TS Inter 2nd Year

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 9 Probability to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Probability Important Questions

Question 1.
A Number x is drawn arbitrarily from the set [1, 2, 3 ……………. 100]. Find the probability that \(\left(x+\frac{100}{x}\right)>29\)
Solution:
The total points of the sample space are 100.
Let A be the event that an x selected (drawn) at random from the set
S = {1, 2, 3 …………… 100} has the property.
\(\left(x+\frac{100}{x}\right)>29\)
Now x +\(\frac{100}{x}\) > 29 ⇔ x² – 29x + 100 >0
(x – 4)(x – 25) > 0 ⇔ x >25 or x< 4
Since x ∈ S. it follows that
A = {1,2,3,26.27. ……………………. 100}
Thus the numbers of cases favourable to A is 78.
∴ The required probability: \(\frac{1}{18}\)
P(A)=\(\frac{78}{100}\)=0.78

Question 2.
Two squares are chosen at random on a chess board. Show that the probability that they have a side in common is \(\frac{1}{18}\).
Solution:
The number of ways of choosing the first square is 64 and that of the second is 63. Therefore the number of ways of choosing the first and the second square is 64 x 63. Let E be the event that these squares have a side in common. We shall find the number of cases favourable to E. If the first square happens to be one of the squares in the four corners of the chess hoard. the second square (with common side) can be chosen in 2 ways.

lf the first square happens to be any one of the remaining 24 squares along the four sides of the chess board other than the corner, the second square can be chosen in 3 ways. If the first square happens to be any one of the remaining 36 Inner squares, then the second square can he choose in 4 ways. Hence the number of cases favourable to E
Is (4 x 2) + (24 x 3) + (36 x 4) = 224
Therefore the required probability is
\(\frac{224}{64 \times 63}=\frac{1}{18}\)

Question 3.
A fair coin is tossed 200 times. Find the probability of getting a head an odd number of time.
Solution:
The total number of points in the sample space is 2²⁰⁰. Let E be event getting a head an odd number of times. Then the number of cases favourable to E is

Question 4.
A and B are among 20 persons who sit at random along a round table. Find the probability that there are any six persons between A and B.
Solution:
Let A occupy any seat at the round table.
Then there are 19 seats left for B. But if six persons are to be seated between A and B, then B has only two ways to sit. Thus the required probability is \(\frac{2}{19}\).

Question 5.
Out of 30 consecutive Integers, two integers are (Irawil at random. Find the probability that their sum is odd.
Solution:
The total number of ways of choosing 2 integers out of 30 is ³⁰C₂. Out of the 30 numbers, 15 are even and 15 are odd. If the sum of the two numbers is to be odd, one should be even and the other odd. Hence the number of cases favourable to the required event is ¹⁵C₁ x ¹⁵C₁
∴ The required probability

Question 6.
Out of 1,00,000 new born babies, 77,181 survived till the age of 20. Find the probability that a new born baby survives till 20 year of age.
Solution:
Here m = 77,181; n = 1,00,000
Required Probability = \(\frac{77,181}{1,00,000} \) = 0.77181

Question 7.
Find the probability of throwing a total score of 7 with 2 dice.
Solution:
The sample space S of this experiment is given by
S = {(1, 1), (1, 2),…………………, (1, 6),
(2, 2), (2, 2),…………………, (2, 6).
(6, 1), (6, 2),…………………, (6, 6)}
In a typical element, the first coordinate represents the score on the first die and the second coordinate represents the score on the second die. There are 36 points in S and all the points are equally likely. Let E be the event of getting a total score of 7. Then E has the following 6 elements.
{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2),(6, 1)}
∴ P(E) = \(\frac{6}{36}=\frac{1}{6}\)

Question 8.
Find the probability of obtaining two tails and one head when 3 coins are tossed.
Solution:
For this experiment of tossing three coins, the sample space can be seen to be
S = {HHH, HHT, HTH, THH, HTT, THT, TTH,TTT}
Let E be the event of obtaining two tails and a head.
Then E = {HTT, THT, TTH}
∴ P(E) = \(\frac{3}{8}\)

Question 9.
A page is opened at random from a book containing 200 pages. What is the probability fly that the number on the page is a perfect square.
Solution:
The sample space S of the experiment in a question is given by S = {1, 2,3 ……………… 200}, so that the number of points in the sample space n(S) 200. Let E be the event of drawing a page whose number is a perfect square. Then
E = {1, 4, 9 , 196) so that n(E) = 14.
∴ \(\frac{n(E)}{n(S)}=\frac{14}{200}=\frac{7}{100}\)

Question 10.
Find the probability of drawing an Ace or a Spade from a well-shuffled pack of 52 playing cards. A pack of cards contains a total of 52 cards. These 52 cards are divided into 4 sets namely Hearts, Diamonds, Spades and Clubs. Each set consists of 13 cards, namely:
A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K
(Here A : Ace, K: King, Q : Queen, J: Jack)
Solution:
Let E₁ be the event of drawing a Spade and E₂ be the event of drawing an Ace. Observe here that E₁, E₂ are not mutually exclusive.
Now n(E₁) = 13, n(E₂) = 4 and n (E₁∩E₂) = 1
∴ P(E₁∪E₂) = P(E₁) P(E₂) – P (E₁∩E₂)
\(=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\).

Question 11.
If A and B are two events then show that
(i) P(A∩B^c)P(A)-P(A∩B) and (ii) the probability that one of them occurs is given by P(A)+P(B) – 2P(AB).
Solution:

Question 12.
A and B are events with P(A) = 0.5, P(B)=0.4 and P(A∩B)=0.3. find the probability that (i) A does not occur (ii) neither A nor B occurs.
Solution:
(i) We know that A^c denotes the event: A does not occur and (A∪B)^c denotes the event: neither A nor B occurs. Then
P(A^c) 1 – P(A) = 1 – 0.5 = 0.5

(ii) By addition theorem
P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.5 + 0.4 – 0.3 = 0.6
∴ P((A∪B)^c ) = 1 – P(A∪B)
= 1 – 0.6 = 0.4

Question 13.
If A, B, C are three events show that
P(A∪B∪C)= P(A)+ P(B)+ P(C) – P(A∩B)-P(B∩C) – P(C∩A)+P(A∩B∩C)
Solution:

Question 14.
Suppose there are 12 boys and 4 girls in a class. If we choose three children one after another in succession at random, find the probability that all three are boys
Solution:
Let E₁ be the event of choosing a male child
In i^th trial, I = 1, 2, 3. We have to find P(E₁∩E₂∩E₃).

Question 15.
A speaks truth in 75% of the cases and B in 80% cases. What is the probability that their statements about an incident do not match.
Solution:
Let E₁, E₂ be the events that A and B respectively speak truth about an incident. Then

If E be the event that their statements do not match about the incident, then this happens in two mutually exclusive ways:
i) A speaks truth and B tells lie
ii) A tells lie and B speaks truth
These two events can be represented by

Question 16.
A problem in Calculus is given to two students A and B whose chances of solving it are \(\frac{1}{3}\) and \(\frac{1}{4}\) respectively. Find the probability of the problem being solved if both of them try independently.
Solution:
Let E₁ and E₂ denote the events that the problem is solved by A and B respectively.
We are given that P(E₁)= \(\frac{1}{3}\) and P(E₂) = \(\frac{1}{4}\)
We have to find P(E₁∪E₂)
By Addition theroem,

Question 17.
A and B toss a coin 50 times each simultaneously. Find the probability that both of the in will not get tails at the same toss.
Solution:
Let E be the event that A and B both will not get tails at the same toss. In each toss we have the following four choices:
(i) A: Head, B: Head
(ii) A: Head, B: Tail
(iii) A: Tail, B: Head
(iv) A: Tail. B:Tail
Since there are 50 trails, the total number of choices is 450
But out of the four choices listed above, only (i), (ii) and (ii) are favourable to the occurrence of the required event E.
∴ P(E)=\(\frac{3^{50}}{4^{50}}=\left(\frac{3}{4}\right)^{50}\)

Question 18.
If A and B are independent events of a random experiment, show that Ac and W are also independent
Solution:
If A and B are independent, then
P(A∩B) = P(A) P(B).

Question 19.
Three boxes B₁, B₂ and B₃ contain balls with different colours as shown below:

A die is thrown, B₁ is chosen if either 1 or 2 turns up. B₂ is chosen if 3 or 4 tunis up and B₃ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball drawn is found to be red, find the probability that ills drawn from box B₂.
Solution:
Let E₁ be the event of choosing the box B_i and P(E_i) be the probability of choosing the box B_i. i = 1, 2. 3. Then
P(E₁) = P(E₂) = P(E₃) = \(\frac{2}{6}=\frac{1}{3}\)
P(R|E₁) = \(\frac{2}{5}\), P(R|E₂) =\(\frac{4}{9}\), P(R|E₃) = \(\frac{2}{9}\)
We have to find the probability P(E₂| R).
By Bayes theorem P(E₂| R)

Note: In all the above problems the sample space is finite. In the following case, the sample space is countably infinite.

Question 20.
An urn contains w white balls and b black balls. Two players Q and R alternatively draw a ball with replacement from time urn. The player that draws a white ball first wins the game. If Q begins the game, find the probability of his minimizing the game.
Solution:
Let W denote the event of drawing a white ball in any draw and B that of a black ball.

TS Inter 2nd Year Maths 2B Important Questions with Solutions Pdf 2022 | Maths 2B Important Questions 2022 TS

TS Inter Second Year Maths 2B Important Questions | Maths 2B Important Questions Pdf 2022 TS

1. Maths 2B Circles Important Questions
2. Maths 2B System of Circles Important Questions
3. Maths 2B Parabola Important Questions
4. Maths 2B Ellipse Important Questions
5. Maths 2B Hyperbola Important Questions
6. Maths 2B Integration Important Questions
7. Maths 2B Definite Integrals Important Questions
8. Maths 2B Differential Equations Important Questions

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(e)

I.
Find the I.F. of the following differential equations by transforming them into linear form.

Question 1.
x \(\frac{d y}{d x}\) – y = 2x² sec² 2x.
Solution:
The given equation can be expressed as
\(\frac{d y}{d x}-\frac{y}{x}\) = 2x sec² 2x
This is of the form \(\frac{d y}{d x}\) + Py = Q where the
Integrating factor I.F = e^{∫ P dx}, P = – \(\frac{1}{x}\)
= e^{– ∫ \(\frac{1}{x}\) dx}
= e^{– log x}
= e^{log x-1} = \(\frac{1}{x}\)

Question 2.
y \(\frac{d y}{d x}\) – x = 2y³
Solution:
The given equation can be written as
\(\frac{d x}{d y}-\frac{x}{y}\) = 2y²
and the integrating factor I.F = e^{∫ P dy}
= e^{– ∫ \(\frac{1}{y}\) dy}
= e^{– log y}
= e^{log y-1} = \(\frac{1}{y}\).

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}\) + y tan x = cos³ x
Solution:
Given \(\frac{d y}{d x}\) + y tan x = cos³ x
which is of the form \(\frac{d y}{d x}\) + Py = Q where
P = tan x and Q = cos³ x
∴ Integrating Factor I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= e^{log sec x} = sec x
General solution is y. sec x = ∫ Q (I.F.) dx
= ∫ cos³ x sec x dx
= ∫ cos² x dx
= \(\int \frac{1+\cos 2 x}{2} d x=\frac{1}{2} x+\frac{1}{4} \sin 2 x\)
= \(\frac{1}{2}\) (x + sin x cos x) + c
\(\frac{y}{\cos x}\) = \(\frac{1}{2}\) (x + sin x cos x) + c
⇒ 2y = cos x (x + sin x cos x) + c cos x
= x cos x + sin x cos² x + c cos x is the solution.

Question 2.
\(\frac{d y}{d x}\) + y sec x = tan x
Solution:
This is of the form \(\frac{d y}{d x}\) + Py = Q
where P = sec x and Q = tan x
∴ Integrating Factor I.F. = e^{∫ sec x dx}
= e^{log (sec x + tan x)}
= sec x + tan x
∴ General solution is
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
∴ y (sec x + tan x) = ∫ tan x (sec x + tan x) + c
= ∫ sec x tan x dx . ∫ tan² x dx
= sec x + ∫ (sec² x – 1) dx + c
= sec x + tan x – x + c
∴ y (sec x + tan x) sec x + tan x – x + c is the solution.

