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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 1.
Solve the following system of linear equations by matrix inversion method:
(i) 2x + 5y = -2, x + 2y = -3
(ii) 2x – y = 8, 3x + 2y = -2
(iii) 2x + 3y – z = 9, x + y + z = 9, 3x – y – z = -1
(iv) x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
Solution:
(i) 2x + 5y = -2, x + 2y = -3
The matrix form of the above equations is \(\left(\begin{array}{ll}{2} & {5} \\ {1} & {2}\end{array}\right)\left(\begin{array}{l}{x} \\ {y}\end{array}\right)=\left(\begin{array}{l}{-2} \\ {-3}\end{array}\right)\)
(i.e) AX = B

Question 2.
If A = \(\left[\begin{array}{ccc}{-5} & {1} & {3} \\ {7} & {1} & {-5} \\ {1} & {-1} & {1}\end{array}\right]\) and B = \(\left[\begin{array}{lll}{1} & {1} & {2} \\ {3} & {2} & {1} \\ {2} & {1} & {3}\end{array}\right]\) find the products AB and BA and hence solve the system of equations x + y + 2z = 1, 3x + 2y + z = 7, 2x + y + 3z = 2.
Solution:

Question 3.
A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was ₹ 19,800 per month at the end of the first month after 3 years of service and ₹ 23,400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method to solve the problem.)
Solution:
Let his monthly salary be ₹ x and his annual increment be ₹ y
Given x + 3y = 19800 and x + 9y = 23400
Writing the above equations in matrix form, we get


x = ₹ 18000 and y = ₹ 600
(i.e) monthly salary = ₹ 18000 and annual increment = ₹ 600

Question 4.
4 men and 4 women can finish a piece of work jointly in 3 days while 2 men and 5 women can finish the same work jointly in 4 days. Find the time taken by one man alone and that of one woman alone to finish the same work by using matrix inversion method.
Solution:
Let the work done by man in 1 day be x and the work done by a woman in 1 day be y
Now we are given
4x + 4y = \(\frac { 1 }{ 3 }\) ⇒ 12x + 12y = 1
and 2x + 5y = \(\frac { 1 }{ 4 }\) ⇒ 8x + 20y = 1
The matrix form of the above equations is \(\left(\begin{array}{cc}{12} & {12} \\ {8} & {20}\end{array}\right)\left(\begin{array}{l}{x} \\ {y}\end{array}\right)=\left(\begin{array}{l}{1} \\ {1}\end{array}\right)\)
(i.e) AX = B ⇒ X = A^-1B
Here


(i.e) The work done by a man in 1 day = x = \(\frac{1}{18}\)
The time taken by a man to finish the work = 18 days.
The work done by a woman in 1 day = y = \(\frac{1}{36}\)
The time is taken by a woman to finish the work = 36 days.

Question 5.
The prices of three commodities A, B and C are ₹ x, y and z per units respectively. A person P purchases 4 units of B and sells two units of A and 5 units of C. Person Q purchases 2 units of C and sells 3 units of A and one unit of B . Person R purchases one unit of A and sells 3 unit of B and one unit of C. In the process, P, Q and R earn ₹ 15,000, ₹ 1,000 and ₹ 4,000 respectively. Find the prices per unit of A,B and C. (Use matrix inversion method to solve the problem.)
Solution:
Price of A = ₹ x /unit
Price of B = ₹ y /unit
Price of C = ₹ z /unit
We are given
2x – 4y + 5z = 15000
3x + y – 2z = 1000
-x + 3y + z = 4000
The matrix form of the above equations is \(\left(\begin{array}{rrr}{2} & {-4} & {5} \\ {3} & {1} & {-2} \\ {-1} & {3} & {1}\end{array}\right)\left(\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right)=\left(\begin{array}{r}{15000} \\ {1000} \\ {4000}\end{array}\right)\)
(i.e) AX = B ⇒ X = A^-1B
Here

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3 Additional Problems

Question 1.
Using matrix method, solve the following system of equations:
x + 2y + z = 7, x + 3z = 11, 2x – 3y = 1
Solution:
The system of equations can be written in the form AX = B, where


= 9 + 12 – 3 = 18 ≠ 0 Non-singular matrix.
A₁₁ = (0 + 9) = 9, A₁₂ = -(0 – 6) = 6,
A₁₃ = (-3 – 0) = -3
A₂₁ = – (0 + 3) = -3, A₂₂ = (0 – 2) = -2
A₂₃ = -(-3 – 4) = 7
A₃₁ = (6 – 0) = 6, A₃₂ = -(-3 – 1) = -2,
A₃₃ = (0 – 2) = -2

Thus x = 2 ,y = 1 and z = 3.

Question 2.
Using matrices, solve the following system of linear equations: x + 2y – 3z = -4, 2x + 3y + 2z = 2, 3x – 3y – 4z = 11.
Solution:
The system of equations can be written in the form AX = B, where


Thus x = 3,
y = -2 and z = 1

Question 3.
If Find A^-1. Hence using A^-1 solve the system of equations.
2x – 3y + 5z = 11,
3x + 2y – 4z = -5,
x + y – 2z = -3.
Solution:

Question 4.
Use product to solve the system of equations
x + 3z = 9,
-x + 2y – 2z = 4,
2x – 3y + 4z = -3
Solution:

Question 5.
Use product to solve the system of equations
x – y + 2z = 1,
2y – 3z = 1,
3x – 2y + 4z = 2
Solution:

