus required probability that selected student failed in both exams = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\)
= 1 – P (A ∪ B) = 1 – \(\frac{80}{100}\) = \(\frac{20}{100}\) = 0.2

Question 16.
(i) A card is drawn from a deck of 52 can is. Find the probability of getting an ace or a spade card.
(ii) From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
Solution:
(i) Total no. of exhaustive cases = 52
Let A : event of getting an ace card ∴ n (A) = 4
B : getting a spade card ∴ n (B) = 13
A ∩ B : getting an ace of spade
∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) n(S) = Total no. of ways of drawing 4 cards out of 52 = ⁵²C₄
All drawing cards are of same colour means all cards are either red or black.
Let A : drawing four red cards.
∴ n (A) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Let B : drawing four black cards
∴ n (B) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Thus required probability =P (A) + P (B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{2 \times{ }^{26} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}\)
= 2 × \(\frac{26 \times 25 \times 24 \times 23}{52 \times 51 \times 50 \times 49}\) = \(\frac{92}{833}\)

Question 17.
(i) A card is drawn at random from a well shuffled pack of cards. What is the probability that it is a heart or a queen ?
(ii) A card is drawn at random from a well shuffled pack of 52 cards. Find the probability it is neither a king nor a heart ?
Solution:
(i) Given total exhaustive cases = 52 = n (S)
Let A : event that heart card is drawing ∴ n (A) = 13
and B : drawing a queen card ∴ n (B) = 4
A ∩ B : drawing a queen of heart ∴ n (A ∩ B) = 1
Thus required probability = P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{13}{52}\) + \(\frac{4}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) Total no. of exhaustive cards = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
and B : drawing a heart card ∴ n (B) = 13
A ∩ B : drawing a king card of heart ∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac{4}{13}\) = \(\frac{9}{13}\)

Question 18.
A card is drawn from a pack of 52 cars. Find the probability of getting a king or a heart or a red card.
Solution:
Total no. of outcomes = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
B : drawing a heart card ∴ n (B) = 13
C : drawing a red card ∴ n (C) = 26
A ∩ B : drawing a king of heart card
∴ n( A ∩ B) = 1
B ∩ C : drawing a heart card
∴ n( B ∩ C) = 13
A ∩ C : drawing a king of red card
∴ n( A ∩ C) = 2
∴ A ∩ B ∩ C : drawing a king of heart card
∴ n(A ∩ B ∩ C) = 1
Required probability = P(A ∪ B ∪ C) = P (A) + P (B) + P(C) – P (A ∩ B) -P (B ∩ C) – P (A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) + \(\frac{26}{52}\) –\(\frac{1}{52}\) – \(\frac{13}{52}\) – \(\frac{2}{52}\) + \(\frac{1}{52}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\)

Question 19.
There are three events A, B, C one of which must and only one can happen. The odds are 8 to 3 against A, 5 to 2 against B ; find the odds against C.
Solution:
Given odds against A be 8 : 3.
∴ P (A) = \(\frac{3}{8+3}\) + \(\frac{3}{11}\)
Also, odds against B be 5 : 2
Since out of three events A, B and C, one of which must and only one can happen ∴ A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
⇒ P (C) = 1 – \(\frac{3}{11}\) – \(\frac{2}{7}\) = \(\frac{77-21-22}{77}\) = \(\frac{34}{77}\)
\(\mathrm{P}(\overline{\mathrm{C}})\) = 1 – P (C) = 1 – \(\frac{34}{77}\) = \(\frac{43}{77}\)
Thus required odds against C be \(\mathrm{P}(\overline{\mathrm{C}})\) : P (C) i.e. \(\frac{43}{77}\) : \(\frac{34}{77}\) i.e. 43 : 34.