Question 3.
\(\frac{d y}{d x}\) – y tan x = e^x sec x.
Solution:
This is of the form \(\frac{d y}{d x}\) + Py = Q where
P = – tan x and Q = e^x sec x.
∴ Integrating Factor IF. = e^{∫ P dx} dx
= e^{∫ tan x} dx
= e^{log cos x} = cos x
∴ General solution is y . e^{∫ P dx} dx = ∫ Q . e^{∫ P dx} dx + c
∴ y . cos x = ∫ e^x sec x cos x dx + c
= ∫ e^x dx + c
= e^x dx + c
y = e^x sec x + c sec x. is the solution.

Question 4.
x \(\frac{d y}{d x}\) + 2y = log x.
Solution:
The equation can be written as
\(\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2}{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{x}}{\mathrm{x}}\)
This is of the form \(\frac{d y}{d x}\) + Py = Q where
I.F. = e^{∫ P dx} where P = \(\frac{2}{x}\), and Q = \(\frac{\pi}{2}\)
= e^{∫ \(\frac{2}{x}\)}
= e^{log x2} = x².
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c

Question 5.
(1 + x²)\(\frac{d y}{d x}\) + y = e^tan-1 x
Solution:
The equation can be written as
\(\frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+x^2}\)

Question 6.
\(\frac{d y}{d x}+\frac{2 y}{x}\) = 2x²
Solution:
The given equation can be written as
\(\frac{d y}{d x}+\frac{2 y}{x}\) = 2x²
∴ I.F.= e^{∫ P dx}
= e^{∫ \(\frac{2}{x}\) dx}
= e^{2 log x}
= e^{log x2} = x²
∴ Solution is y . x² = ∫ 2x² . x² dx
= 2 ∫ x⁴ dx
= 2 \(\frac{x^5}{5}\) + c.

Question 7.
\(\frac{d y}{d x}+\frac{4 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)
Solution:
This is of the form \(\frac{d y}{d x}\) + Py = Q where
P = \(\frac{4 x}{1+x^2}\) and Q = \(\frac{1}{\left(1+x^2\right)^2}\)
∴ I.F = e^{∫ P dx}
= \(e^{\int \frac{4 x}{1+x^2} d x}\)
= e^{2 log (1 + x2)}
= (1 + x²)² dx + c
= x + c is the solution.

Question 8.
x \(\frac{d y}{d x}\) + y = (1 + x) e^x
Solution:
The given equation can be written as
\(\frac{d y}{d x}+\frac{y}{x}=\frac{1+x}{x} e^x\)
∴ I.F = e^{∫ P dx}
= e^{∫ \(\frac{1}{x}\) dx}
= e^{log x} = x
∴ Solution is
y . x = ∫ \(\frac{(1+x)}{x}\) e^x . x dx + c
= ∫ (1 + x) e^x dx + c
= ∫ e^x dx + ∫ x e^x dx + c
= e^x + x e^x – e^x + c
= x e^x + c is the solution.

Question 9.
\(\frac{d y}{d x}+\frac{3 x^2}{1+x^3} y=\frac{1+x^2}{1+x^3}\)
Solution:

Question 10.
\(\frac{d y}{d x}\) – y = – 2 e^-x
Solution:
Here P = – 1 and Q = – 2 e^-x
∴ I.F. = e^{∫ P dx} dx
= e^{∫ – 1 dx} dx
∴ Solution is
y . e^-x = ∫ – 2 e^-x e^-x dx
= – 2 ∫ e^-2x dx
= \(\frac{(-2)}{(-2)}\) e^-2x + c
= e^-2x + c
∴ y = e^-x + C e^x is the solution.

Question 11.
(1 + x²) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y = Tan^-1 x
Solution:
The equation can be written as

= ∫ t e^t dt where t = tan^-1 x
= t e^t – e^t
= e^{tan-1 x (tan-1 – 1)} + c
y = (tan^-1 x – 1) + c e^{– tan-1 x} is the solution.

Question 12.
\(\frac{d y}{d x}\) + y tan x = sin x
Solution:
We have P = tan x and Q = sin x
∴ I.F = e^{∫ P dx} dx
= e^{∫ tan x dx} dx
= e^{log sec x} dx = sec x
∴ Solution is
y sec x = ∫ sin x . sec x dx + c
= ∫ tan x dx
= log |sec x| + c is the solution.

III. Solve the following differential equations.

Question 1.
cos x + y sin x = sec² x
Solution:
The given equation can be written as
\(\frac{d y}{d x}\) + sin x sec x =sec x
Here P = sin x sec x – tan x and Q = sec³ x
∴ I.F. = e^{∫ P dx} dx
= e^{∫ tan x dx} dx
= e^{log (sec x)} = sec x
∴ Solution is
y sec x = ∫ sec⁴ x dx + c
= ∫ (1 + tan² x) sec² x dx + c
= tan x + \(\frac{\tan ^3 x}{3}\) + c.

Question 2.
sec x dy = (y + sin x) dx
Solution:
The given equation can be written as
sec x \(\frac{d y}{d x}\) = y + sin x
⇒ \(\frac{d y}{d x}=\frac{y}{\sec x}+\frac{\sin x}{\sec x}\)
= y cos x + sin x cos x
\(\frac{d y}{d x}\) – y cos x = sin x cos x
I.F. = e^{∫ P dx}
= e^{– ∫ cos x dx}
= e^{– sin x}
∴ Solution is y e^{– sin x}
= ∫ sin x cos x e^{– sin x} dx
= ∫ t e^{– t} dt where t = sin x
= t (- e^{– t}) + ∫ e^{– t} dt
= – e^{– t} (t + 1) + c
= – e^{– sin x} (sin x + 1) + c
y = – (sin x + 1) + c e^{sin x} is the solution.

Question 3.
x log x \(\frac{d y}{d x}\) + y = 2 log x
Solution:
The equation can be written as

Question 4.
(x + y + 1) \(\frac{d y}{d x}\) = 1
Solution:
From the given equation
\(\frac{d y}{d x}=\frac{1}{x+y+1}\)
\(\frac{d y}{d x}\) = x + y + 1
∴ \(\frac{d y}{d x}\) – x = (y + 1)
This is of the form \(\frac{d x}{d y}\) + x = Q(y)
P = – 1 and Q = (y + 1)
∴ I.F. = e^{∫ P dy} = e^{– y}
∴ Solution is
x e^{– y} = ∫ (y + 1) e^{– y} dy
= ∫ e^{– y} y dy + ∫ e^{– y} dy
= – y e^{– y} + ∫ e^{– y} dy – e^{– y}
= – y e^{– y} – e^{– y} – e^{– y} + c
= – y e^{– y} – 2 e^{– y} + c
x = – y – 2 + ce^y
= – (y + 2) ce^y is the solution.

Question 5.
x (x – 1) \(\frac{d y}{d x}\) – y = x³ (x – 1)³
Solution:
The equation can be written as

Question 6.
(x + 2y³) \(\frac{d y}{d x}\) = y.
Solution:
The given differential equation is
(x + 2y³) \(\frac{d y}{d x}\) = y

Question 7.
(1 – x²) \(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-\mathbf{x}^2}\)
Solution:
Dividing by (1 – x²) both sides

Question 8.
x (x – 1) \(\frac{d y}{d x}\) – (x – 2) y = x³ (2x – 1)
Solution:
Dividing the given equation by x(x – 1) we get
\(\frac{d y}{d x}+\frac{(-x+2)}{x(x-1)} y=\frac{x^2(2 x-1)}{x-1}\)
Now \(\frac{2-x}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\)
∴ 2 – x = A (x – 1) + Bx
Put x = 1 both sides, 1 = B
and A + B = – 1
⇒ A = – 2
∴ \(\frac{2-x}{x(x-1)}=\frac{-2}{x}+\frac{1}{x-1}\)
∴ I.F = \(e^{\int \frac{2-x}{x(x-1)} d x}=e^{\int\left(\frac{-2}{x}+\frac{1}{x-1}\right) d x}\)
= e^{log (x – 1) – 2 log x}
= e^{log (x – 1) – log x2}
= \(\frac{x-1}{x^2}\)
∴ General Solution is \(y\left(\frac{x-1}{x^2}\right)=\int \frac{x^2(2 x-1)}{x-1}\left(\frac{x-1}{x^2}\right) d x\)
= ∫ (2x – 1) dx + c
= x² – x + c
∴ y (x – 1) = x² (x² – x + c) is the solution of the differential equation.

Question 9.
\(\frac{d y}{d x}\) (x² y³ + xy) = 1
Solution:

Question 10.
\(\frac{d y}{d x}\) + x sin 2y = x³ cos² y
Solution:
Dividing by cos² y we get
sec² y \(\frac{d y}{d x}\) + 2x tan y = x³ …………….(1)
[∴ \(\frac{\sin 2 y}{\cos ^2 y}=\frac{2 \sin y \cos y}{\cos ^2 y}\) = 2 tan y]
Let tan y = t then sec² y \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
∴ \(\frac{d t}{d x}\) + 2xt = x³
Here, P = 2x and Q = x³
∴ I.F = e^{∫ P dx} = e^x2
∴ t . e^x2 = ∫ x³ e^x2 dx
= ∫ x² . x . e^x2 dx
= \(\frac{1}{2}\) ∫ z . e^z dz
where z = x²
= \(\frac{1}{2}\) e^z (z – 1) + c
= \(\frac{1}{2}\) e^x2 (x² – 1) + c
∴ tan y e^x2 = \(\frac{1}{2}\) e^x2 (x² – 1) + c
∴ Solution of the given differential equation is tan y = \(\frac{1}{2}\) (x² – 1) + c e-^x2

Question 11.
y² + (x – \(\frac{1}{y}\)) \(\frac{d y}{d x}\) = 0.
Solution:

Where P = y^-2 and Q = y^-3
∴ I.F. = e^{∫ P dx}
= e∫^{y-2 dy}
= \(e^{-\frac{1}{y}}\)
∴ Solution is
x \(e^{-\frac{1}{y}}\) = ∫ \(e^{-\frac{1}{y}}\) y^-3 dy + c
= ∫ e^-y-1 y^-3 dy + c
= ∫ e^-y-1 y^-2 y^-1 dy + c
Let y^-1 = t then – y^-2 dy = dt
∴ x \(e^{-\frac{1}{y}}\) = – ∫ t e^-t dt + c
= – [- t e^-t + ∫ e^-t dt] + c
= t e^-t + e^-t + c
= e^-t (t + 1) + c
= \(e^{-\frac{1}{y}}\) (\(\frac{1}{y}\) + 1) + c
∴ x = (\(\frac{1}{y}\) + 1) + c \(e^{\frac{1}{y}}\)
⇒ xy = 1 + y + y . c\(e^{\frac{1}{y}}\)
∴ Solution of the given differential equation is xy = 1 + y + cy \(e^{\frac{1}{y}}\).

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 8 Measures of Dispersion to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Measures of Dispersion Important Questions

Question 1.
Find the mean deviation from the mean of the following data, using the step deviation method.

Solution:

Question 2.
The following table gives the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.

Solution:
We shall solve this problem using the step deviation method, since the mid points of the class intervals are numerically large.

Question 3.
An analysis of monthly wages paid to the workers of two firms A and B belonging to the same industry gives the following data.

i) Which firm A or B, has greater variability in individual wages?
ii) Which firm lias a larger wage bill?
Solution:

Since C.V. of firm B is greater than CV. of firm A, we can say that firm B has greater variability in individual wages.

(ii) Firm A has number of workers i.e., wage earners (n₁) = 500
Its average daily wage, say \(\overline{\mathbf{x}}_1\) Rs. 186
Since Average daily wage = \(\frac{Total Wages Paid}{no.of Workers}\), it follows that total wages paid to the workers
n₁\(\overline{\mathbf{x}}_1\) = 500 x 186 – Rs.93,000
Firm B has number of wage earners (n₂) – 600
Average daily wage, say \(\overline{\mathbf{x}}_2\) = Rs 175
∴ Total daily wages paid to the workers n₂\(\overline{\mathbf{x}}_2\) 600 x 175 = Rs. 1,05,000
Hence we see that firm B has larger wage bill.

Question 4.
The variance of 20 observations is 5. If each of the observations is multiplied by 2, find the variance of the resulting observations.
Solution:
Let the given observations be x₁, x₂ ……………….. x₂₀ and \(\overline{\mathbf{x}}\) be their mean.
Given that n = 20 and variance = 5

Note: From this example we note that, if each observation in a data is multiplied by a constant k, then the variance of the resulting observations is k² time that of the variance of original observations.

Question 5.
If each of the observations x₁, x₂, ………………….., x_n is increased by k, where k is a positive or negative number, their show that the variance remains unchanged.
Solution:
Let \(\overline{\mathbf{x}}\) be the mean of x₁, x₂, ………………….., x_n. Then their variance is given by
\(\sigma_1^2=\frac{1}{n} \sum_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)
If to each observation we add a constant k, then the new (changed) observations will be y_i = x_i+k …………….. (1)

Thus the variance of the new observations is the same as that of the original observations.
Note: We note that adding (or subtracting) a positive number to (or form) each of the given set of observations does not affect the variance.