The system of equations can be written in the form AX = C, where

Question 6.
An amount of ₹ 7000 is invested in three types of investments x, y and z at the rate of 3%, 4% and 5% interest respectively. The total annual income is ₹ 280. If the combined income from x and y is ₹ 80 more than that from z, then
(i) Represent the above situation in form of linear equations .
(ii) Is it possible to frame the given linear equations in the form of matrix to obtain the three values x, y and z using matrix multiplication? If yes, find.
(iii) Which value is more beneficial to invest?
Solution:

⇒ -40 + 30 + 0 = -10 ≠ 0
Cofactors C₁₁ = +(-20 – 20) = -40
C₂₁ = -(5 – 4) = 9
C₃₁ = +(5 – 4) = 1
C₁₂ = +(-15 – 15) = -30
C₂₂ = +(-5 – 3) = -8
C₃₂ = -(5 – 3) = -2
C₁₃ = +(12 – 12) = 0
C₂₃ = -(4 – 3) = -1
C₃₃ = +(4 – 3) = 1

Thus, the three values x = ₹ 2000, y = ₹ 3000 and z = ₹ 2000
(iii) The value of y is more beneficial to invest

Question 7.

Solution:

In matrix form

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7

Question 1.
If A + B + C = 180°, prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

(iii) sin² A + sin² B + sin² C = 2 + 2 cos A cos B cos C
(iv) sin² A + sin² B – sin² C = 2 sin A sin B cos C

(vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C.
Solution:
(i)
LHS = (sin 2A + sin 2B) + sin 2C
= 2 sin (A + B) cos (A – B) + 2 sin C cos C
[sin (A + B) = sin (180° – C) = sin C]
= 2 sin C cos (A – B) + 2 sin C cos C
= 2 sin C [ cos (A – B) + cos C]
{cos C = cos [180° – (A + B)] = – cos (A + B)}
= 2 sin C [cos (A – B) – cos (A + B)]

(ii)

(iii)

[cos (180° –  C) – cos C – cos C]
= 2 + cos C [cos (A – B) + cos (A + B)]
= 2+ cos C[2 cos A cos B]
= 2 + 2 cos A cos B cos C = RHS

(iv)

(v)

(vi)

(vii)
Now A + B + C = 180°
So B + C = 180° – A
sin (B + C – A) = sin (180° – A – A)
= sin(180° – 2A) = sin 2A
Now LHS = sin 2A + sin 2B + sin 2C
= 4 sin A sin B sin C (from (i) ans) = RHS

Question 2.
If A + B + C = 2s, then prove that sin(s – A) sin(s – B) + sin s sin(s – C) = sin A sin B.
Solution:

Question 3.

Solution:

⇒ A+B+C = 180°
⇒ A + B = 180° – C
multiply 2 on both sides ⇒ 2A + 2B = 360° – 2C
2(A + B) =360° – 2C
⇒ tan(2A + 2B) = tan(360° – 2C) = – tan 2C

⇒ tan 2A + tan 2B = -tan2C[1 – tan 2A tan 2B]
⇒ tan 2A + tan 2B = -tan 2C + tan 2A tan 2B tan 2C
⇒ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

Question 4.
If A + B + C = \(\frac{\pi}{2}\), prove the following
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
(ii) COS 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.
Solution:
(i) LHS = (sin 2A + sin 2B) + sin 2C
= 2 sin (A + B) cos (A – B) + 2 sin C cos C = 2 sin (90° – C) cos (A – B) + 2 sin C cos C
= 2 cos C [cos (A – B) + sin C] + cos (A + B) ( ∴ A + B = π/2 – C)
= 2 cos C [cos (A – B) + cos (A + B)]
= 2 cos C [2 cos A cos B]
= 4 cos A cos B cos C = RHS

(ii) LHS = (cos 2A + cos 2B) + cos 2C
= 2 cos (A + B) cos (A – B) + 1 – 2 sin² C
= 1 + 2 sin C (cos (A – B) – 2 sin² C)
{∴ cos (A + B) = cos (90° – C) = sin C}
= 1 + 2 sin C [cos (A- B) – sin C]
= 1 + 2 sin C [cos (A – B) – cos (A + B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C = RHS

Question 5.
If ∆ABC is a right triangle and if ∠A = \(\frac{\pi}{2}\), then prove that
(i) cos² B + cos² C = 1
(ii) sin² B + sin² C = 1

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7 Additional Questions

Question 1.
A + B + C = π, prove that sin 2A – sin 2B + sin 2C = 4 cos A sin B cos C
Solution:
LHS = sin 2A – sin 2B + sin 2C
= sin 2A + sin 2C – sin 2B

= 2 sin (A + C) cos (A – C) – 2 sin B cos B
= 2 sin (180° – B) cos (A-C) – 2 sin B cos B
= 2 sin B cos (A – C) – 2 sin B cos B
= 2 sin B [cos (A – C) – cos B]
= 2 sin B [cos (A – C) – cos (180° – (A + C))]
= 2 sin B [cos (A – C) + cos (A + C)]
= 2 sin B [2 cos A cos C]
= 4 cos A sin B cos C = RHS

Question 2.