Question 20.
In a group of students, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
Given total no. of students = 3 + 3 = 6
∴ n (S) = Total no. of ways of selecting 4 students out of 6 = ⁶C₄
Let A : selecting 3 boys and 1 girl
∴ n (A) = Total no. of ways of selecting 3 boys out of 3 and 1 girl out of 3 = ³C₃ × ³C₁
Let B : selecting 1 boy and 3 girls
∴ n (B) = Total no. of ways of selecting 1 boy out of 3 and 3 girls out of 3 = ³C₁ × ³C₃
Thus required probability = P (A ∪ B) = P
(A) + P (B) = \(\frac{{ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_3}{{ }^6 \mathrm{C}_4}\) = \(\frac{1 \times 3+3 \times 1}{\frac{6 \times 5}{2}}\) = \(\frac{6}{15}\) =  \(\frac{2}{5}\)

Students often turn to S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(e) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(e)

Question 1.
Sameer throws an ordinary die. What is the probability that he throws
(i) 2
(ii) 5
(iii) 2 or 5 ?
Solution:
(i) When a die is thrown then total no. of outcomes = n (S) = 6 where S = {1, 2,3, 4, 5, 6}
∴ P (getting 2) = \(\frac { 1 }{ 6 }\)
(ii) P (getting 5) = \(\frac { 1 }{ 6 }\)
(iii) P (getting 2 or 5) = P (2) + P (5) = \(\frac { 1 }{ 6 }\) + \(\frac { 1 }{ 6 }\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

Question 2.
Kavita draws a card from a pack of cards. What is the probability that she draws
(i) a heart
(ii) a club
(iii) a heart or a club ?
Solution:
Total no. of cards = n (S) = 52
(i) since there are 13 heart cards ∴ P (heart) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(ii) p (a club) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(iii) P (a heart or a club) = P (heart) + P (club)  = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

Question 3.
Anurag draws a card from a pack of cards. What is the probability that he draws one of the following numbers ?
(i) 3
(ii) 7
(iii) 3 or 7
Solution:
Here n (S) = 52
(i) Since there are four 3’s (one each from heart, diamond, club and spade)
∴ required probability of getting 3 = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(ii) P (getting 7) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(iii) P (3 or 7) = P (3) + P (7) = \(\frac { 1 }{ 13 }\) = \(\frac { 1 }{ 13 }\) = \(\frac { 2 }{ 13 }\)

Question 4.
A letter is chosen at random from the letters in the word PROBABILITY. What is the probability that the letter will be
(i) B
(ii) a vowel
(iii) B or a vowel ?
Solution:
Total no. of letters in word PROBABILITY = 11
(i) P (getting a letter B) = \(\frac { 2 }{ 11 }\) [since there are 2B’s]

(ii) since there are 4 vowels O, A, I, I
∴ P (getting a vowel) = \(\frac { 4 }{ 11 }\)

(iii) P (B or a vowel) = P (B) + P (a vowel) = \(\frac { 2 }{ 11 }\) + \(\frac { 4 }{ 11 }\) = \(\frac { 6 }{ 11 }\)

Question 5.
A bag contains 7 white balls, 9 green balls and 10 yellow balls. A ball is drawn at random from the bag. What is the probability that it will be
(i) white
(ii) green
(iii) green or white
(iv) not yellow
(v) neither yellow nor green ?
Solution:
Given no. of white balls = 7, no. of green balls = 9 and no. of yellow balls = 10
∴ Total no. of balls = 7 + 9 + 10 = 26

(i) P(white ball) = \(\frac { 7 }{ 26 }\)

(ii) P(green ball) = \(\frac { 9 }{ 26 }\)

(iii) P(green or white) = P(green) + P(white) = \(\frac { 9 }{ 26 }\) + \(\frac { 7 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)

(iv) P(yellow ball) = \(\frac { 10 }{ 26 }\) ∴ P(getting not yellow ball) = 1 – \(\frac { 10 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)
Thus, P (neither yellow nor green) = P (white ball) = \(\frac { 7 }{ 26 }\)

Question 6.
Suyash needs his calculator for his mathematics lesson. It is either in his pocket, bag or locker. The probability it is in his pocket is 0.20, the probability it is in his bag is 0.58. What is the probability that (i) he will have the calculator for the lesson, (ii) it is in his locker ?