Question 6.
The scores of two cricketers A and B in 10 innings are given below. Find who is a better run getter and who is a more consistent player.

Solution:

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 7 Partial Fractions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Partial Fractions Important Questions

Question 1.
Resolve \(\frac{5 x+1}{(x+2)(x-1)} \) into partial fraction.
Solution:
Let \(\frac{5 x+1}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\)
where A and B are non-zero numbers to be determined.
Then \(\frac{5 x+1}{(x+2)(x-1)}=\frac{A(x-1)+B(x+2)}{(x+2)(x-1)}\)
∴ A(x – 1)+B(x+2)5x+1
Putting x = 1 in (1), we get
3B = 5 +1 i.e.., B = 2
Putting x = – 2 in (1) we get
– 3A=-.9 i.e., A=3
∴ \(\frac{5 x+1}{(x+2)(x-1)}=\frac{3}{x+2}+\frac{2}{x-1}\)

Question 2.
Resolve \(\frac{2 x+3}{5(x+2)(2 x+1)}\) into partial fraction.
Solution:
Let \(\frac{2 x+3}{(x+2)(2 x+1)}=\frac{A}{x+2}+\frac{B}{2 x+1}\)
where A and B are non-zero real numbers, to be determined.

Question 3.
Resolve \(\frac{13 x+43}{2 x^2+17 x+30}\) into partial fraction.
Solution:
We have 2x² + 17x + 34) = (2x + 5) (X + 6)

∴ 13x+43 = A(x+6) + B(2x+ 5)
= (A + 2B)x + (6A + 5B)
Comparing the coefficients of like powers of x, we have
A+2B=13 and 6A+5B = 43
Solving these two equations.
we get A = 3 and R = 5
∴ \(\frac{13 x+43}{2 x^2+17 x+30}=\frac{3}{2 x+5}+\frac{5}{x+6}\)

Question 4.
Resolve \(\frac{x^2+5 x+7}{(x-3)^3}\) into partial fraction
Solution:

where A, B and C are constants determined.
∴ \(\frac{\mathrm{x}^2+5 \mathrm{x}+7}{(\mathrm{x}-3)^3}=\frac{\mathrm{A}(\mathrm{x}-3)^2+\mathrm{B}(\mathrm{x}-3)+\mathrm{C}}{(\mathrm{x}-3)^3}\)
∴ x²+ 5x + 7 = A x² +(B – 6A)x  + (9A – 3B + C)
Now comparing the cçefficients of like powers of x in (1), we get
A=1,B – 6A = 5, 9A – 3B +C = 7
Solving these equations, we get
A = 1, B = 11, C = 31

Question 5.
Resolve \(\frac{x^2+13 x+15}{(2 x+3)(x+3)^2}\) into partial fraction
Solution:
Here (2x + 3) is a linear factor and (x + 3) is, a repeated linear factor. We apply Rules I and-II, and write
\(\frac{x^2+13 x+15}{(2 x+3)(x+3)^2}\)

Question 6.
Resolve \(\frac{1}{(x-1)^2(x-2)}\) into partial fraction
Solution:

Equating the corresponding coefficients, we have
A+C=0, – 3A+ B – 2C = 0 2A-2B+C= 1
Solving these equations, we get
A= -1, B = – 1, C = 1

Question 7.
Resolve \(\frac{3 x-18}{x^3(x+3)}\) into partial fraction
Solution:

Equating the corresponding coefficients, we have
A+D = 0, 3AB=0, 3B+C=3, 3C=-18
Solving these equations, we get
A=-1,B=3,C=-6, D=1
∴ \(\frac{3 x-18}{x^3(x+3)}=\frac{-1}{x}+\frac{3}{x^2}-\frac{6}{x^3}+\frac{1}{(x+3)}\)

Question 8.
Resolve \(\frac{x-1}{(x+1)(x-2)^2}\) into partial fraction
Solution:

Question 9.
Resolve \(\frac{2 x^2+1}{x^3-1}\) into partial fraction
Solution:

∴ 2x²+1=(A+B)x²+(A-B+C)x+(A-C)
∴ Comparing the corresponding coefficients, we have
A B = 2, A – B + C = 0, A-C = 1
Solving these equations, we get
A = 1, B = 1, C = 0
∴ \(\frac{2 x^2+1}{\left(x^3-1\right)}=\frac{1}{x-1}+\frac{x}{x^2+x+1}\)

Question 10.
Resolve \(\frac{x^3+x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}\) into partial fraction.
Solution:

∴ x³+x²+1(A+C)x³+(B+D)x²+(3A+2C)x+(3B+2D) = 1
Comparing the corresponding coefficients, we have
A+C = 1, B+D = 1, 3A+2C = 0,3B+2D = 1
Solving these equations, we get
A=-2, B=-1,C=3,D= 2

Question 11.
Resolve \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}\) into partial fraction.
Solution:
x⁴ +x² +1=(x +1) – x
(x²+x+ t)(x²-x+ 1)

= (Ax+B)(x²-x+1)+(Cx+D)(x²+x+1)
=3x – 2x – 1 ……………….. (1)
Comparing the coefficients of
x³, x² , x and constants in (1), we get
A+C = 3 …………………….. (2)
– A+B+C+D – 2 …………………. (3)
A – B+C+D = 0 …………………… (4)
B+D = – 1 …………………… (5)
From (2), C = 3 – A …………………… (6)
and from (5), D = – 1 – B …………………… (7)
Putting these values in (3) we get
– A+ B + 3 – A – 1 – B  – 2
∴  – 2A = – 4 or A = 2
Similarly from (4) we get
A-B + 3 – A – 1 -B= 0
∴ – 2B=- 2 or B = 1
∴ From (6); C = 3 – A= 3 – 2= 1
and from(7); D= – 1 – B= – 1  – 1 = – 2
∴ \(\frac{3 x^3-2 x^2-1}{x^4+x^2+1}=\frac{2 x+1}{x^2+x+1}+\frac{x-2}{x^2-x+1}\)
To find the partial fractions of \(\frac{f(x)}{g(x)}\), when g(x) contains repeated irreducible factor, we use the following rule.

Question 12.
Resolve \(\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}\) into partial fraction.
Solution:

Comparing the corresponding coefficients.
we have
A = 0, B = 1. 2A C =0, 2B + D = 24,
A+C+E = 0 , B + D+F=28 .
Solving these equations, we get
A = 0,B = 1, C = 0, D= 22, E = 0, F=,5

Question 13.
Resolve \(\frac{x+3}{(1-x)^2\left(1+x^2\right)}\) into partial fraction.
Solution:

Question 14.
Resolve  \(\frac{x^3}{(2 x-1)(x+2)(x-3)}\) into partial fraction.
Solution:
The given fraction is improper with degree of the numerator equal to degree of denominator.

Question 15.
Resolve \(\frac{x^4}{(x-1)(x-2)}\) into partial fraction.
Solution:
By dividing x⁴ with (x – 1) (x -2) we get

Question 16.
Find the coefficient of x⁴ in the expansion of  \(\frac{3 x}{(x-2)(x+1)}\) in powers of specifying the interval in which the expansion is valid.
Solution:
Resolving the given fraction into partial fractions we get

Question 17.
Find the coefficient of xⁿ in the power series of expansion of \(\frac{x}{(x-1)^2(x-2)}\) specifying the region in which the expansion is valid.
Solution:
Resolving the given fraction into partial fractions, we get

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 10 Random Variables and Probability Distributions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

Question 1.
A cubical die is thrown. Find the mean and variance of X, giving the number on the face that shows up.
Solution:
Let S be the sample space and X be the random variable associated with S, where P(X) is given by the following table.

Question 2.
The probability distribution of a random variable X is given below:

Find the value of k and the mean and variance of X.
Solution:

Question 3.
If x is a random variable with probability distribution p(X=k)=\(\frac{(k+1) c}{2^k}\) k = 0,1,2 ………… then find c.
Solution:
Since p(X=k)=\(\frac{(k+1) c}{2^k}\) k = 0,1,2 ……….. is the probability distribution of x

Question 4.
Let X be a random variable such that
P(x=-2) = P(X = -1) = P(X=2)
P(X = 1) = \(\frac{1}{6}\) and P(X = 0) \(\frac{1}{3}\)
Find the mean and variance of X.
Solution:

Question 5.
Two dice are roiled at random. Find the probability distribution of the sum of the numbers on them. Find the mean of the random variable.
Solution:
When two dice are rolled, the sample space S consists of 6 x 6 = 36 sample points
S = ((1, 1), (1, 2), (1, 6), (2, 1), (2, 6),(6, 6)).
Let X denote the sum of the numbers on the two dice. Then the range of X = {(2, 3, 4 , 12)}
The probability distribution for X is given here under:

Question 6.
8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.
Solution:
In the experiment of tossing a coin, the probability of getting a head \(\frac{1}{2}\) and the probability of getting a tail \(\frac{1}{2}\). The probability of getting r heads in a random throw of 8 coins is

Question 7.
The mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X≥1).
Solution:
Here x = B(n,p) is specified by np = 4 = μ and npq = σ² = 3

Question 8.
The probability that a person chosen at random is left handed (in hand wilting) is 0.1. What is the probability that in a group of 10 people, there is one who is left handed.
Solution:
Heye n = 10, find p =\(\frac{1}{10}\) = 0.1.
Hence q = 0.9
We have to find P(X = 1); the probability that
exactly one out of 10 is left handed
P(X = 1) = ¹⁰C₁ p¹ q^10-1
= 10 x 0.1 x (0.9)⁹ = (0.9)⁹

Question 9.
In a book of 450 pages, there are 400 typo graphical errors. Assuming that the number of errors per page follow the Poisson law, find the probability that a random sample of 5 pages will contain no typo graphical error.
Solution:
The average number of errors per page in the book is \(\lambda=\frac{400}{450}=\frac{8}{9}\)
The probability that there are r errors per page:

Hence P(X=0) = e^-8/9
The required probability that a random sample of 5 pages will contain no error is [P(X) = 0)]⁵ = (e^-8/9)⁵

Question 10.
The deficiency of red cells In the blood cells is determined by examining a specimen of blood under a microscope. Suppose a small fixed volume contains on an average 20 red cells for normal persons. Using the Poisson distribution, find the probability that a specimen of blood taken from a normal person will contain less than 15 red cells.
Solution:
Here λ = 20.
Let P(X = r) denote the probability that a specimen taken from a normal person will contain r red cells.
Then we have P(X < 15)

Question 11.
A Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5).
Solution:

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(c)

I.
Question 1.
Express x dy – y dx = \(\sqrt{x^2+y^2}\) dx in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x dy – y dx = \(\sqrt{x^2+y^2}\) dx

Question 2.
Express (x – y tan^-1 \(\frac{y}{x}\)) dx + x tan^-1 \(\frac{y}{x}\) dy = 0 in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is

Question 3.
Express x \(\frac{d y}{d x}\) = y (log y – log x + 1) in the form \(F\left(\frac{y}{x}\right)=\frac{d y}{d x}\).
Solution:
Given equation is x \(\frac{d y}{d x}\) = y (log y – log x + 1)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\left[\log \left(\frac{y}{x}\right)+1\right]=F\left(\frac{y}{x}\right)\)

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{x-y}{x+y}\)
Solution:
Let y = vx then \(\frac{d y}{d x}\) = v + x . \(\frac{d v}{d x}\)
∴ The given differential equation is

Question 2.
(x² + y²) dy = 2xy dx.
Solution:
The given differential equation is (x² + y²) dy = 2xy dx

∴ 1 + v² = A (1 – v²) + Bv (1 + v) + Cv (1 – v)
take v = – 1, we get
2C = – 2
⇒ C = – 1
coefficient of v² gives
⇒ – A + B – C = 1
⇒ – A + B = 0
coefficient of v gives B + C = 0
⇒ B = 1
∴ A = 1
∴ \(\frac{1+v^2}{v-v^3}=\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\)
∴ From (1)
\(\int\left(\frac{1}{v}+\frac{1}{1-v}-\frac{1}{1+v}\right) \mathrm{d} v=\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
⇒ log v – log (1 – v – log (1 + v) – log x + log c
⇒ log v – log(1 – v) – log (1 + v) = log cx
⇒ log v – [Iog(1 – v²)] = log cx

⇒ yx = cx (x² – y²)
⇒ y = c (x² – y²).