Solution:


substitute in (1) we get,

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Find the rank of the following matrices by the minor method:


Solution:

Question 2.
Find the rank of the folowing matrices by row reduction method:

Solution:
(i) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. So ρ(A) = 3
(ii) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. ρ(A) = 3
(iii) Let

The last equivalent matrix is in row-echelon form. It has three non zero rows. ρ(A) = 3

Question 3.
Find the inverse of each of the following by Gauss-Jordan method:

Solution:
(i) Let \(A=\left(\begin{array}{cc}{2} & {-1} \\ {5} & {-2}\end{array}\right)\)
Applying Gauss-Jordan method we get

(ii) Let


(iii) Let

Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2 Additional Problems

Question 1.
Find the rank of the following matrices.
Solution:

A has at least one non-zero minor of order 2. \(\rho(\mathrm{A})\) = 2

Question 2.
Find the rank of the following matrices.
Solution:

The last equivalent matrix is in the echelon form. It has three non-zero rows.
∴ \(\rho(\mathrm{A})\) = 3; Here A is of order 3 × 4

Question 3.
Find the rank of the following matrices.
Solution:


The last equivalent matrix is in the echelon form. The number of non-zero rows in this matrix is two. A is a matrix of order 3 × 4. ∴ \(\rho(\mathrm{A})\) = 2

Question 4.
Using elementary transformations find the inverse of the following matrix
Solution:

Question 5.
Using elementary transformations find the inverse of the following matrices
Solution:

Question 6.
Using elementary transformations find the inverse of the following matrices
Solution:

Question 7.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 8.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 9.
Using elementary transformations, find the inverse of the following matrices
Solution:

Question 10.
Using elementary transformations, find the inverse of the following matrices
Solution:

Since R₂ has all numbers zero, Thus inverse of matrix A does not exist.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 1.
Express each of the following as a sum or difference
(i) sin 35° cos 28°
(ii) sin 4x cos 2x
(iii) 2 sin 10θ cos 2θ
(iv) cos 5θ cos 2θ
(v) sin 5θ sin 4θ.
Solution:

Question 2.
Express each of the following as a product
(i) sin 75° – sin 35°
(ii) cos 65° + cos 15°
(iii) sin 50° + sin 40°
(iv) cos 35° – cos 75°.
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x).
Solution:

Question 8.

Solution:

Question 9.
Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x.
Solution:

LHS = 1 + cos 2x + cos 4x + cos 6x
= (1 + cos 6x) + (cos 2x + cos 4x)
= 2cos² 3x + 2cos 3x cos x
= 2 cos 3x (cos 3x + cos x)

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

Question 14.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6 Additional Questions

Question 1.

Solution:
Let cos 20° cos 40° cos 60° cos 80° = x
Multiply by 2 sin 20° on both sides.

Question 2.

Solution:

Question 3.
Prove that sin 50° – sin 70° + cos 80° = 0.
Solution:
LHS = sin 50° – sin 70° + cos 80°

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.
Prove that tan 70° – tan 20° – 2 tan 40° = 4 tan 10°
Solution:

Question 9.

Solution:

Question 10.

Solution:
Multiplying numerator and denominator by 2, we have

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 1.
Find the value of cos 2A, A lies in the first quadrant, when

Solution:

Question 2.
If θ is an acute angle, then find

Solution:

Question 3.

Solution:

Question 4.
Prove that cos ⁵ θ = 16 cos⁵ θ – 20 cos³ θ + 5 cos θ.
Solution:
cos ⁵ θ = cos(2θ + 3θ) = cos 2θ cos 3θ – sin 2θ sin 3θ
= (2 cos² θ – 1) (4 cos³ θ – 3 cos θ) – 2 sin θ cos θ (3 sin θ – 4 sin³ θ)
= 8cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 sin² θ cos θ + 8 cos θ sin⁴ θ
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6(1 – cos² θ) cos θ + 8 cos θ (1 – cos² θ)²
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos 0(1+ cos⁴ θ – 2 cos² θ)
= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos θ + 8 cos⁵ θ – 16 cos³ θ
= 16 cos⁵ θ – 20 cos³ θ + 5 cos θ = RHS

Question 5.

Solution:

Question 6.
If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2.
Solution:

Now LHS = (1 + tan A) (1 + tan B)
= tan A + tan B + tan A tan B + 1
= (1 – tan A tan B) + (tan A tan B + 1) from (1)
= 2 = RHS

Question 7.
Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°)… (1 + tan 44°) is a multiple of 4.
Solution:
1 + tan 44° = 1 + tan (45° – 1°)

(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.

Question 8.

Solution:

Question 9.

Solution:

Question 10.
Prove that (1 + sec 2θ)(1 + sec 4θ)….. (1 + sec 2ⁿθ) = tan 2ⁿθ
Solution:

Question 11.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.
Show that 4 sin A sin (60° + A). sin(60° – A) = sin 3A
Solution:
LHS = 4sinAsin(60° + A). sin(60° – A)
= 4 sin A{sin (60° + A). sin (60° – A)}
= 4 sin A {sin² 60° – sin² A)}

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.4

Question 1.
A factory has two Machines-I and II. Machine-I produces 60% of items and Machine-II produces 40% of the items of the total output. Further, 2% of the items produced by Machine-I are defective whereas 4% produced by Machine-II are defective. If an item is drawn at random what is the probability that it is defective?
Solution:

Question 2.
There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. An urn is chosen at random and a ball is drawn from it, (z) find the probability that the ball is black (ii) if the ball is black, what is the probability that it is from the first urn?
Solution:

Question 3.
A firm manufactures PVC pipes in three plants viz, X, Y and Z. The daily production volumes from the three firms X, Y and Z are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that 3% of the output from plant X, 4% from plant Y and 2% from plant Z are defective. A pipe is selected at random from a day’s total production,
(i) find the probability that the selected pipe is a defective one.
(ii) if the selected pipe is a defective, then what is the probability that it was produced by plant Y?
Solution:

Question 4.
The changes of A, B and C becoming manager of a certain company are 5 : 3 : 2. The probabilities that the office canteen will be improved if A, B, and C become managers are 0.4, 0.5 and 0.3 respectively. If the office canteen has been improved, what is the probability that B was appointed as the manager?
Solution:
Given A : B : C = 5 : 3 : 2

Question 5.
An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television 60% of the time. It has also been determined that when the wife is watching television, 40% of the time the husband is also watching. When the wife is not watching the television, 30% of the time husband is watching the television. Find the probability that (i) the husband is watching the television during the prime time of television (ii) if the husband is watching the television, the wife is also watching the television.
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.4 Additional Problems

Question 1.
A factory has two Machines-I and II. Machines-I produces 25% of items and Machine-II produces 75% of the items of the total output. Further 3% of the items produces by Machine-I are defective whereas 4% produced by Machine-II are defective. If an item is drawn at random what is the probability that it is defective?
Solution:
Let A₁ be the event that the items are produced by Machine-I and A₂ be the event that the items are produced by Machine-II.
Let B the event of drawing a defective item

Question 2.
There are two identical boxes containing respectively 5 white and 3 red balls, 4 white and 6 rpd balls. A box is chosen at random and a ball is drawn from it (i) find the probability that the ball is white (ii) if the ball is white, what is the probability that it from the first box?
Solution:
Let A₁ be the event of selecting the first box and A₂ be the event of selecting the second box. Let B be the event of selecting a white ball.

Question 3.
In a factory, Machine-I produces 45% of the output and Machine-II produces 55% of the output. On the average 10% items produced by I and 5% of the items produced by II are defective. An item is drawn at random from a day’s output, (i) Find the probability that it is a defective item (ii) If it is defective, what is the probability that it was produced by Machine-II?
Solution:
Let A₁ and A₂ be the events that the items produced by Machine-I and II respectively.
Let B be the event of selecting a defective item

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.3

Question 1.
Can two events be mutually exclusive and independent simultaneously?
Solution:
When A and B are independent
P(A ∩ B) = P(A) P(B)
But when A and B are mutually
Exclusive P(A ∩ B) = 0

Question 2.
If A and B are two events such that P(A ∪ B) = 0.7, P(A ∩ B) = 0.2, and P(B) = 0.5 then show that A and B are independent.
Solution:
GivenP(A ∪ B) = 0.7, P(A ∩ B)= 0.2 and P(B) = 0.5
To find P(A)
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) 0.7 = P(A) + 0.5 – 0.2
⇒ 0.7 – 0.5 + 0.2 = P(A)
(i.e.,) P(A) = 0.4
Now P(A ∩ B) = 0.2 …………. (i)
P(A) P(B) = 0.4 × 0.5 = 0.2 ………… (ii)
(1) = (2) ⇒ P(A ∩ B) = P(A) P(B)
⇒ A and B are independent.

Question 3.
If A and B are two independent events such that P(A ∪ B) = 0.6, P(A) = 0.2, find P(B).
Solution:
Given A and B are independent.
⇒ P(A ∪ B) = P(A).P(B)
Here P(A ∪ B) = 0.6 and P(A) = 0.2
To find P(B):
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
(i.e.,) 0.6 = 0.2 + P(B) (1 – 0.2)
P(B) (0.8) = 0.4
⇒ P(B) = \(\frac{0.4}{0.8}=\frac{4}{8}=\frac{1}{2}\) = 0.5

Question 4.
If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find (P(A/B)) and P(A ∪ B)
Solution:
Given P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8 to find P(A/B) & P(A ∪ B)

So, P(A/B) =0.5 and P(A ∪ B) = 0.9.

Question 5.
If for two events A and B, P(A) = \(\frac{3}{4}\), P(B) = \(\frac{2}{5}\) and A ∪ B = S (sample space), find the conditional probability P(A/B).
Solution:

Question 6.
A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{3}, \frac{1}{4}\) and\(\frac{1}{5}\)
(i) What is the probability that the problem is solved?
(ii) What is the probability that exactly one of them will solve it?
Solution:

Question 7.
The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40, and the probability that both the oil and filter need changing is 0.15.
(i) If the oil had to be changed, what is the probability that a new oil filter is needed?
(ii) If a new oil filter is needed, what is the probability that the oil has to be changed?
Solution:
Given P(A) = 0.3, P(B) = 0.4 and P(A ∩ B) = 0.15

Question 8.
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that (i) both are white (ii) both are black (iii) one white and one black.
Solution:

Question 9.
Two thirds of students in a class are boys and rest girls. It is known that the probability of a girl getting a first grade is 0.85 and that of boys is 0.70. Find the probability that a students chosen at random will get first grade marks.
Solution:

Question 10.
Given P(A) = 0.4 and P(A ∪ B) = 0.7. Find P(B) if
(i) A and B are mutually exclusive
(ii) A and B are independent events
(iii) P(A/B) = 0.4
(iv) P(B/A) = 0.5
Solution:
P(A) = 0.4, P(A ∪ B) = 0.7
(i) When A and B are mutually exclusive
P(A ∪ B) = P(A) P(B)
(i.e.,) 0.7 = 0.4 + P(B)
0.7 – 0.4 = P(B)
(i.e.,) P(B) = 0.3

(ii) Given A and B are independent
⇒ P(A ∩ B) = P(A). P(B)
Now, P(A ∪ B) = P(A) + P(B) – P (A ∩ B)
(i.e.,) 0.7 = 0.4 + P(B) – (0.4) (P(B))
(i.e.,) 0.7 – 0.4 = P(B) (1 – 0.4)
0.3 = P (B) 0.6
⇒ P(B) = \(\frac{0 \cdot 3}{0 \cdot 6}=\frac{3}{6}\) = 0.5