Solution:
(i) Given P (calculator is in his pocket) = 0.20
P (calculator is in his bag) = 0.58
∴ required probability = P (calculator in his pocket) + P (calculator in his bag) = 0.20 + 0.58 = 0.78
(iii) required probability that calculator is in his locker = 1 – P (calculator is in his pocket) – P (calculator is in his bag) = 1 – 0.20 – 0.58 = 0.22

Question 7.
A spinner has numbers and colours on it, as shown in the diagram. Their probabilities are given in the tables. When the spinner is spun what is the probability of each of the following ?

(i) red or green
(ii) 2 or 3
(iii) 3 or green
(iv) 2 or green
(v) Explain why the answer to P (1 or red) is not 0.9.
Solution:
Given P (red) = 0.5 ; P (1) = 0.4
P (green) = 0.25 ; P (2) = 0.35
P (blue) = 0.25 ; P (3) = 0.25
(i) P (red or green) = P (red) + P (green) = 0.5 + 0.25 = 0.75
(ii) P (2 or 3) = P (2) + P (3) = 0.35 + 0.25 = 0.6
(iii) P (3 or green) = P (3) + P (green) = 0.25 + 0.25 = 0.50
(iv) P (2 or green) = P (2) + P (green) = 0.35 + 0.25 = 0.60
(v) Here 1 and red are not mutually exclusive event
∴ P (1 and red) ≠ 0 [∵ 1 ∩ red ≠ Φ]
∴ P (1 or red) ≠ P(1) + P(red) = 0.9

The availability of step-by-step S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(d) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(d)

Question 1.
Four digit numbers are formed by using the digits 1, 2, 3, 4 and 5 without repeating the digit. Find the probability that a number, chose at random, is an odd number.
Solution:
Total no. of 4 digit numbers formed by using given digits 1, 2, 3, 4 and 5 without repetition

∴ Total no. of exhaustive cases = 120
Total no. of favourable cases = total no. of 4 digit odd number = 2 × 3 × 4 × 3 = 72

Question 2.
What is the probability of getting 3 white balls in a draw of 3 balls from a box containing 6 white and 4 red balls ?
Solution:
Given Total no. of balls = 6 + 4 = 10
Total no. of cases = Total no. of drawing 3 balls out of 10 balls = ¹⁰C₃
Favourable no. of cases = Total no. of ways of drawing 3 white balls out of 6 = ⁶C₃

Question 3.
A bag contains 6 white, 7 red and 5 black balls, three balls are drawn at random. Find the probability that they will be white.
Solution:
Total no. of balls = 6 + 7 + 5 = 18
∴ Total no. of exhaustive cases = Total no. of ways of drawing 3 balls out of 18 = ¹⁸C₃
Total no. of favourable cases = no. of ways of drawing 3 white balls out of 6 = ⁶C₃

Question 4.
A bag contains 2 white marbles, 4 blue marbles, and 6 red marbles. A marble is drawn at random from the bag. What is the probability’ that
(i) it is white ?
(ii) it is blue ?
(iii) it is red ?
(iv) it is not white ?
(v) it is not blue ?
(vi) it is black ?
Solution:
Total number of marbles = 6 + 4 + 2=12
∴ total no. of Exhaustive cases = 12
(i) Required probability of drawing a white ball = \(\frac { 2 }{ 12 }\) = \(\frac { 1 }{ 6 }\)

(ii) Required probability of drawing a blue ball = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)

(iii) Required probability of drawing a red ball = \(\frac { 6 }{ 12 }\) = \(\frac { 1 }{ 2 }\)

(iv) Required probability of drawing not a white ball = probability of drawing either red or blue ball = \(\frac { 6+4 }{ 12 }\) = \(\frac { 10 }{ 12 }\) = \(\frac { 5 }{ 6 }\)

(v) drawing ball is not blue ∴ it is either white or red
∴ Total no. of favourable cases = 2 + 6 = 8
Thus required probability = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)

(vi) Since there is no black ball
∴ Total no. of favourable cases = 0
Thus, required probability = \(\frac { 0 }{ 12 }\) = 0