Question 3.
\(\frac{d y}{d x}=\frac{-\left(x^2+3 y^2\right)}{3 x^2+y^2}\)
Solution:

Question 4.
y² dx + (x² – xy) dy = 0
Solution:

⇒ u – log v = log x + log c
⇒ v = log (cx v)
⇒ \(\frac{y}{x}\) = log (cx \(\frac{y}{x}\)) = log (cy)
⇒ cy = e^y/x is the solution.

Question 5.
\(\frac{d y}{d x}=\frac{(x+y)^2}{2 x^2}\)
Solution:
Let y = vx
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

2 tan^-1 v = log x + log c
2 tan^-1 (\(\frac{y}{x}\)) = log cx is the solution of the differential equation.

Question 6.
(x² – y²) dx – xy dy = 0
Solution:

Question 7.
(x²y – 2xy²) dx = (x³ – 3x²y) dy
Solution:

Question 8.
y² dx + (x² – xy + y²) dy = 0
Solution:

∴ 1 + v + v² = A (1 + v²) + (Bv + C) v
Comparing coefficient of v²,
A + B = 1
Also A = 1,
∴ B = 0.
Comparing coefficient of v,
C = – 1
∴ \(\frac{1-v+v^2}{v\left(1+v^2\right)}=\frac{1}{v}-\frac{1}{1+v^2}\)
∴ From (1)
\(\int \frac{1}{v} \mathrm{~d} v-\int \frac{1}{1+v^2} \mathrm{~d} v=-\int \frac{\mathrm{dx}}{\mathrm{x}}+\log \mathrm{c}\)
log v – tan^-1 v = – log x + log c

Question 9.
(y² – 2xy) dx + (2xy – x²) dy = 0
Solution:
The given equation is (y² – 2xy) dx = – (2xy – x²) dy

2v – 1 = A (1 – v) + 3Bv
Put v = 1,
1 = 3B
⇒ B = \(\frac{1}{3}\)
Also, – A + 3B = 2
⇒ 3B = 2 + A
⇒ 1 = 2 + A
⇒A = – 1
∴ \(\frac{2 v-1}{3 v(1-v)}=-\frac{1}{3 v}+\frac{1}{3} \frac{1}{1-v}\)
∴ From (1)

Question 10.
\(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
Given \(\frac{d y}{d x}+\frac{y}{x}=\frac{y^2}{x^2}\)

v – 2 = c² x² . v
\(\frac{y}{x}\) – 2 = c²x² \(\frac{y}{x}\)
y – 2x = c²x³ \(\frac{y}{x}\)
= c²x²y
= kx²y
where c² = k
∴ Solution of the given equation is y – 2x= kx²y.

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\) dx
Solution:
x dy = (y + \(\sqrt{x^2+y^2}\)) dx

Question 12.
(2x – y) dy = (2y – x) dx
Solution:
Given \(\frac{d y}{d x}=\frac{2 y-x}{2 x-y}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

⇒ (y – x)² = c² (y + x)² (y² – x²)
⇒ y – x = c² (y + x)³
⇒ (x + y)³ = c (x – y) where c = \(-\frac{1}{c^2}\) (constant)
Solution of the given differential equation is (x + y)³ = c (x – y)

Question 13.
(x² – y²) \(\frac{d y}{d x}\) = xy.
Solution:
The given equation is \(\frac{d y}{d x}\) = \(\frac{x y}{x^2-y^2}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
v + x \(\frac{d v}{d x}\) = \(\frac{\mathrm{x}(v \mathrm{x})}{\mathrm{x}^2-v^2 \mathrm{x}^2}=\frac{v}{1-v^2}\)

∴ x² = – 2y² [log c + log y]
⇒ x² + 2y² (c + log y) = 0 is the solution of the given equation where log c = c.

Question 14.
2 \(\frac{d y}{d x}\) = \(\frac{y}{x}+\frac{y^2}{x^2}\)
then \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ 2 [v + x \(\frac{d v}{d x}\)] = v + v²
∴ 2v + 2x \(\frac{d v}{d x}\) = v + v²
⇒ 2x \(\frac{d v}{d x}\) = v + v² – 2v = v² – v
⇒ \(\frac{\mathrm{d} v}{v^2-v}=\frac{\mathrm{dx}}{2 \mathrm{x}}\) ………..(1)
writing in variable separable lorm
⇒ \(\frac{1}{v^2-v}=\frac{1}{v(v-1)}=\frac{\mathrm{A}}{v}+\frac{\mathrm{B}}{v-1}\)
∴ 1 = A (v – 1) + Bv
Put v = 1 then B = 1, and A + B = 0
⇒ A = – B = – 1
∴ \(\frac{1}{v^2-v}=-\frac{1}{v}+\frac{1}{v-1}\)
∴ From (1)

⇒ (y – x)² = y²xc²
⇒ c₁ (x – y)² = y²x
where c₁ = \(\frac{1}{c^2}\)
∴ The solution of the given differential equation is y²x = c₁ (x – y)².

III.
Question 1.
Solve: (1 + e^x/y) dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0
Solution:
The given equation is
(1 + e^x/y) dx + e^x/y (1 – \(\frac{x}{y}\)) dy = 0
⇒ (1 + e^x/y) \(\frac{d x}{d y}\) + e^x/y (1 – \(\frac{x}{y}\)) = 0
Let \(\frac{x}{y}\) = v then x = vy
∴ \(\frac{d x}{d y}\) = v + y \(\frac{d v}{d y}\)
∴ From (1)
(1 + e^y) (v + y \(\frac{d v}{d y}\) ) + e^v (1 – v) = 0

⇒ log (e^v + v) = – log y + log c
⇒ e^v + v = \(\frac{c}{y}\)
⇒ e^x/y + \(\frac{x}{y}\) = \(\frac{c}{y}\)
⇒ ye^x/y + x = c is the solution of the given equation.

Question 2.
Solve x sin \(\frac{y}{x}\) . \(\frac{d y}{d x}\) = y sin \(\frac{4}{4}\) – x.
Solution:

⇒ – cos v = – log x + log c
⇒ cos v = log x + log c = log (cx)
⇒ cos (\(\frac{y}{x}\)) = log (cx)
∴ The solution of the given equation is cx = e^cos(y/x).

Question 3.
x dy = (y + x cos² \(\frac{y}{x}\)) dx.
Solution:
Given x dy = (y + x cos² \(\frac{y}{x}\)) dx

⇒ ∫ sec² v dv = ∫ \(\frac{\mathrm{dx}}{\mathrm{x}}\) + c
⇒ tan v = log x + c
⇒ tan (\(\frac{y}{x}\)) = log x + c is the solution.

Question 4.
Solve (x – y log y + y log x) dx + x (log y – log x) dy = 0.
Solution:
Given (x – y log y + y log x) dx + x (log y – log x) dy = 0
⇒ x (log y – log x) dy = – (x – y log y + log x) dx

Writing in variable separable form we get
log v dv = – \(\frac{d x}{x}\)
∴ ∫ log v dv = – ∫ \(\frac{d x}{x}\) + c
⇒ v log v – v = – log x + c
⇒ \(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
∴ Solution of the given differential equation is
\(\frac{y}{x} \log \left(\frac{y}{x}\right)-\frac{y}{x}\) = – log x + c
⇒ \(\frac{y}{x}\) (log y – log x) – \(\frac{y}{x}\) = – log x + c
⇒ y log y – y log x – y – x log x + cx
⇒ y log y + log x [(x – y)] = y + Cx.

Question 5.
Solve (y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))
Solution:
Given equation is
(y dx + x dy) x cos (\(\frac{y}{x}\)) = (x dy – y dx) y sin (\(\frac{y}{x}\))

Question 6.
Find the equation of a curve whose gradient is \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\) where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Given gradient of the curve as \(\frac{d y}{d x}=\frac{y}{x}-\cos ^2 \frac{y}{x}\)
Let y = vx then
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ v + x \(\frac{d v}{d x}\) = v – cos² v
⇒ x \(\frac{d v}{d x}\) = – cos² v
⇒ x \(\frac{d v}{d x}\) = – cos² v
⇒ ∫ sec² v dv = – ∫ \(\frac{d x}{x}\)
⇒ tan v = – log x + log c
⇒ \(\tan \left(\frac{y}{x}\right)=\log \left(\frac{\mathrm{c}}{\mathrm{x}}\right)\)
Given curve passes through (1, \(\frac{\pi}{4}\)) we have
tan (\(\frac{\pi}{4}\)) = log (c)
⇒ c = e
∴ Solution of the given differential equation is
\(\tan \left(\frac{y}{x}\right)=\log \left(\frac{e}{x}\right)\)
= log e – log x
∴ Equation of the required curve is
tan \(\frac{y}{x}\) = 1 – log x.

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(b)

I.
Question 1.
Find the general solution of \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0.
Solution:
Given equation is \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0

⇒ sin^-1 y = – sin^-1 x + c
⇒ sin^-1 x + sin^-1 y + c is the general solution.

Question 2.
Find the general solution of \(\frac{d y}{d x}=\frac{2 y}{x}\).
?Solution:
The given equation can be written in variable separable form as \(\frac{d y}{d x}=\frac{2 y}{x}\).
⇒ \(\frac{\mathrm{dy}}{\mathrm{y}}=2\left(\frac{\mathrm{dx}}{x}\right)\)
⇒ log |y| = 2 log |x| + log c₁
⇒ log y = log x² + log c
⇒ log \(\left(\frac{y}{x^2}\right)\) = log c
⇒ y = cx²
⇒ x² = \(\frac{1}{c}\) y
⇒ x² = c₁y where c₁ is a constant is the general solution.

II. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
Solution:
The given equation can be written in variable seperable form as
\(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
∴ \(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\)
⇒ tan^-1 y = tan^-1 x + tan^-1 c
⇒ tan^-1 y = tan^-1 x + tan^-1 c is the solution of the given differential equation.

Question 2.
\(\frac{d y}{d x}\) = e^y-x
Solution:
The given equation can be written in variable separable form as
\(\frac{d y}{d x}\) = e^y-x
⇒ \(\frac{d y}{e^y}=\frac{d x}{e^x}\)
⇒ ∫ e^-y dy = ∫ e^-x dx
⇒ – e^-y dy = – e^-x + c
⇒ e^-x – e^-y = c is the solution 0f the given differential equation.

Question 3.
(e^x + 1) y dy + (y + 1) dx = 0.
Solution:
The given differential equation can be written as (e^x + 1)y dy = – (y + 1) dx
⇒ \(\frac{y d y}{y+1}=-\frac{d x}{e^x+1}\)
⇒ \(\left[\frac{(y+1)-1}{y+1}\right] d y=\frac{-e^{-x}}{1+e^{-x}} d x\)
∴ ∫ dy – ∫ \(\frac{d y}{y+1}\) dx = ∫ \(\frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}}\) dx
⇒ y – log (y + 1) = log (1 + e^-x)
⇒ y = log (y + 1) + log (1 + e^-x) + log c
= log [(y + 1) (e^-x + 1) c]
∴ e^y = c(y + 1) (e^-x + 1)
∴ The solution of the given equation is
e^y = c (y + 1) (1 + e^-x).

Question 4.
\(\frac{d y}{d x}\) = e^x-y + x²e^-y
Solution:
\(\frac{d y}{d x}\) = e^x-y + x²e^-y
= e^-y (e^x + x²)
Writing in variable separable form we get
\(\frac{d y}{d x}\) = \(\frac{1}{y}\) (e^x + x²)
⇒ ∫ e^y dy = ∫ (e^x + x²) dx
⇒ e^y = e^x + \(\frac{x^3}{3}\) + c
The solution of the given equation is e^y = e^x + \(\frac{x^3}{3}\) + c.

Question 5.
tan y dx + tan x dy = 0.
Solution:
The given equation can be written as
\(\frac{d x}{\tan x}+\frac{d y}{\tan y}\) = 0
⇒ \(\int \frac{\mathrm{dx}}{\tan x}+\int \frac{\mathrm{dy}}{\tan y}\) = 0
⇒ ∫ cot x dx + ∫ cot y dy = 0
⇒ log (sin x) + log (sin y) = log c
⇒ sin x sin y = c is the solution of the given differential equation.