(iii) P(A/B) = 0.4
(i.e.,) \(\frac{P(A \cap B)}{P(B)}\) = 0.4
⇒ P(A ∩ B) = 0.4 [P(B)] …………. (i)
But We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A ∩ B) = 0.4 + P(B) – 0.7
= P(B) – 0.3 …………. (ii)
from (i) and (ii) (equating R.H.S) We get
0.4 [P(B)] = P(B) – 0.3
0.3 = P(B) (1 – 0.4)
0.6 (P(B)) = 0.3 ⇒ P(B) = \(\frac{0.3}{06}=\frac{3}{6}\) = 0.5

(iv) P(B/A) = 0.5
(i.e.,) \(\frac{P(A \cap B)}{P(A)}\) = 0.5
(i.e.,) P(A ∩ B) = 0.5 × P(A)
= 0.5 × 0.4 = 0.2
Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.4 + P(B) – 0.2
⇒ 0.7 = P(B) + 0.2
⇒ P(B) = 0.7 – 0.2 = 0.5

Question 11.
A years is selected at random. What is the probability that (i) it contains 53 Sundays (ii) it is a leap year which contains 53 Sundays?
Solution:
(i) A non-leap year contains 365 \(\frac{1}{4}\) days 365\(\frac{1}{4}\) ÷ 7 = 52 weeks + 1\(\frac{1}{4}\) days. In 52 weeks,
we get 52 Sundays from the remaining 1\(\frac{1}{4}\) days we should get one sunday.
∴ The probability of getting the day as Sunday = \(\frac{5 / 4}{7}=\frac{5}{4 \times 7}=\frac{5}{28}\)
(ii) A leap year has 366 days
\(\frac{366}{7}\) = 52 weeks + 2 days
In 52 weeks, we get 52 Sundays.
From the remaining two days we should get one Sunday, the remaining two days can be any one of the following combinations.
Saturday and Sunday, Sunday and Monday, Monday and Tuesday, Tuesday and Wednes¬day, Wednesday and Thursday, Thursday and Friday, Friday and Saturday of the seven combination two have Sundays.
∴ (Probability of getting a Sunday = \(\frac{2}{7}\)
Selecting a leap year = \(\frac{1}{4}\)
{∴ In every four consecutive years we get one leap year}
∴ Probability of getting 53 Sundays = \(\frac{2}{7} \times \frac{1}{4} \times \frac{1}{14}\)

Question 12.
Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?
Solution:
Given P(X) = 3/4, P(X’) = 1 – 3/4 = 1/4
∴ P(Y) = 4/5, P(Y’) = 1 – 4/5 = 1/5
P(Z) = \(\frac{2}{3}\) ∴ P(Z’) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
P(X ∩ Y ∩ Z’) + P(X ∩ Y’ ∩ Z) + P(X’ ∩ Y ∩ Z)
= \(\frac{3}{4} \times \frac{4}{5} \times \frac{1}{3}+\frac{3}{4} \times \frac{1}{5} \times \frac{2}{3}+\frac{1}{4} \times \frac{4}{5} \times \frac{2}{3}\)
= \(\frac{12}{60}+\frac{6}{60}+\frac{8}{60}=\frac{26}{60}=\frac{13}{30}\)

Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.3 Additional Problems

Question 1.
If P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.5, find P(A /B) and P(A ∪ B).
Solution:
P(B/A) = 0.5 ⇒ \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}\) = 0.5
(i.e.,) \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{0.4}\) = 0.5
∴ P(B ∩ A) = 0.4 × 0.5 = 0.2
(i.e.,) P(A ∩ B)= 0.2
P(A ∪ B)= P(A) + P(B) – P(A ∩ B)
P(A ∪ B) = 0.4 + 0.7 – 0.2 = 0.9
P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}\)

Question 2.
If A and B are two events such that P(A ∪ B) = \(\frac{5}{6}\), P(A ∩ B) = \(\frac{1}{3}\), P(\(\overline{B}\)) = \(\frac{1}{2}\) show that A and B are independent.
Solution:
P(\(\overline{B}\)) = \(\frac{1}{2}\) (i.e.,)1 – P(B) = \(\frac{1}{2}\)

∴ A and B are independent

Question 3.
P(A) = 0.3, P(B) = 0.6 and P(A ∩ B) = 0.25. Find
(i) P(A ∪ B)
(ii) P(A/B)
(iii) P(B/\(\overline{\mathrm{A}}\))
(iv) \(\mathrm{P}(\overline{\mathrm{A}} / \mathrm{B})\)
(v) \(\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})\)
Solution:

Question 4.
Two cards are drawn one by one at random from a deck of 52 playing cards. What is the – probability Of getting two jacks if (i) the first card is replaced before the second card is drawn (ii) the first card is not replaced before the second card is draw?
Solution:

Question 5.
A husband and wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is \(\frac{1}{6}\) and that of wife’s selection is \(\frac{1}{5}\). What is the probability that (i) both of them will be selected, (ii) only one of them will be selected, (iii) none of them will be selected?
Solution:
P(H) = Probability of husband’s selection = \(\frac{1}{6}\)
P(W) = Probability of wife’s selection = \(\frac{1}{5}\)

Question 6.
For a student the probability of getting admission in IIT is 60% and probability of getting admission in Anna university is 75%. Find the probability that (i) getting admission in only one of these, (ii) getting admission in atleast one of these.
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2