Question 5.
In Q. 4, three marbles are drawn from the bag. What is the probability that
(i) they are all blue ?
(ii) they are all red ?
(iii) they are all white ?
(iv) none of them is red ?
(v) not all of them are red ?
(vi) none is black ?
Solution:
Here total no. of exhaustive cases = Total no. of ways of drawing 3 balls out of 12 = ¹²C₃
(i) Total no. of favourable cases = Total no. of ways of drawing 3 out of 4 = ⁴C₃

(iii) Since we have to draw 3 white marbles but we have only 2 white marbles.
∴ Total no. of favourable cses = 0
Thus required probability = \(\frac{0}{{ }^{12} C_3}\) = 0

(iv) Since none of drawing marble be red
∴Total no. of favourable outcomes = Total no. of ways of drawing 3 marbles out of remaining 6 marbles = ⁶C₃

(v) Required probability = 1 – P(all are red) = 1 – \(\frac { 1 }{ 11 }\) = \(\frac { 10 }{ 11 }\)
(vi) Since there is no black ball
∴ required probability that none is black = probability of drawing ball of any colour out of red, blue or white = 1

Question 6.
Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls, one by one without replacement. What is the probability that (i) both balls are of the same colour, (ii) at least one ball is red ?
Solution:
Given no. of white balls = 2, no. of red balls = 3 and no. of black balls = 4
∴ Total no. of balls = 2 + 3 + 4 = 9
(i) Required probability = P (WW) + P (RR) + P (BB)
= \(\frac { 2 }{ 9 }\) × \(\frac { 1 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 8 }\) = \(\frac { 20 }{ 72 }\) = \(\frac { 5 }{ 18 }\)

(ii) Required probability of drawing atleast one ball is red = P (RW, RB, RR, WR, BR) = P (RW) + P (RB) + P (RR) + P (WR) + P (BR)
= \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 4 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 2 }{ 9 }\) × \(\frac { 3 }{ 8 }\) + \(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 8 }\) = \(\frac { 42 }{ 72 }\) = \(\frac { 7 }{ 12 }\)

Question 7.
From a pack of cards, three are drawn at random. Find the chance that they are a king, a queen and a knave.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing three cards out of 52 = ⁵²C₃
Total no. of favourable cases = Total no. of ways of drawing a king, a queen and a knave = ⁴C₁ × ⁴C₁ × ⁴C₁

Question 8.
(i) Two cards are drawn at random from 8 cards numbered from 1 to 8. What is the probability that the sum of the numbers is odd, if the two cards are drawn together ?
(ii) Two cards are drawn at random from a well shuffled pack of 52 cards, show that the chances of drawing two aces is \(\frac { 1 }{ 221 }\).
(iii) From a pack of 52 cards, 3 cards are drawn ‘at random’. Find the probability of drawing exactly two aces.
(iv) Three cards are drawn at a time at random from a well shuffled pack of 52 cards. Find the probability that all the three cards have the same number.
(v) Two cards are drawn from a well shuffled pack of cards without replacement. Find the probability that neither a Jack nor a card of spades is drawn.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing two cards out of 8 = ⁸C₂ = n (S)

(i) Let E : the sum of the numbers is odd
= {(1, 2), (2, 1), (1, 4), (4, 1), (1, 6), (6, 1), (1, 8), (8, 1), (2, 3), (3, 2), (2, 5), (5, 2), (2, 7), (7, 2), (3, 4), (4, 3), (3, 6), (6, 3), (3, 8), (8, 3), (4, 5), (4, 7), (5, 4), (7, 4), (6, 5), (5, 6), (5, 8), (8, 5), (6, 7), (7, 6), (8, 7), (7, 8)}
Since cards are taken together
∴ (a, b) = (b, a)
Thus n (E) = 16
∴ required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{16}{{ }^8 C_2}\) = \(\frac{16}{28}\) = \(\frac{4}{7}\)