Question 6.
\(\sqrt{1+x^2} d x+\sqrt{1+y^2} d y\) = 0
Solution:
The given equation can be written in variable seperable form as

Question 7.
y – x\(\frac{d y}{d x}\) = 5 (y² + \(\frac{d y}{d x}\))
Solution:
The given differential equation is

∴ 1 = A (1 – 5y) + By
⇒ A = 1 and B – 5A = 0
⇒ B = 5
∴ \(\int \frac{1}{y-5 y^2} d y=\int \frac{1}{y} d y+\int \frac{5}{1-5 y} d y\)
= log |y| – log (1 – 5y)
∴ From (1)
log |x + 5| = log |y| – log (1 – 5y) + log c
⇒ x + 5 = \(\frac{c y}{1-5 y}\)
∴ Solution of the given dillerential equation is 5 + x = \(\frac{c y}{1-5 y}\)

Question 8.
\(\frac{d y}{d x}=\frac{y(x+1)}{x(y+1)}\)
Solution:
The given equation \(\frac{d y}{d x}=\frac{y(x+1)}{x(y+1)}\) writing in variable separable form
\(\frac{(y+1) d y}{y}=\frac{(x+1) d x}{x}\)
\(\int\left(\frac{y+1}{y}\right) d y=\int \frac{(x+1) d x}{x}\)
⇒ y + log |y| = x + log |x| + log c
⇒ y – x = |og |x| – log |y| + log c
= log \(\left(\frac{c x}{y}\right)\)
∴ y – x = log \(\left(\frac{c x}{y}\right)\) is the solution.

III. Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}=\frac{1+y^2}{\left(1+x^2\right) x y}\)
Solution:
The given equation is \(\frac{d y}{d x}=\frac{1+y^2}{\left(1+x^2\right) x y}\) which can be written in variable separable form as

⇒ (1 + x²) (1 + y²) = c²x²
= cx² where c² = c is a constant
∴ The solution of the given differential equation is
(1 + x²) (1 + y²) = cx²

Question 2.
\(\frac{d y}{d x}\) + x² = x² e^3y
Solution:
The given equation can be written in variable separable form as
\(\frac{d y}{d x}\) = x² e^3y – x²
= x² (e^3y – 1)
\(\frac{d y}{e^{3 y}-1}\) = xsup>2 dx
∴ ∫ \(\frac{d y}{e^{3 y}-1}\) = ∫ x² dx
⇒ ∫ \(\left(\frac{e^{-3 y}}{1-e^{-3 y}}\right)\) dy = \(\frac{x^3}{3}\) + c
⇒ log (1 – e^-3y) = x³ + c
⇒ 1 – e^-3y = e^x3 + e^c
= k e^x3
∴ The solution of the given differential equation is 1 – e^-3y = k e^x3.

Question 3.
(xy² + x) dx + (yx² + y) dy = 0
Solution:
The given dilferential equation can be written as
x (y² + 1) dx + y(x² + 1) dy = 0 which can be expressed in variable separable form as
\(\frac{1}{2} \int \frac{2 x d x}{x^2+1}+\frac{1}{2} \int \frac{2 y d y}{y^2+1}\) = 0
⇒ \(\frac{1}{2}\) log(x² + 1) + log (y² + 1) = log c
⇒ log \(\sqrt{\mathrm{x}^2+1}\) + log \(\sqrt{\mathrm{y}^2+1}\) = log c
⇒ (1 + x²) (1 + y²) = c².

Question 4.
\(\frac{d y}{d x}\) = 2y tanh x
Solution:
The given equation is \(\frac{d y}{d x}\) = 2y tanh x.
\(\frac{d y}{y}\) = 2 tanh x (variable separable form)
∫ \(\frac{d y}{y}\) = 2 ∫ tanh x dx
⇒ log y = 2 log |cosh x| + log c
= log |cosh² x| + log c
log y = log (c cosh² x)
y = c .cosh² x which is the solution of the given differential equation.

Question 5.
Sin^-1 (\(\frac{d y}{d x}\)) = x + y
Solution:
The given equation is Sin^-1 (\(\frac{d y}{d x}\)) = x + y
⇒ sin (x + y) = \(\frac{d y}{d x}\) …………..(1)

⇒ ∫ sec² dz – ∫ sec z tan x dx = ∫ dx + c
⇒ tan z – sec z = x + c
⇒ tan (x + y) – sec (x + y) = x + c is the solution of the given differential equation.

Question 6.
\(\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}\) = 0
Solution:
Given equation in variable separable form is
\(\frac{d y}{y^2+y+1}=-\frac{d x}{x^2+x+1}\)

Question 7.
\(\frac{d y}{d x}\) = tan² (x + y)
Solution:
Let x + y = z then 1 + \(\frac{d y}{d x}\) = \(\frac{d z}{d x}\)
from the given equation
∴ 1 + \(\frac{d y}{d x}\) = 1 + tan² (x + y)
⇒ \(\frac{d z}{d x}\) = sec² z
⇒ ∫ \(\frac{\mathrm{d} z}{\sec ^2 z}\) = ∫ dx + c
⇒ ∫ cos² z dz = x + c
⇒ ∫ \(\left(\frac{1+\cos 2 z}{2}\right)\) dz = x + c
⇒ \(\frac{1}{2}\) z + \(\frac{1}{4}\) sin 2z = x + c
⇒ 2z + sin 2z = 4x + 4c
⇒ 2 (x + y) + sin 2 (x + y) = 4x + 4c
⇒ sin 2 (x + y) = 2x – 2y + 4c
= 2x – 2y + c₁
where c₁ = 4c.

Students must practice these TS Inter 2nd Year Maths 2A Important Questions Chapter 5 Permutations and Combinations Important Questions to help strengthen their preparations for exams.

TS Inter 2nd Year Maths 2A Permutations and Combinations Important Questions Important Questions

Question 1.
If ⁿP₄= 1680, find n.
Solution:
We know that ‘P4 is the product of 4 consecutive integers of which n is the largest.
That is ⁿP₄ = n(n – 1) (n – 2) (n – 3) and 1680 = 8  x 7 x 6 x 5
on comparing the largest integers, we get n = 8.

Question 2.
If ¹²P_r = 1320, find r.
Solution:
1320 = 12 x 11 x 10= ¹²P₃ .
Thus r = 3.

Question 3.
If ⁽ⁿ⁺¹⁾P₅ : ⁿP₅ = 3 : 2, find n.
Solution:

Question 4.
If ⁵⁶_(r+6) : ⁵⁴P_(r+3) = 30800 : I, find r.
Solution:

Question 5.
In how many ways 9 mathematics papers can be arranged so that the best and the worst
(i) may come together
(ii) may not come together?
Solution:
(i) lf the best and worst papers are treated as one unit, then we have 9 – 2 + 1 = 7+ 1+ 8 papers.
Now these can be arranged in (7+1) ! ways and the best and worst papers between themselves can be permuted in 2 ! ways. Therefore the number of arrangements in which best and worst papers come together is 8 ! 2 !

(ii) Total number of ways of arranging 9 mathematics papers is 9! . The best and worst papers come together in 8! 2! ways. Therefore the number of ways they may not come together is 9! – 8! 2! = 8!(9-2)= 8 ! × 7.

Question 6.
Find the number of ways of arranging 6 boys and 6 girls In a row. In how many of these arrangements
(i) all the girls are together
(ii) no two girls are together
(iii) boys and girls come alternately?
Solution:
6 boys + 6 girls = 12 persons. They can be arranged in a row in (12) ! ways.
(i) Treat the 6 girls as one unit. Then we have 6 boys + 1 unit of girls. They can be arranged in 7! ways. Now, the 6 girls among themselves can be permuted in 6! ways. Hence, by the fundamental principle, the number of arrangements in which all 6 girls are together 7! x 6!.

(ii) First we arrange 6 boys in a row in 6! ways. The girls can be arranged in the 7 gaps between the boys. These gaps are shown below by the letter X.

Now, the girls can be arranged In these 7 gaps in ⁷P₆ ways. Hence, by the fundamental principle, the number of
arrangements in which no two girls come together is 6! x ⁷P₆ = 6! x 7! = 7 x 6! x 6!.

(iii) Let us take 12 places. The row may begin with either a boy or a girl. That is, 2 ways. If it begins with a boy, then all odd places (1, 3, 5, 7, 9, Ii) will be occupied by boys and the even places (2, 4, 6, 8. 10, 12) by girls. The 6 boys can be arranged in the 6 odd places in 6! ways and the 6 girls can be arranged in the 6 even places in 6! ways. Thus the number of arrangements in which boys and girls come alternately is 2 x 6! x 6!.

Note: In the above, one may think that ques tions (ii) and (iii) are same. But they are not (as evident Irom the answers). In Question (ii), after arranging 6 boys, we found 7 gaps and 6 girls are arranged in these 7 gaps. Hence one place remains vacant. It can be any one of the 7 gaps. But in Question (iii), the vacant place should either be at the beginning or at the ending but not in between. Thus, only 2 choices for the vacant place.

Question 7.
Find the number of 4 letter words that can he formed using the letters of the word
MIRACLE. How many of them
(i) begin with an vowel
(ii) begin and end with vowels
(iii) end with a consonant?
Solution:
The word MIRACLE has 7 letters. Hence the number of 4 letter worlds that can be formed using these letters is
⁷P₄ = 7 x 6 x 5 x 4 = 840. Let us take 4 blanks.

(i) We can fill the first place with one of the 3 vowels (1, A, E) ³P₁ days. Now, the remaining 3 places can be filled using the remaining 6 letters in ⁶P₃ 120 ways. Thus the number of 4 letter words that begin with an vowel is 3 x 120 360.

(ii) Fill the first and last places with 2 vowels in ⁶P₂ 6 ways. The remaining 2 places can be filled with the remaining 5 letters in ⁵P₂ = 20 ways. Thus the number of 4 letter words that begin and end with vowels is 6 x 20= 120.

(iii) We can fill the last place with one of the 4 consonants (M, R, C, L) in ⁴P₁ = ways. The remaining 3 places can be filled with the letters in ⁶P₃ ways. Thus the number of 4 letter words that end with an vowel is 4 x ⁶P₃ = 4 x 120 = 480.

Question 8.
Find the number of ways of permuting the letters of the word PICTURE so that
(i) all vowels come together
(ii) no two vowels come together.
(iii) the relative positions of vowels and consonants are not disturbed.
Solution:
The word PICTURE has 3 vowels (I, U, E) and 4 consonants (P, C, T, R).
(i) Treat the 3 vowels as one unit. Then we can arrange 4 consonants + 1 unit of vowels in 5! ways. Now the 3 vowels among themselves can be permuted in 3! ways. Hence the number of permutations in which the 3 vowels come together is 5! x 3! 720.

(ii) First arrange the 4 consonants in 4! ways. Then in between the vowels, in the beginning and in the ending, there are 5 gaps as shown below by the letter X.

In these 5 places we can arrange the 3 vowels 5P3 ways. Thus the number of words in which rio two vowels come
to gether is 4! x ⁵P₃ = 24 x 60 = 1440.

(iii)The three vowels can be arranged in thier relative positions in 3’ ways and the 4 consonants can be arranged in their relative positions in 4 ways. The required number of arrangements is 3! . 4! = 144.

Note: In the above problem, from (i) we get that the number of permutations in which the vowels do not come together is = Total number of permutations – number of permutations in which 3 vowels come together.
7! – 5!. 3! = 5040 – 720 = 4320.

But the number of permutations in which no two vowels come together is only 1440. In the remaining 2880 permutations, two vowels come together and third appears away from these.

Question 9.
If the letters of the word PRISON are permuted in all possible ways and the words thus formed are arranged In dictionary order, find the rank of the word PRISON.
Solution:
The letters of the given word in dictionary order is
N  O  P  R  S
In the dictionary order, first we write all words that begin with I. If we fill the first place with I, the remaining 5 places can be filled with the remaining 5 letters in 5! ways. That is, there are 5! words that begin with I. Proceeding like this, after writing all words that begin with I, N, O, we have to write the words begin with P. Among them first come the words with first two letters P, I. As above there are 4! such words. On proceeding like this, we get

Hence the rank of the word PRISON Is
3×5! + 3×4! + 2×2! + 1! + 1
= 360+72+4+1 + 1= 438

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8 (without repetition). How many of them are divisible by
(i) 2 (ii) 3 (iii) 4 (iv) 5 (v)25
Solution:
The number of 4 digit numbers that can be formed using the 5 digit 2, 3, 5, 6, 8 is = 120.
(i) Divisible by 2: For a number to be divisible by 2, the units place should be filled with an even digit. This can be done in 3 ways (2 or 6 or 8).

Now, the remaining 3 places can be filled with the remaining 4 digits in = 24 ways. Hence, the number of 4-digit numbers divisible by 2 is 3 x 24 = 72.

(ii) Divisible by 3: A number is divisible by 3 ii the sum of the digits in it is a multiple of 3. Since the sum of the given 5 digits is 24, we have to leave either 3 or 6 and use the digits 2, 5, 6, 8 or 2, 3, 5, 8. In each cae, we can permute them In 4! ways. Thus the number of 4 – digit numbers divisible by 3 is 2 x 4! = 48.