Question 1.
If A and B are mutually exclusive events P(A) = \(\frac{3}{8}\) and P (B) = \(\frac{1}{8}\), then find
(i) P\((\overline{\mathrm{A}})\)
(ii) P(A ∪ B)
(iii) P(\(\overline{\mathrm{A}}\) ∩ B
(iv) P\((\overline{\mathrm{A}} \cup \overline{\mathrm{B}})\)
Solution:

Question 2.
If A and B are two events associated with a random experiment for which P(A) = 0.35, P(A or B) = 0.85, and P(A and B) = 0.15.
Find (i) P(only B)
(ii) P\((\overline{\mathrm{B}})\)
(iii) P(only A)
Solution:
Given P(A) = 0.35
P(A ∪ B) = 0.85
P(A ∩ B) = 0.15
We know P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) 0.85 = 0.35 + P(B) – 0.15
⇒ 0.85 – 0.2 = P(B)
(i.e.,) P(B) = 0.65

(i) P(only B ) = P(B) – P(A ∩ B)
= 0.65 – 0.15 = 0.50
(ii) P\((\overline{\mathrm{B}})\) = 1 – P(B) = 1 – 0.65 = 0.35
(iii) P(A only) = P(A) – P(A ∩ B) = 0.35 – 0.15 = 0.20

Question 3.
A die is thrown twice. Let Abe the event, ‘First die shows 5’ and B be the event, ‘second die shows 5’. Find P(A ∪ B).
Solution:
When a die is throw twice
n(s) = 6² = 36
Let A be the event that first die shows 5 and B be the event that second die shows 5 Now A = {(5, 1), (5, 2) (5, 3), (5, 4), (5, 5) (5, 6}
n(A) = 6 ⇒ P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}\)
and B = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
n(B) = 6 ⇒ P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}\)
Also A ∩ B = {(5, 5)} ⇒ n (A ∩ B) = 1
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}\)

Question 4.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of
(i) P(A ∪ B)
(ii) P(A ∩ \(\overline{B}\))
(iii)P(\(\overline{A}\) ∩ B)
Solution:
P(A) = 0.5, P(B) = 0.3
Here A and B are mutually exclusive.
(i) P(A ∪ B) = P(A) + P(B)
= 0.5 + 0.3 = 0.8
(ii) P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 0.5 + 0.3 – 0.8
P(A ∩ B) = 0
P(A ∩ \(\overline{B}\)) = P(A) – P(A ∩ B) = 0.5 – 0 = 0.5
(iii) P(\(\overline{A}\) ∩ B) = P(B) – P(A ∩ B) = 0.3 – 0 = 0.3

Question 5.
A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96.
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?

Solution:
(i) P(atleast one engine is available) = (1 – probability of no engine available)
= 1 – P(A’ ∩ B’) = 1 – P (A’) P(B’)
= 1 – (0.04) (0.04) = 1 – 0.0016 = 0.9984
(ii) P (A’ ∩ B’) = P (A’) P(B’)
= 0.04 × 0.04
= 0.0016

Question 6.
The probability that a new railway bridge will get an award for its design is 0.48, the probability that it will get an award for the efficient use of materials is 0.36, and that it will get both awards is 0.2. What is the probability, that (i) it will get atleast one of the two awards 00 it will get only one of the awards.
Solution:
Given P(A) = 0.48, P(B) = 0.36 and P(A ∩ B) = 0.2

(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.48 + 0.36 – 0.2 = 0.64.
(ii) P (Getting only one award)
= P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
= (0.48 – 0.2) + (0.36 – 0.2)
= 0.28 + 0.16 = 0.44.

Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.2 Additional Problems

Question 1.
A and B are two events associated with random experiment for which P(A) = 0.36, P(A or B) = 0.90 and P(A and B) = 0.25. Find
(i) P(B)
(ii) P\((\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\)
Solution:
(i) Given P(A) = 0.36, P(A ∪ B) = 0.09, P(A ∩ B) = 0.25
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) 0.90 = 0.36 + P(B) – 0.25
0.90 = 0.11 + P(B)
∴ P(B) = 0.90 – 0.11 = 0.79

(ii) P\((\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\) = P{(A’ ∪ B)’} (Demorgan Law)
P(A ∪ B)’ = 1 – P(A ∪ B) = 1 – 0.90 = 0.1.

Question 2.
Given P(A) = 0.5, P(B) = 0.6 and P(A ∩ B) = 0.24. Find
(i) P(A ∪ B)
(ii) P(\(\overline{\mathrm{A}}\) ∩ B)
(iii) P(A ∩ \(\overline{\mathrm{B}}\))
(iv) P(\(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})\)
(v) P\((\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\)
Solution:
(i) P(A) = 0.5, P(B) = 0.6, P(A ∩ B) = 0.24
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
(i.e.,) P(A ∪ B) = 0.5 + 0.6 – 0.24
= 1.1 – 0.24 = 0.86
∴ P(A ∪ B)= 0.86

(ii) P(\(\overline{\mathrm{A}}\) ∩ B) = P(B) – P(A ∩ B)
= 0.6 – 0.24 = 0.36

(iii) P(A ∩ \(\overline{\mathrm{B}}\)) = P(A) – P(A ∩ B)
= 0.5 – 0.24 = 0.26

(iv) P(\(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})\) = P {(A ∩ B)’} = 1 – P(A ∩ B)
= 1 – 0.24 = 0.76

(v) P\((\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\) = P{A ∪ B)’} = 1 – P(A ∪ B)
= 1 – 0.86 = 0.14.