(ii) Total no. of exhaustive cases = Total no. of ways of drawing 2 cards out of 52 = ⁵²C₂
Total no. of favourable outcomes = Total no. of ways of drawing 2 out of 4 = ⁴C₂
∴ required probability = \(\frac{{ }^4 C_2}{{ }^{52} C_2}\) = \(\frac{\frac{4 !}{2 ! 2 !}}{\frac{52 !}{50 ! 2 !}}\) = \(\frac{6}{26 \times 51}\) = \(\frac{1}{13 \times 17}\) = \(\frac { 1 }{ 221 }\)

(iii) Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
∴ Total no. of favourable outcomes = Total no. of ways of drawing exactly two aces = ⁴C₂ × ⁴⁸C₁

(iv) Total no. of exhaustive case = total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
Let E : all the three cards are having the same number.
Since there are 13 such groups each containing 4 same cards
∴ n (E) = 13 × ⁴C₃ = 13 × 4 = 52

(v) Total no. of exhaustive cases = Total no. of ways of drawing 2 cards out of 52 without replacement = ⁵²C₂ = n (S)
E : drawing card is neither a Jack nor a card of spades and there are (52 – 13 – 3) = 36 such cards
∴ n (E) = Total no. of ways of drawing 2 cards out of 36 = ³⁶C₂

Question 9.
Three cards are drawn from a deck of 52 cards. What is the probability that
(i) they are all spades ?
(ii) they are all red cards ?
(iii) none of them is a club ?
(iv) all of them are aces ?
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
(i) Since there are 13 spade cards.
Then total no. of favourable cases = ¹³C₃
∴ required Probability = \(\frac{{ }^{13} C_3}{{ }^{52} C_2}\) = \(\frac{\frac{13 !}{10 ! 3 !}}{\frac{52 !}{49 ! 3 !}}\) = \(\frac{13 \times 12 \times 11}{52 \times 51 \times 50}\) = \(\frac{11}{850}\)

(ii) Since there are 26 red cards
∴ Total no. of ways of drawing 3 cards out of 26 = ²⁶C₃
∴ required probability = \(\frac{{ }^{26} C_3}{{ }^{52} C_3}\) = \(\frac{\frac{26 !}{3 ! 23 !}}{\frac{52 !}{49 ! 3 !}}\) = \(\frac{26 \times 26 \times 24}{52 \times 51 \times 50}\) = \(\frac{2}{17}\)

(iii) Since none of the drawing card is club card so the drawing cards can be out of (52 – 13) = 39 cards
∴ Total no. of favourable cases = No. of ways of drawing 3 cards out of 39 = ³⁹C₃

Question 10.
In Q. 7 what are the odds that
(i) all three cards are spades?
(ii) they are all red cards?
(iii) none is a club?
(iv) all of them are aces?
Solution:
Total no. of exhaustive outcomes = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃

(i) Since there are 13 spade cards
∴ Total no. of ways of drawing 3 cards out of 13 = ¹³C₃

(ii) Total no. of favourable cases = Total no. of ways of drawing 3 cards out of 26 = ²⁶C₃
∴ required probability = \(\frac{{ }^{26} C_3}{{ }^{52} C_3}\) = \(\frac{26 \times 25 \times 24}{52 \times 51 \times 50}\) = \(\frac{2}{17}\)
Thus required odds = \(\frac { 2 }{ 17 }\) : 1 – \(\frac { 2 }{ 17 }\) = \(\frac { 2 }{ 17 }\) : \(\frac { 15 }{ 17 }\) = \(\frac { 2 }{ 15 }\)

(iii) Total no. of favourable cases = Total no. of ways drawing 3 cards out of (52 – 13) = 39 cards = ³⁹C₃
∴ required probability of drawing none is a club card out of 3 = \(\frac{{ }^{39} C_3}{{ }^{52} C_3}\) = \(\frac{39 \times 38 \times 37}{52 \times 51 \times 50}\) = \(\frac{703}{1700}\)
Thus required odd = \(\frac{703}{1700}\) : 1 – \(\frac{703}{1700}\) = \(\frac{703}{1700}\) : \(\frac{997}{1700}\) = 703 : 997