(iii) Divisible by 4 : A number is divisible by 4 if the number formed by the digit in the last two places (tens and units places) is a multiple of 4.

Thus we fill the last two places (as shown in the figure) with one of 28,32,36,52,56,68 That is done in 6 ways. After filling the last two places, we can fill the remaining two places with the remaining 3 digits in ³P₂ ways.
Thus, the number of 4 – digit numbers divisible by 4 is 6 ×6=36.

(v) Divisible by 5 : After filling the units place with 5 (one way), the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ = 24 ways. Hence the number of 4 digit numbers divisible by 5 is 24.

(vi) Divisible by 25 : Here also we have to fill. the last two places (that is, units and tens place) with 25 (one way) as shown below.

Now the remaining 2 places can be filled with the remaining 3 digits in ³P₂ = 6 ways. Hence the number of 4 digit numbers divisible by 25 is 6.

Question 11.
Find the sum of all 4-digit numbers that can be fonned using the digits 1, 3, 5, 7, 9.
Solution:
We know that the number of 4-digit numbers that can be formed using the given ⁵P₄ digits is = 120. Now we find their sum. We first find the sum of the digits in the unit place of all these 120 numbers. If we fill the units place with 1 as shown below

then the remaining 3 places can be filled with the remaining 4 digits in ⁴P₃ ways. This means, the number of 4 digit numbers having 1 in units place is ⁴P₃ . Similarly, each of the digits 3, 5, 7, 9 appear 24 times in units place. By adding aB these digits we get the sum of the digits in units place of all 120 numbers as

Similarly, we get the sum of the digits in tens place as ⁴P₃ x 25.
Since it is in 10’s place, its value is = ⁴P₃ x 25 x 10.
Similarly, the value of the sum of the digits in 100s place and 1000s place are ⁴P₃ x 25 x 100 and ⁴P₃ x 25 x 1000
respectively. Hence the sum of the 4 digit numbers formed by using the digits 1, 3,5, 7, 9 is.
⁴P₃ x 25 x 1+⁴P₃ x 25 x 10 + ⁴P₃ x 25 x 100
= ⁴P₃ x 25 x 1000
= ⁴P₃ x 25 x 1111 ……………………. (*)
= 24 x 25 x  1111 = 6,66,600

Note:
1. From (*) in the above example, we can derive that the sum of all r-digit numbers that can be formed using the given ‘n non-zero digits (1 ≤ r ≤ n ≤ 9) is
^(n-1)P_^(r-1) x sum of the given digits x 111 …. 1 (r times)

2. In the above, if ‘0’ is one digit among the given n digits, then we get the sum of the r – digit numbers that can be formed using thegiven n digits (including ‘0’)

Question 12.
How many four digited numbers can be formed using the digits 1, 2, 5,7, 8, 9? How many of them begin with 9 and end with 2?
Solution:
The number of four digited numbers that can be formed using the given digits 1, 2, 5, 7, 8, 9 is ⁶P₄ = 360. Now, the first place and last place can be filled with 9 and 2 in one way.

The remaining 2 places can be filled by the remaining 4 digits 1, 5, 7, 8. Therefore these two places can be filled in ⁴P₂ ways. Hence, the required number of ways = 1 . ⁴P₂ = 12.

Question 13.
Find the number of injections of a set A with 5 elements to a set B with 7 elements.
Solution:
The first element of A can be mapped to any one of the 7 elements in 7 ways. The second element of A can be mapped to any one of the remaining 6 elements in 6 ways. Proceeding like this we get the number of injections from
A to B as ⁷P₅ = 2520.

Note : If a set A has m elements and set B has n elements, then the number of injections from A into B is ⁿP_m if m≤n and 0 if m > n.

Question 14.
Find the number of ways in which 4 letters can be put in 4 addressed envelopes so that no letter goes into the envelope meant for it.
Solution:
Required number of ways is
\(4 !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)=12-4+1=9\)

Note : If there are n things is a row, a permutation of these n things such that none of them occupies its original position is called a derangement of n things. The number of derangements of n distinct things is
\(\mathrm{n} !\left(\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots . .+(-1)^{\mathrm{n}} \frac{1}{\mathrm{n} !}\right)=9\)

Question 15.
Find the number of 5-letter words that can be formed using the letters of the word ‘ MIXTURE which begin with an vowel when repetitions are allowed.
Solution:
We have to fill up 5 blanks using the letters of the word MIXTURE having 7 letters among which there are 3 vowels. Fill the first place with one of the vowels (I or U or E) in 3 ways as shown below.

Each of the remaining 4 places can be filled in 7 ways (since we can use all 7 letters each time). Thus the number of 5 letter words is 3 x 7 x 7 x 7 x 7 3 x 7⁴.

Question 16.
Find the number of functions from a set A with in elements to a set B with n elements.
Solution:
Let A {a₁,a₂, ……………….. a_m} and B {b₁, b₂,…., b_n} To define the image of a₁ we have n choices (any element of B). Then we can define the image of a₂ again in n ways (since a₁, a₂ can have same image). Thus we can define the image of each of the m elements in n ways. Therefore the number of functions from A to B is n x n x …………x n (m times) = nm.

Question 17.
Find the number of surjections from a set A with n elements to a set B with 2 elements when n > i.
Solution:
Let A {a₁,a₂, ……………….. a_n}and B = {x, y}. From the above problem. the total number of functions from A onto B is 2. For a function to be a surjection its range should contain both x, y. Observe that the number of functions which are not surjections that is, the functions which contain x or y alone in the range is 2. Hence the number of surjections from A to B is 2ⁿ – 2.

Note: In the above problem. even if B has more than 2 elements also we can derive a formula to find the number of surjections from A to B. But this result is beyond the scope of this book and hence it is not included here.

Question 18.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 that ate divisible by
(i) 2
(ii) 3 when repetition is allowed.
Solution:
(i) Numbers divisible by 2:
Take 4 blanks. For a number to be divisible by 2, the units place should be filled with an even digit. This can be done in 3
ways (2 or 4 or 6).

Now, each of the remaining 3 places can be filled in 6 ways. Hence the number of 4 digit numbers that are divisible by 2 is 3 x 6³ = 3 x 216 = 648.

(ii) Numbers divisible by 3:
Fill the first 3 places with the given 6 digits in 6³ ways.

Now, after filling up the first 3 places with three digits, if we fill up the units place in 6 ways, we get 6 consecutive positive integers. Out of any six consecutive integers exactly two are divisible by 3. Hence the units place can be filled in 2 ways. hence the number of 4 digit numbers divisible by 3 is 2 x 216 = 432.

Question 19.
Find the number of 5- letter words that can be formed using the letters of the word Explain that begin and end with a vowel when repetitions are allowed.
Solution:
We can fill the first and last places with vowels each in 3 ways (E or A or I).

Now each of the remaining 3 places can be filled in 7 ways (using any letter of given 7 letters). Hence the number of 5 letter words which begin and end with vowels is 3² x 7³ = 9 x 343 = 3087.

Question 20.
Find the number of ways of arranging 8 men and 4 women around a circular table. In how many of them
(i) all the women come together
(ii) no two women come together.
Solution:
Total number of persons = 12 (8 men + 4 women)
Therefore, the number of circular permutations is (1 1)

(i) Treat the 4 women as single unit. Then we have 8 men. 1 unit of women = 9 entities. They can be arranged around a circular table In 8! ways. Now the 4 women among themselves can be arranged in 4! ways. Hence by the Fundamental principle, the required number of arrangements is 8! x 4!.’

(ii)

First we arrange 8 men around the circular table in 7! ways. There are 8 places In between them as shown In figure by the symbol x. (one place in between any two consecutive men).

Now we can arrange the 4 women in these 8 places in ⁸P₄ ways. Thus, the number of circular permutations in which no
two women come together is 7! x ⁸P₄.

Question 21.
Find the number of ways of seating 5 indians, 4 Americans and 3 Russians at a round table so that
i) all Indians sit together
ii) no two Russians sit together
iii) persons of same nationality sit together.
Solution:
(i) Treat the 5 indians as one unit. Then we have 4 Americans + 3 Russians + 1 unit of Indians = 8 entities.
They can be arranged at a round table in (8 – 1)! = 7! ways. Now, the 5 Indians among themselves can be arranged in 5! ways. Hence, the required number of arrangements is 7! x 5!.

(ii) First we arrange the 5 Indians + 4 Americans around the table in (9 – 1)! = 8! ways. Now, there are 9 gaps in between these 9 persons (one gap between any two consecutive persons). The 3 Russians can be arranged in these 9 gaps in ⁹P₃ ways. Hence, the required number of arrangments is 8! x ⁹P₃.

(iii)Treat the 5 Indinas as one unit, the 4 Americans as one unit and the 3 Russians as one unit. These 3 units can be
arranged at round table in (3 – 1)! = 2! ways.

Now, the 5 Indians among themselves can be permuted in 5! ways. Similarly, the 4 Americans in 4! ways and 3 Russians in 3! ways. Hence, the required number of arrangments is 2! x 5! x 4! x 3!.

Question 22.
Find the number of different chains that can be prepared using 7 different coloured beads.
Solution:
We know that the number of circular permutations of hanging type that can be formed using n things is \(\frac{1}{2}\) ((n -1)!). Hence the number of different ways of preparing the chains = \(\frac{1}{2}\{(7-1) !\}=\frac{6 !}{2}=360\)

Question 23.
Find the number of different ways of preparing a garland using 7 distinct red roses and 4 distinct yellow roses such that no two yellow roses come together.
Solution:
First we arrange 7 red roses in a circular form (garland form) in (7 – 1)! = 6! ways. Now, there are 7 gaps in between the red roses and we can arrange the 4 yellow roses in these 7 gaps ⁷P₄ ways. Thus the total number of circular permutations is 6! x ⁷P₄. But, this being the case of garlands, clock wise and anti-clock-wise arrangements look a like. Hence the required number of ways is \(\frac{1}{2}\) (6! x ⁷P₄)

Question 24.
Find the number of ways of arranging the letters of the word SINGING so that
i) they begin and end with l
ii) the two G’s come together
iii) relative positions of vowels and consonants are not disturbed.
Solution:
(i) First we fill the first and last places with
I’s in \(\frac{2 !}{2 !}\) = 1 way as shown below

Now, we fill the remaining 5 places with the remaining 5 letters S, N, G, N, G in
\(\frac{5 !}{2 ! 2 !}\)  = 30 ways.
Hence, the number of required permutations is 30.

(ii) Treat the two G’s as one unit. Then we have 6 letters In which there are 2I’s and 2N’s.
Hence they can be arranged in
\(\frac{5 !}{2 ! 2 !}\) = 180 ways
Now, the two G’s among themselves can be arranged in \(\frac{2 !}{2 !}\) = 1 way. Hence the number of required permutations is 180.

(iii) In the word SINGING, there are 2 vowels which are alike i.e., 1, and there are 5 consonants of which 2Ns and 2Gs are
alike and one S is different.
C   V   C   C  V  C  C
The two vowels can be interchanged among themselves in \(\frac{2 !}{2 !}\) = 1 way. Now, the 5 consonants can be arranged in the remaining 5 places in \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
∴ Number of required arrangements = 1 x 30 = 30.

Question 25.
Find the number of ways of arranging the letters of the word a⁴b³c⁵ in its expanded form.
Solution:
The expanded form of a⁴b³c⁵ is
aaaa  bbb  ccccc
This word has 12 letters in which there are 4 a’s, 4 b’s and 5c’s. By Theorem 5.5.2, they can be arranged in ways.
\(\frac{12 !}{4 ! 3 ! 5 !}\) ways.

Question 26.
Find the number of 5 – digit numbers that can be formed using the digit 1, 1, 2, 2, 3. How many of them are even?
Solution:
In the given 5 digits, there are two l’s and two 2’s. Hence they can be arranged in 5!
\(\frac{5 !}{2 ! 2 !}\) = 30 ways.

Now, to find even numbers fill the units place by 2. Now the remaining 4 places can be filled using the remaining digits 1, 1, 2, 3, in
\(\frac{4 !}{2 !}\) = 12 ways.
Thus the number of 5 – digit even numbers that can be formed using the digits 1, 1, 2, 2, 3 is 12.

Question 27.
There are 4 copies (alike) each of 3 different books. Find the number of ways of arranging these 12 books in a shelf in a single row.
Solution:
We have 12 books in which 4 books are alike of one kind, 4 books are alike of second kind and 4 books are alike of third kind. Hence, by Therorem 5.5.2., they can be arranged in a shelf in a row in \(\frac{12 !}{4 ! 4 ! 4 !}\) ways.

In problem 9 of solved problems 5.2.12, we have calculated the rank of the word PRISION. In the following problem we find the rank of a word when it contains repreated letters.