Question 3.
The probability of an event A occurring is 0.5 and B occurring is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
Given A and B are mutually exclusive and P(A) = 0.5, P(B) = 0.3
∴ P(A ∪ B) = P(A) + P(B) = 0.5 + 0.3 = 0.8
So, P(A’ ∩ B’) = P{(A ∪ B)’} = 1 – P(A ∪ B)
= 1 – 0.8 = 0.2

Question 4.
The probability that a new ship will get an award for its design is 0.25, the probability that it will get an award for the efficient use of materials is 0.35 and that it will get both awards is 0.15. What is the probability, that (/) it will get atleast one of the two awards (ii) it will get only one of the awards?
Solution:
Probability of getting the award for its design = P(A) = 0.25
Probability of getting the award for the efficient use of materials = P(B) = 0.35
Probability of getting both awards = P(A ∩ B) = 0.15
Now P(A) =0.25; P(B) = 0.35 and P(A ∩ B) = 0.15
∴ (i) P(A ∪ B)= P(A) + P(B) – P(A ∩ B)
= 0.25 + 0.35 – 0.15 = 0.60 – 0.15 = 0.45
(ii) P(A’ ∩ B’ or B ∩ A’) = P(A ∩ B’) + P(A’ ∩ B)
P(A ∩ B’) = P(A) – P(A ∩ B)
= 0.25 – 0.15 = 0.10
P(A’ ∩ B) = P(B) – P(A ∩ B)
= 0.35 – 0.15 = 0.20
∴ P(A ∩ B’) + P(A’ ∩ B) = 0.10 + 0.20 = 0.30.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.1

Question 1.
An experiment has the four possible mutually exclusive and exhaustive outcomes A, B, C and D. Check whether the following assignments of probability are permissible.
(i) P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12
(ii) P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34
(iii) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = \(-\frac{1}{5}\), P(D) = \(\frac{1}{5}\)
Solution:
When A, B, C, D are the possible exclusive and exhaustive events the P(A) + P(B) + P(C) + P(D) = 1.
(i) P(A) = 0.15, P(B) = 0.30, P(C) = 0.43, P(D) = 0.12
Now P(A) + P(B) + P(C) + P(D) = 0.15 + 0.30 + 0.43 + 0.12 = 1
0.15 + 0.30 + 0.43 + 0.12 = 1
∴ The assignment of probability is permissible

(ii) P(A) = 0.22, P(B) = 0.38, P(C) = 0.16, P(D) = 0.34
Now P(A) + P(B) + P(C) + P(D) = 1
0.22 + 0.38 + 0.16 + 0.34 = 1.10 = ≠1
∴ The assignment of probability is not permissible.

(iii) P(A) = \(\frac{2}{5}\), P(B) = \(\frac{3}{5}\), P(C) = \(-\frac{1}{5}\), P(D) = \(\frac{1}{5}\)
P(C) = \(-\frac{1}{5}\) which is not possible
(i.e.) for any event A, (0 ≤ P(A) ≤ 1)
∴ The assignment of probability is not permissible.

Question 2.
If two coins are tossed simultaneously, then find the probability of getting
(i) one head and one tail
(ii) at most two tails
Solution:
When two coins are tossed the sample space will be
S = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
(i) probability of getting 1 head and one tail = \(\frac{2}{4}=\frac{1}{2}\)
(ii) Probability of getting atmost two tails = \(\frac{4}{4}\) = 1

Question 3.
Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that (i) one is a mango and the other is an apple (ii) both are of the same variety.
Solution:
(i) Mangoes (M) = 5
Apples (A) = 4 Total = 5 + 4=9
P(mango) = P(M) = 5/9
P(A) = 4/9.
When two Suits are chosen at random
P(one mango and one Apple) = P(MA or AM)
= P(M)P (A) × 2!
= \(\frac{5}{9} \times \frac{4}{8} \times 2 !=\frac{5}{9}\)
(PCM) = \(\frac{5}{9}\) (Sell from 5 mangoes and set 1 from a fruit)
P(A) = \(\frac{4}{8}\) (Sel 1 from 4 apples and set 1 from the remaining 8 fruits)

(ii) P(MMorAA)
= P(M) P(M) + P(A) P(A)

Question 4.
What is the chance that (i) non-leap year (ii) leap year should have fifty three Sundays?
Solution:
(i) Non leap year
No of days =365
= \(\frac{365}{7}\) weeks = 52 weeks + 1 day
In 52 week we have 52 Sundays. So we have to find the probability of getting the remaining one day as Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
(i.e.,) n(S) = 7
In the n (Sunday) = A {Saturday to Sunday or Sunday to Monday}
(i.e.,) n(A) = 1
So, P(A) = 1.
∴ Probability of getting 53 Sundays \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{1}{7}\)

(ii) Leap Year:
In 52 weeks we have 52 Sundays. We have to find the probability of getting one Sunday form the remaining 2 days the remaining 2 days can be a combination of the following S = {Saturday to Sunday, Sunday to Monday, Monday, to Tuesday, Tuesday to wednes¬day, Wednesday to Thursday, Thursday to Friday, Friday and Saturday}.
(i.e) n(s) =7
In this = A {Saturday to Sunday, Sunday to Monday}
(i.e) n(A) = 2
So, P(A) = \(\frac{2}{7}\)