(iv) Total no. of ways of drawing 3 ace cards out of 4 = ⁴C₃ = 4
∴ required probability = \(\frac{{ }^4 C_3}{{ }^{52} C_3}\) = \(\frac{4}{\frac{52 \times 51 \times 50}{6}}\) = \(\frac{1}{5525}\)
Thus required odds in favour of favourable event = \(\frac{1}{5525}\) : 1 – \(\frac{1}{5525}\) = 1 : 5524

Question 11.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally like to be drawn. Find the probability that the card drawn is (i) an ace, (ii) red, (iii) either red or king, (iv) red and a king.
Solution:
Total no. of exhaustive cases = 52
(i) since there are 4 ace cards
∴ required probability = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(ii) since there are 26 red cards
∴ required probability = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(iii) Since there are 26 red cards that includes 2 red kings and 2 black kings
∴ Total no. of such cards = 26 + 2 = 28
∴ required probability = \(\frac{28}{52}\) = \(\frac{7}{13}\)

(iv) Since there are 2 red kings in 52 cards
Thus required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 12.
Four cards are drawn at random from a pack of 52 playing cards, Find the probability of getting
(i) all the four cards of the same suit;
(ii) all the four cards of the same number;
(iii) one card from each suit;
(iv) two red cards and two black cards ;
(v) all cards of the same colour ;
(vi) all face cards ; (King, Queen, Jack)
(vii) four honours of the same suit.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₄
(i) Total no. of favourable cases = Total no. of ways of drawing 4 cards from same suit = 4 × ¹³C₄ [since there are 4 suits and each suit containing 13 cards]

(ii) Let E : getting all the four cards of the same number
∴ n (E) = 13 × ⁴C₄ = 13
[There are 13 such groups and each group containing 4 same cards each one from each suit]
Thus required probability = \(\frac{13}{{ }^{52} C_4}\) = \(\frac{\frac{13}{52 \times 51 \times 50 \times 49}}{24}\) = \(\frac{13}{270725}\)

(iii) Total no. of ways of drawing one card from each suit = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ [since each suit contains 13 cards each]
∴ required probability = \(\frac { favourable cases }{ total cases }\) = \(\frac{{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_4}\) =\(\frac{13 \times 13 \times 13 \times 13}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{2197}{20825}\)

(iv) Total no. of favourable cases = Total no. of ways of drawing 2 red and 2 black cards out of 26 cards each = ²⁶C₂ × ²⁶C₂
Thus required probability = \(\frac{{ }^{26} C_2 \times{ }^{26} C_2}{{ }^{52} C_4}\) = \(\frac{\frac{26 \times 25}{2} \times \frac{26 \times 25}{2}}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{325}{833}\)

(v) drawing 4 cards of same colour means either all 4 cards are red or black.
∴ Total favourable cases = Total no. of ways of drawing all four cards of same colour = ²⁶C₄ × ²⁶C₄
∴ required probability = \(\frac{{ }^{26} C_4 \times{ }^{26} C_4}{{ }^{52} C_4}\)

(vi) since there are 12 face cards (4 kings, 4 queens and 4 jacks)
Thus total no. of ways of drawing 4 cards from 12 cards = ¹²C₄
Therefore required probability = \(\frac{{ }^{12} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}\) = \(\frac{\frac{12 !}{4 ! 8 !}}{\frac{52 !}{4 ! 48 !}}\) = \(\frac{12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49}\) = \(\frac{99}{54145}\)

(vii) since there are 16 honours cards including 4 aces, 4 kings, 4 queens and 4 jacks i.e. 4 suits.
Total no. of ways of drawing 4 honour cards of same suit = 4
∴ Required probability = \(\frac{4}{{ }^{52} \mathrm{C}_4}\) = \(\frac{4}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{4}{270725}\)

Question 13.
In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. Find the probability that
(i) both the tickets drawn have prime numbers ;
(ii) none of the tickets drawn has prime number ;
(iii) a ticket has prime number.
Solution:
Total exhaustive cases = Total no. of ways of drawing 2 tickets out of 50 = ⁵⁰C₂
(i) Prime numbers from 1 to 50 are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
∴ Total no. of prime numbers from 1 to 50 = 15
Thus total favourable cases = Total no. of ways of drawing 2 tickets out of 15 = ¹⁵C₂
∴ required probability = \(\frac { favourable cases }{ Exhaustive cases }\) = \(\frac{{ }^{15} C_2}{{ }^{50} C_2}\) = \(\frac{15 \times 14}{50 \times 49}\) = \(\frac{3}{35}\)