Question 28.
If the letters of the word EAMCET are permuted in ail possible ways and If the words thus formed are arranged in the dictionary order, find the rank of the word EAMCET.
Solution:
The dictionary order of the letters of given word is A C E E M T
In the dictionary order the words which begin with the letter A come first. If we fill the first place with A, remaining 5 letters can be arranged \(\frac{5 !}{2 !}\) ways (since there are two Es).

On proceeding like this (as in problem 9 or 5.2.12) we get

Question 29.
Find the number of ways of selecting 4 boys and 3 girls from a group of 8 boys and 5 girls.
Solution:
4 boys can be selected from the given 8 boys in ⁵C₄ ways and 3 girls can be selected from the given 5 girls in ⁵C₃ ways. Hence, by the Fundamental principle, the number of required selections is
⁸C₄ x ⁵C₃ = 70 x 10 = 700

Question 30.
Find the number of ways of selecting
4 English, 3 Telugu and 2 Hindi books out of 7 English, 6 Telugu and 5 Hindi books.
Solution:
The number of ways of selecting
4 English books out of 7 books = ⁷C₄
3 Telugu books out of 6 books = ⁶C₃
2 HIndi books out of 5 books = ⁵C₂
Hence, the number of required ways
⁷C₄ x ⁶C₃ x ⁵C₂  = 35 x 20 x 10 = 7000

Question 31.
Find the number of ways of forming a committee of 4 members out of 6 boys and 4 girls such that there is least one girl in the committee.
Solution:
The number of ways of forming a committee of 4 members out of 10 members (6 boys + 4 girls) is ¹⁰C₄ . Out of these, the number of ways of forming the committee having no girl is ⁶C₄ (we select all 4 members from boys). Therefore, the number of ways of forming the committees having atleast one girl is ¹⁰C₄^{– 6}C₄ = 210 – 15 = 195.

Question 32.
Find the number of ways of selecting 11 member cricket team from 7 batsmen, 6 bowlers and 2 wicket-keepers so taht the team contaIns 2 wicket-keepers and atleast 4 bowlers.
Solution:
The required cricket team can have the following compositions.

Therefore, the number of ways of selecting the required cricket team = 315 + 210 + 35 = 560

Question 33.
If a set of rn’ parallel lines intersect another set of ‘n’ parallel lines (not parallel to the lines in the first set), then find the number of parallelograms formed In this lattice structure.

Solution:
Whenever we select 2 lines from the first set of m lines ad 2 lines from the second set of n lines, one parallelogram is formed as shown in the figure. Thus, the number of parallelogram formed ^mC₂ x ^mC₂

Question 34.
There are rn’ points in a plane out of which ‘p’ points are colinear and no three of the points are collinear unless all the three are from these p points. Find the number of different
(i) straight lines passing through pairs of distinct points.
(ii) triangles formed by joining these points (by line segments).
Solution:
(i) From the given ‘m’ points, by drawing straight lines passing through 2 distinct points at a time, we are supposed to get ^mC₂ number of lines. But, since p’ out of these ‘m’ points are coil mear, by forming lines passing through these p points 2 at a time we get only one line instead of getting ^pC₂. Therefore, the number of different lines as required is
^mC₂ – ^pC₂ + 1.

(ii) From the given m points, by joining 3 at a time, we are supposed to get ^mC₃ number of triangles. Since p points out of these m point are collinear, by joining these p points 3 at a time we do not get any triangle (we get only a Line) when we are supposed to get number of triangles. Hence the number of triangles formed by joining the given m points is
^mC₃ – ^pC₃

Note : The number of diagonals in an n-sided polygon = \({ }^n C_2-n=\frac{n(n-3)}{2}\)

Question 35.
A teacher wants to take 10 students to a park. He can take exactly 3 students at a time and will not take the same group of 3 students more than once. Find the number of times (i) each student can go to the park (ii) the teacher can go to the part.
Solution:
i) To find the number of times a student can go to the park, we have to select 2 more students from the remaining 9
students. This can be done in ^pC₂ ways. Hence, each student can go to park = 36 times.

ii) The number of times the teacher can go to park = The number of different ways of selecting 3 students out of 10
= ¹⁰C₃ = 120

Question 36.
A double decker minibus has 8 seats in the lower deck and 10 seats On the upper deck. Find the number of ways of arranging 18 persons in the bus If 3 children want to go to the upper deck and 4 old people can not go to the upper deck.
Solution:
Allowing 3 children, to the upper deck and 4 old people to the lower deck we are left with li people and 11 seats (7 in the upper deck and 4 in the lower deck). We can select 7 people for the upper deck out of the 11 people in ¹¹C₇ ways. The remaining 4 persons go to lower deck. Now we can arrange 10 persons (3 children and 7 others) in the upper deck and 8 persons (4 old people and 4 others) in the lower deck in 10! and 8! ways respectively. Hence, the required number arrangements = ¹¹C₇ x 10! 8!

Question 37.
Prove that

Solution:

Question 38.
(i) If \({ }^{12} C_{s+1)}={ }^{12} C_{(2 s-5)}\), find s
(ii) If \({ }^n C_{21}={ }^n C_{27} \text {, find }{ }^{50} C_n\)
Solution:

Question 39.
14 persons are seated at a round table. Find the number of ways of selecting two persons out of them who are not seated adjacent to each other.
Solution:

Let the seating arrangement of given 14 persons at the round table be as shown in figure.
Number of ways of selecting 2 persons out of 14 persons ¹⁴C₂ = 91.
In the above arrangement two persons sitting adjacent to each other can be selected in 14 ways
(they are a₁, a₂, a₃, a_13,a₁₄, a₁₅ a₁).
Therefore, the required number of ways = 91 – 14 = 77

Students must practice this TS Intermediate Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 8 Differential Equations Ex 8(a)

I.
Question 1.
Find the order of the differential equation obtained by eliminating the arbitrary constants b and c from xy = ce^x+ be^-x + x².
Solution:
Given equation is xy = ce^x + be^-x + x² ………….(1)
Differentiating (1) w.r.t x, we get
xy₁ + y = ce^x – be^-x + x²
Again differentiating w.r.t x, we get
xy₂ + y₁ + y₁ = ce^x + be^-x + 2
= (xy – x²) + 2
∴ xy₂ + 2y₁ – (xy – x²) – 2 = 0 ………………(2)
Arbitrary constants a and b are eliminated in the differential equation (2).
The order of the differential equation (2) is 2.

Question 2.
Find the order of the differential equation of the family of all circles with their cen¬tres at the origin.
Solution:
The equation of circle with centre 0 is given by x² + y² = a² where a is any constant.
Differentiating w.r.t x we get 2x + 2yy₁ = 0
⇒ x + yy₁ = 0
Which is the required differential equation of family of all circles with their centres at origin.
The order of the above differential equation is 1.

II.
Question 1.
Form the differential equations of the following family of curves where parameters are given in brackets.
(i) y = c (x – c)²; (c)
Given y = c(x – c)² ……………..(1)
Differentiating w.r.t ‘x’ we have
y₁ = 2c(x – c) ………….(2)

∴ y . y₁³ = (xy₁ – 2y) 4y²
⇒ y₁³ = (xy₁ – 2y) 4y
⇒ y₁³ = 4xyy₁ – 8y²
⇒ y₁³ – 4xyy₁ + 8y² = 0
⇒ \(\left(\frac{d y}{d x}\right)^3\) – 4xy \(\frac{d y}{d x}\) + 8y² = 0
This the differential equation in which c is eliminated.

ii) xy = ae^x + be^-x ; (a, b)
Solution:
Given xy = ae^x + be^-x and ………….(1)
Differentiating (1) w.r.t x
xy₁ + y = ae^x – be^-x
Again differentiating w.r.t x,
xy₂ + y₁ + y₁ = ae_x + be_-x = xy
= xy₂ + 2y₁ – xy = 0
which is the required equation obtained on the elimination of a and b.

(iii) y = (a + bx) e^kx; (a, b)
Solution:
Given y = (a + bx) e^kx …………(1)
and Differentiating (1) w.r.t x, we get
y₁ = (a + bx) ke^kx + e^kx . b
= ky + e^kx . b
∴ y₁ – ky = be^kx ………….(2)
Again differentiating w.r.t x,
y₂ – ky₁ = kbe^kx
= k(y₁ – ky)
⇒ y₂ – 2ky₁ + k²y = 0
⇒ \(\frac{d^2 y}{d x^2}-2 \mathrm{k} \frac{d y}{d x}\) + k²y = 0
is the required equation obtained on the elimination of a, b.

v) y = a cos (nx + b); (a, b)
Solution:
Given equation is y = a cos (nx + b)
∴ y₁ = – an sin (nx + b)
= – an² cos (nx + b)
= – n²y
∴ y₁ + n²y = 0
⇒ \(\frac{d^2 \mathrm{y}}{d x^2}\) + n²y = 0
is the required differential equation obtained on elimination of a and b.

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
(i) The rectangular hyperbolas which have the coordinate axes as asymptotes.
Solution:
Equation of rectangular hyperbolas which have the coordinate axes as asymptotes is
xy = c².
Differentiating w.r.t x,
xy₁ + y = 0
⇒ x\(\frac{d y}{d x}\) + y = 0 is the required equation.

(ii) The ellipses with centres at the origin and having coordinate axes as axes.
Solution:
Equation of ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Differentiating w.r.t x’ we get,

⇒ \(\frac{2}{\mathrm{~b}^2}\) [yy₁ – xyy₂ – xy₁²] = 0
⇒ yy₁ – xyy₂ – xy₁² = 0
⇒ xyy₂ + xy₁ – yy₁ = 0
⇒ \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\) is the required differential equation.

III.
Question 1.
Form the differential equations of the following family of curves whose parameters are given in brackets.
(i) y = ae^3x + be^4x; (a, b)
Solution:
Given y = ae^3x + be^4x
Differentiating w.r.t. ‘X’
y₁ = 3ae^3x + 4be^4x
⇒ y₁ – 3ae^3x = 4be^4x
= 4 [y – ae^3x]
⇒ y₁ – 4y = ae^3x ………….(1)
Again differentiating w.r.t. x,
y₂ – 4y₁ = – 3ae^3x
⇒ y₂ – 4y₁ = 3 (y₁ – 4y)
⇒ y₂ – 7y₁ + 12y = 0 is the required differential equation.

(ii) y = ax² + bx, (a, b)
Solution:
Given equation is
y = ax² + bx …………..(1)
and dill erentiating w.r.t. x
y₁ = 2ax + b …………(2)
Again differentiating w.r.t. x,
y₂ = 2a
⇒ x²y₂ = 2ax² …………..(3)
Also from (2)
– 2xy₁ = – 4x²a – 2bx …………..(4)
From (1)
2y = 2ax² + 2bx …………(5)
Adding (3), (4), (5) we get
x²y₂ – 2xy₁ + 2y = 2ax² – 4ax² – 2bx + 2ax² + 2bx = 0
∴ \(x^2 \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=0\) is the required differential equation in which a, b are eliminated.

(iii) ax² + by² = 1; (a, b)
Solution:
Given equation of the curve is
ax² + by² = 1 ……………(1)
Differentiating (1) w.r.t. ‘x we get
2ax + 2by \(\frac{d y}{d x}\) = 0 and
by² = 1 – ax² from (1)
⇒ 2ax + 2byy₁ = 0 ……………(2)
⇒ b(2yy₁) = – 2ax …………….(3)
From (3) + (2) we get

⇒ – xay = y₁ (1 – ax²)
⇒ – axy = y₁ – ax²y₁
⇒ y₁ = ax (xy₁ – y)
⇒ a = \(\frac{y_1}{x\left(x y_1-y\right)}\)
Differentiating w.r.t x,
0 = \(\frac{d}{d x}\left[\frac{y_1}{x\left(x y_1-y\right)}\right]\)
= \(\frac{y_2\left(x^2 y_1-x y\right)-y_1\left(\frac{d}{d x}\left(x^2 y_1-x y\right)\right)}{x^2\left(x y_1-y\right)^2}\)
⇒ (x²y₁ – xy)y₂ – y₁(x²y₂ + 2xy₁ – xy₁ – y) = 0
⇒ x²y₁y₂ – xyy₁ – x²y₁y₂ – 2xy₁² + xy₁² + yy₁ = 0
⇒ xyy₂ + xy₁² – yy₁ = 0
⇒ \(x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y\left(\frac{d y}{d x}\right)\) = 0 is the required differential equation obtained on elimination of constants a and b.