(ii) Total numbers which are not prime from 1 to 50 = 50 – 15 = 35
Thus total favourable cases = Total no. ot ways of drawing 2 tickets when none of tickets drawn has prime number = ³⁵C₂
∴ required probability = \(\frac{{ }^{35} \mathrm{C}_2}{{ }^{50} \mathrm{C}_2}\) = \(\frac{35 \times 34}{50 \times 49}\) = \(\frac{17}{35}\)

(iii) Total no. of ways of drawing 2 tickets in which one ticket containing prime no. = ¹⁵C₁ × ³⁵C₁
∴ required probability = \(\frac{{ }^{15} C_1 \times{ }^{35} C_1}{{ }^{50} C_2}\) = \(\frac{15 \times 35 \times 2}{50 \times 49}\) = \(\frac{3}{7}\)

Question 14.
(i) Out of 9 outstanding students in a college, there are 4 boys and 5 girls. A team of 4 students is to be selected for a quiz programme. Find the probability that 2 are girls and 2 are boys.
(ii) Four people are chosen at random from a group consisting of 3 men, 2 women, and 3 children. Find the probability that out of four chosen people, exactly 2 are children ?
Solution:
(i) Given total no. of outstanding students = 9
∴ Total no. of ways of selecting 4 students out of 9 = ⁹C₄
Total no. of ways of selecting 2 boys out of 4 be ⁴C₂ and no. of ways of selecting 2 girls out of 5 be ⁵C₂
∴ Total no. of ways of drawing 2 girls and 2 boys = ⁴C₂ × ⁵C₂

(ii) Total no. of peoples =3 + 2 + 3 = 8
∴ Total no. of ways of choosing 4 people out of 8 = ⁸C₄
Total no. of ways of choosing exactly 2 children = ³C₂ × ⁵C₂ = 3 × \(\frac { 5×4 }{ 2 }\) = 30
∴ required probability = \(\frac{30}{{ }^8 C_4}\) = \(\frac{30}{\frac{8 \times 7 \times 6 \times 5}{24}}\) = \(\frac { 3 }{ 7 }\)

Question 15.
A committee of 5 principals is to be selected from a group of 6 grant principals and 8 lady principals. If the selected is made randomly, find the probability that there are 3 lady principals and 2 gent principals.
Solution:
Total no. of principals = 6 + 8 = 14
∴ Total no. of ways of selecting 5 principals out of 14 = ¹⁴C₅
Total no. of ways of selecting 3 lady principals from 8 and 2 gents principals out of 6 = ⁸C₃ × ⁶C₂
∴ required probability = \(\frac{{ }^8 C_3 \times{ }^6 C_2}{{ }^{14} C_5}\) = \(\frac{\frac{8 \times 7 \times 6}{6} \times \frac{6 \times 5}{2}}{\frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2}}\) = \(\frac { 60 }{ 143 }\)

Question 16.
A bag contains tickets numbered 1 to 20. Two tickets are drawn. Find the probability that both numbers are odd.
Solution:
Total no. of ways of drawing 2 tickets out of 20 = ²⁰C₂
since there are 10 odd numbers from 1 to 20.
∴ Total no. of ways of drawing 2 tickets out of 10 = ¹⁰C₂
Thus required probability = \(\frac{{ }^{10} C_2}{{ }^{20} C_2}\) = \(\frac{10 \times 9}{20 \times 19}\) = \(\frac{9}{38}\)

Question 17.
A bag contains tickets numbered 1 to 30 . Three tickets are drawn at random from the bag. What is the probability that the maximum number of the selected tickets exceeds 25 ?
Solution:
Total no. of ways of drawing 3 tickets out of 30 = ³⁰C₃
Let A : event that none of the selected ticket bear number exceeding 25

Question 18.
(i) A room has 3 lamps. From a collection of 10 light bulbs of which 6 are no good, a person selects 3 at random and puts them in a socket. What is the probability, that he will have light ?
(ii) A has 3 shares in a lottery containing 3 prizes and 9 blanks ; B has two shares in lottery containing 2 prizes and 6 blanks. Compare their chances of success.
Solution:
(i) Total no. of light bulbs = 10
no. of no good bulbs = 6
Thus, no. of good bulbs = 10 – 6 = 4
Total exhaustive cases = total no. of ways of selecting 3 out of 10 = ¹⁰C₃
∴ required probability = 1 – P (no good bulb)

Question 19.
(i) There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelope ?
(ii) Three letters are written to different persons, and the addresses on the three envelopes are also written. Without looking at the addresses, find the probability that the letters go into the right envelopes.
Solution:
(i) Total no. of ways of planning n letters in n envelopes = n!
There is only one way in which all the letters placed correctly.
∴ probability of planning the letters in right envelopes = \(\frac{1}{n !}\)
Thus, required probability that all letters are not placed in right envelopes =1 – \(\frac{1}{n !}\)

(ii) The total number of ways of placing 3 letters in 3 envelopes = 3! = 6
There is only 1 way in which all the three letters placed correctly.
∴ required probability that the letters placed in right envelopes = \(\frac{1}{6}\)

Question 20.
The letters of word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ?
Solution:
The total no. of ways in which all the letters of the word SOCIETY are placed in a row = 7!
Total no. of vowels in word society =3 {O, I, E}
When all three vowels come together they form a group. ∴ 1 group and 4 consonants can be arranged in 5! ways and 3 vowels arranged themselves in 3! ways.
Thus total no. of ways in which three vowels come together = \(\lfloor 5\) × \(\lfloor 3\)
∴ required probability = \(\frac{3 ! \times 5 !}{7 !}\) = \(\frac{6}{7 \times 6}\) = \(\frac{1}{7}\)

Question 21.
In a random arrangement of the letters of the word “COMMERCE”, find the probability that all vowels come together.
Solution:
In word COMMERCE, there are 2E’s, 2M’s, 2C’s, 1O and 1R
∴ Total no. of ways in which letters of word COMMERCE arranged themselves = \(\frac{8 !}{2 ! 2 ! 2 !}\)
since there are 3 vowels (O, E, E) and came together to form a group. Thus, this group and 5 consonants arranged themselves in \(\frac{6 !}{2 ! 2 !}\) and 3 vowels can be arranged themselves in \(\frac{3 !}{2 !}\) .
Thus total no. of favourable cases = \(\frac{6 !}{2 ! 2 !} \times \frac{3 !}{2 !}\)
∴ required probability = \(\frac{\frac{6 !}{2 ! 2 !} \times \frac{3 !}{2 !}}{\frac{8 !}{2 ! 2 ! 2 !}}\) = \(\frac{6 ! \times 3 !}{8 !}\) = \(\frac{3}{28}\)

Question 22.
Given a group of 4 persons, find the probability that
(i) no two of them have their birthdays on the same day,
(ii) ail of them have same birthday. [Ignore the existence of a leap year]
Solution:
(i) The first person can have birthday on any of 365 days.
The second person can also have birthday on any of 365 days and so on.
Thus the total no. of ways of choosing the birthday for the four persons = (365)⁴
(i) Since no two persons have their birthday on the same day. Thus for the first person we have 365 choices and second person have 364, third person have 363 and 4th person have 362 choices.
Thus, the total no. of ways in which all the four persons have their birthdays on different days = 365 × 364 × 363 × 362
∴ required probability = \(\frac{365 \times 364 \times 363 \times 362}{(365)^4}\) = \(\frac{{ }^{364} \mathrm{P}_3}{(365)^2}\)

(ii) Since all the four persons have birthday on same day and that day can be any day out of365 days
∴ Total no. of such ways = 365
∴ Required probability = \(\frac{365}{(365)^4}\) = \(\frac{1}{(365)^3}\)