(iv) xy = ax² + \(\frac{b}{x}\); (a, b)
Solution:
Given equation is x²y = ax³ + b ………….(1)
Differentiating (1) w.r.t. ‘x’
2xy + x²y₁ = 3ax² ………….(2)
Again differentiating w.r.t x,
x²y₂ + 2xy₁ + 2xy₁ + 2y = 6ax
⇒ x²y₂ + 4xy₁ + 2y = 6ax
⇒ x³y₂ + 4x2y₁ + 2xy = 6ax²
= 2(3ax²)
= 2 [2xy + x²y₁]
= 2x²y₁ + 4xy
⇒ x³y² + 2x²y₁ – 2xy = 0
⇒ x²y₂ + 2xy₁ – 2y = 0
⇒ \(x^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}\) – 2y = 0 which is the required differential equation on elimination of constants a and b from (1).

Question 2.
Obtain the differential equation which corresponds to each of the following family of curves.
(i) The circles which touch the Y – axis at the origin.
Solution:
The cquation of circle which touch the Y-axis at the origin is x² + y² + 2gx = 0 ………..(1)
Differentiating wr.t. x we get
2x + 2yy₁ + 2g = 0
⇒ g = – (x + yy₁)
Hence from (1)
x² + y² + 2x [- (x + yy₁)] = 0]
x² + y² – 2x² – 2xyy₁ = 0
⇒ – y² – x² = 2xy . \(\frac{d y}{d x}\) which is the required differential equation obtained on elimination of ‘g’ from (1).

(ii) The parabola each of which has a laws rectum 4a and whose axis are parallel to X- axis.
Solution:
Equation of parabola which has latus rectum 4a and whose axes are parallel to X-axis is
(y – k)² = 4a(x – h) ………….(1)
Differentiating w.r.t ‘x’
2 (y – k) y₁ = 4a
⇒ (y – k) y₁ = 2a ……………(2)
Differentiating again w.r.t ‘x’
(y – k) y₂ + y₁² = 0
From (2)
y – k = \(\frac{2 a}{y_1}\)
∴ From (3)
\(\frac{2 a}{y_1}\) y₂ + y₁² = 0
⇒ 2ay₂ + y₁³ = 0
⇒ 2a \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\) = 0 which is the required differential equation obtained on elimination of constants h, k from (1).

(iii) The parabolas having their focli at the origin and axis along the X-axis.
Solution:
Equation of parabola having focii at origin and axis is along X-axis is given by
y² = 4a(x + a) ……….(1)
Differentiating w.r.t x
2yy₁ = 4a
a = \(\frac{\mathrm{yy}_1}{2}\)
∴ From (1)
y² = 4a(x + a)
= 4 \(\frac{\mathrm{yy}_1}{2}\) (x + \(\frac{\mathrm{yy}_1}{2}\))
= 2yy₁ (x + \(\frac{\mathrm{yy}_1}{2}\))
= 2xyy₁ + y²y₁²
⇒ y = 2xy₁ + yy₁²
⇒ yy₁² + 2xy₁ – y = 0
⇒ \(y\left(\frac{d y}{d x}\right)^2+2 x\left(\frac{d y}{d x}\right)\) – y = 0
which is the required differential equation obtained on elimination of ‘a’ from (1).

Students must practice this TS Intermediate Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 10 Random Variables and Probability Distributions Ex 10(b)

I.
Question 1.
In the experiment of tossing a coin n times, If the variable X denotes the number of heads and P (X = 4), P (X = 5), p (X = 6) are in arithmetic progression then find n.
Solution:
X follows binomial distribution with p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
P (X = 4), P (X = 5), P (X = 6) are in A.P.
∴ 2P (X = 5) = P (X = 4) + P (X = 6)

⇒ n² – 21n + 98 = 0
⇒ (n – 7) (n – 14) = 0
⇒ n = 7 or 14.

Question 2.
Find the minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is 0.8.
Solution:
When a coin is tossed we have p = \(\frac{1}{2}\), q = \(\frac{1}{2}\)
P(X = x) = ⁿC_x p^xq^n-x
P(X = 0) = ⁿC₀ p⁰q^n-0
Given probability 0f getting atleast one head is atleast 0.8
P(X ≥ 1 ) ≥ 0.8
⇒ 1 – P (X = 0) ≥ 0.8
⇒ 1 – \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) ≥ 0.8
⇒ – \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) ≥ – 0.2
⇒ \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) < (0.2). Which is true for n = 3.

Question 3.
The probability of a bomb hitting a bridge is 1/2 and three direct hits (not necessarily consecutive) are needed to destroy it. Find the mmiinuni number of bombs required so that the probability of bridge being destroyed is greater than 0.9.
Solution:
Given probability of a bomb hitting a bridge is \(\frac{1}{2}\).
∴ p = \(\frac{1}{2}\);
q + p = 1 ⇒ q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let X represents the minimum number of bombs to be dropped so that the bridge can be destroyed. Given that P (X ≥ 3) > 0.9
⇒ 1 – P(X < 3) > 0.9
⇒ 1 – [ P(X = 0) + P(X = 1) + P (X = 2)] > 0.9
⇒ 1 – [\({ }^n \mathrm{C}_0\left(\frac{1}{2}\right)^{\mathrm{n}}+{ }^{\mathrm{n}} \mathrm{C}_1\left(\frac{1}{2}\right)^{\mathrm{n}-1}\left(\frac{1}{2}\right)\) + \({ }^{\mathrm{n}} C_2\left(\frac{1}{2}\right)^{\mathrm{n}-2}\left(\frac{1}{2}\right)^2\)] > 0.9

By substituting the values for ‘n’ we see that n = 9, 10, 11. satisfies the above in equation.
∴ The minimum number of bombs to be dropped so that the bridge Is to be destroyed is n = 9.

Question 4.
If the difference between the mean and the variance of a binomial variate is 5/9 then find the probability for the event of 2 successes when the experiment is conducted 5 times.
Solution:
Given n = 5 and
given that mean – variance of a Binomial variate is \(\frac{5}{9}\).
∴ np – npq = \(\frac{5}{9}\)
np(1 – q) = \(\frac{5}{9}\)
⇒ npq = \(\frac{5}{9}\)
⇒ 5p (1 – q) = \(\frac{5}{9}\)
⇒ 5p (p) = \(\frac{5}{9}\)
⇒ p² = \(\frac{5}{9}\)
⇒ p = \(\frac{1}{3}\)
∴ q = \(\frac{2}{3}\)
∴ P(X = 2) = ⁵C₂ \(\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{5-2}\)
= 10 × \(\frac{1}{9} \times \frac{8}{27}=\frac{80}{243}\)

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

i) Probability ¡or atleast one will arrive safely
P (X ≥ 1) = 1 – P(X = 0)
= 1 – \({ }^6 C_0\left(\frac{8}{9}\right)^0\left(\frac{1}{9}\right)^{6-0}\)
= 1 – \(\frac{1}{9^6}\)

ii) Exactly, three will arrive safely
P (X = 3) = ⁶C₃ \(\left(\frac{8}{9}\right)^3\left(\frac{1}{9}\right)^{6-3}\)
= \({ }^6 C_3 \frac{8^3}{9^3} \frac{1}{9^3}={ }^6 C_3 \frac{8^3}{9^6}\)
= \(20\left(\frac{8^3}{9^6}\right)\).

Question 6.
If the mean and variance of a binomial variate X are 2.4 and 1.44 respectively, find P (1 < X ≤ 4).
Solution:
Mean = np = 2.4;
Variance = npq = 1.44
∴ \(\frac{\mathrm{npq}}{\mathrm{np}}=\frac{1.44}{2.4}\) = 0.6
⇒ q = 0.6
and p = 1 – 0.6 = 0.4
∴ np = 2.4
⇒ n (0.4) = 2.4
⇒ n = 6
∴ P(1 < X ≤ 4) = P (X = 2) + P(X = 3) + P (X = 4)

Question 7.
If is given that 10% of the electric bulbs manufactured by a company are defective. In a sample of 20 bulbs, find the probability that more than 2 are defective.
Solution:
Probability for an electric bulb to be defective, p = \(\frac{10}{100}\) = 0.1
∴ Probability for a non defective bulb
q = 1 – p = 1 – 0.1 = 0.9
Probability to have more than 2 are defectives
⇒ P(X > 2) = 1 – P(X ≤ 2)
1 – [P (X = 0) + P(X = 1) + P(X = 2)]
We have
P(X = x) = ⁿC_x p^xq^n-x

Question 8.
On an average, rain falls on 12 days in every 30 days, find the probability that, rain will fall on just 3 days of a given week.
Solution:
Given the probability for the day to be rainy = \(\frac{12}{30}=\frac{2}{5}\)
∴ p = \(\frac{2}{5}\) and
q = 1 – p
= 1 – \(\frac{2}{5}\) = \(\frac{3}{5}\)
Also, n = 7
∴ Probability for the rain to fall on just 3 days of a given week
P(X = 3) = ⁷C₃ \(\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^{7-3}\)
= ⁷C₃ \(\left(\frac{2}{5}\right)^3\left(\frac{3}{5}\right)^4\)
= \(\frac{35 \times 2^3 \times 3^4}{5^7}\)

Question 9.
For a binomial distribution with mean 6 and variance 2, find the first two terms of the distribution.
Solution:
Given mean = np = 6
and variance = npq = 2
∴ q = \(\frac{\mathrm{npq}}{\mathrm{np}}=\frac{2}{6}=\frac{1}{3}\)
∴ p = 1 – q
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
∴ np = 6
⇒ \(\frac{2 n}{3}\) = 6
⇒ n = 9
We have P (X = x)
= ⁿC_x p^x q^n-x

Question 10.
In a city 10 accidents take place in a span of 50 days. Assuming that the number of accidents follows the Poisson distribution, find the probability that there will be 3 or more accidents in a day.
Solution:
Here λ = \(\frac{10}{50}=\frac{1}{5}\)
∴ The probability that there will be 3 or more accidents in a day using Poisson variate.
P (X = x) = \(\frac{c^{-\lambda} \lambda^x}{x !}\)
P (X ≥ 3) = 1 – [P (X = 0) + P(X = 1) + P(X = 2)]

II.
Question 1.
Five coins are tossed 320 times. Find the frequencies of the distribution of the number of heads and tabulate the result.
Solution:
When a coin is tossed we have p = \(\frac{1}{2}\) and q = \(\frac{1}{2}\).
Here n = 5;
The frequencies of distribution of the number of heads is given using binomial distribution given by
f_x = N. ⁿC_x p^x q^n-x,
x = 0,1, 2, 3, 4, 5, (Number of heads)
when x = 0

Question 2.
Find the probability of guessing atleast 6 out of 10 of answers in
(i) True or false type examination
(ii) Multiple choice with 4 possible answers.
Solution:
i) Probability of guessing atleast 6 out of 10 answers in
(i) True or false type examination is
P(X ≥ 6) = P(X = 6) P(X = 7) + P (X = 8) + P (X = 9)
Here P (X = x) = ⁿC_x p^x q^n-x and
probability for an answer to be true or false,
p = \(\frac{1}{2}\), q = \(\frac{1}{2}\), n = 10
p – q = n = 10

ii) Probability for a question to guess in multiple choice type with 4 answers is
p = \(\frac{1}{4}\), q = \(\frac{3}{4}\)
∴ P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P (X = 9) + P(X = 10)
where P (X = x) = ⁿC_x p^x q^n-x
= ¹⁰C_x \(\left(\frac{1}{4}\right)^x\left(\frac{3}{4}\right)^{10-x}\)

In the above two cases (i) and (ii).

i) Probability to guess 6 out of 10 questions in true or false examination is = ¹⁰C₆ \(\left(\frac{1}{2}\right)^{10}\)

ii) Probability to guess 6 out of 10 questions in Multiple choice type with 4 answers = ¹⁰C₆ . \(\frac{3^4}{4^{10}}\).

Question 3.
The number of persons joining a cinema ticket counter in a minute has poisson distribution with parameter 6. Find the probability that
(i) no one joins the queue in a particular minute
(ii) two or more persons join in the queue in a minute.
Solution:
Given λ = 6, and poisson distribution function is given by
P (X = x) = \(\frac{\lambda^x e^{-\lambda}}{x !}\), x = 0, 1, 2, 3, ………..

i) Probability that no one joins the queue in a particular minute is
P (X = 0) = \(\frac{\lambda^0 \mathrm{e}^{-6}}{0 !}\) = e^-6.

ii) Probability for two or more persons join in the queue in a minute is
P (X ≥ 2) = 1 – [P (X = 0) + P (X = 1)]
= 1 – \(\left[\mathrm{e}^{-6}+\frac{\lambda \mathrm{e}^{-\lambda}}{1 !}\right]\)
= 1 – e^-6 – 6 e^-6.