Question 5.
Eight coins are tossed once, find the probability of getting
(i) exactly two tails
(ii)at least two tails
(iii) at most two tails
Solution:
When a coin is tossed 8 times or 8 coins are tossed one time n(s) = 2⁸ = 256
(i) Let A be the event of getting exactly 2 tails.
Here n(A) = 8C₂ = \(\frac{8 \times 7}{2 \times 1}\) = 28
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{28}{256}=\frac{7}{64}\)

(ii) Let B be the event of getting at least two facts.
n(B) = 8C₂ + 8C₃ + ……….. + 8C₈
= n( S) – (8C₀ + 8C₁) = 256 – (1 + 8) = 247
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{247}{256}\)

(iii) Let C be the event of getting atmost two facts.
n(C) = 8C₀ + 8C₁ + 8C₀
= 1 + 8 + \(\frac{8 \times 7}{2 \times 1}\) = 1 + 8 + 28 = 37
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{37}{256}\)

Question 6.
An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8?
Solution:
S= {1, 2, 3, …………. 100}
n(S) = 100
Let A be the event of choosing a prime number
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89}
n(A) = 25 So P(A) = \(\frac{25}{100}\)
Let B be the event of getting a number multiple of 8
B = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
n(B)= 12 So P(B) = \(\frac{12}{100}\)
also A ∩ B = ϕ
⇒ A and B are mutually exclusive
∴ P(A ∪ B) = P(A) + P(B) = \(\frac{25}{100}+\frac{12}{100}=\frac{37}{100}\)

Question 7.
A bag contains 7 red and 4 black balls, 3 balls are drawn at random.
Find the probability that (i) all are red (ii) one red and 2 black.
Solution:
No. of Red balls = n(R) = 7
No. of Black balls = n(B) = 4
Total = 7 + 4 = 11 ⇒ n(S) = 11
Three balls are drawn at random

Question 8.
A single card is drawn from a pack of 52 cards. What is the probability that
(i) the card is an ace or a king
(ii) the card will be 6 or smaller
(iii) the card is either a queen or 9?
Solution:
Total No. of cards = 52 = n(S)
No. of ace cards = n(A) = 4
No. of king card = n(k) = 4
(i) P(A or K) = P(A) + P(K)
(∵ A and K are mutually exclusive).
= \(\frac{n(\mathrm{A})}{n(\mathrm{S})}+\frac{n(\mathrm{K})}{n(\mathrm{S})}=\frac{4}{52}+\frac{4}{52}\)
= \(\frac{8}{52}=\frac{2}{13}\)

(ii) Let B be the event of getting a number be 6 or smaller
So the numbers can be 6, 5, 4, 3, 2
There are 4 types of cards
So n(B) = 4 × 5 = 20
and So, p(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{20}{52}=\frac{5}{13}\)

(iii) Let C be the event of getting a queen ⇒ so n(c) = 4
and Let D be the event of getting a number 9 ⇒ n(D) = 4
Now C ∩ D = ϕ
(i.e.,) C and D are mutually exclusive.
∴ P(C ∪ D) = P(C) + P(D) = \(\frac{4}{52}+\frac{4}{52}=\frac{8}{52}\)
= \(\frac{2}{13}\)

Question 9.
A cricket club has 16 members, of whom only 5 can bowl. What is the probability that in a team of 11 members at least 3 bowlers are selected?
Solution:
No. of players = 16
We need to select 11 players which can be done in 16 C₁₁ ways
(i. e) n(S) = 16C₁₁ ways
= 4368
Out of the selection of 11 members there should be a least 3 bowler So we can have 3 or 4 or 5 bowlers and S the remaining will be 8 or 7 or 6 players. So the selection can be done as follows.

Let A be the event of selecting atleast 3 bowlers out of a selection of 11 players.
So n(A) = (5C₃ × 11C₈) + (5C₄ × 11C₇) + (5C₅) (11C₆)
∴ 5C₃ = 10, 5C₄ = 5, 5C₅ = 1

Question 10.
(i) The odds that the event A occurs is 5 to 7, find P(A)
(ii) Suppose P(B) = \(\frac{2}{5}\). Express the odds that the event B occurs.
Solution:
If the probability of an event is P then the odds is favour of its occurrence are P to (1 – P) and the odds against its occurrence are (1 – P) to P.
Here we are given the odds that The event A occurs = 5 to 7
So, the odds that the event B occurs is 2 to 3.
∴ P(A) = \(\frac{5}{5+7}=\frac{5}{12}\)
(ii) We are given P(B) = \(\frac{2}{5}\)
(i.e.,) P(B) = \(\frac{2}{2+3}\)
So, the odds that the event B occurs is 2 to 3.

Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.1 Additional Problems

Question 1.
An experiment has the four possible mutually exclusive outcomes A, B, C and D. Check whether the following assignments of probability are permissible.
P(A) = 0.32, P(B) = 0.28, P(C) = -0.06, P(D) = 0.46
Solution:
Probability of an event cannot be negative. Here P(C) = – 0.06.
∴ the above set of events are not possible.
P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{6}\), P(C) = \(\frac{2}{9}\), P(D) = \(\frac{5}{18}\)

Question 2.
In a single throw of two dice, find the probability of obtaining
(i) sum of less than 5
(ii) a sum of greater than 10
(iii) a sum of 9 or 11.
Solution:
The sample space when throwing two dice once =
{(1, 1), (1, 2), …………. (1, 6)
(2, 1), ………… (2, 6)
:
:
(6, 1), ……….. (6, 6)}
n(S) = 6² = 36
(i) Let A be the event of getting a sum less than 5.
Then A = {(1, 1), (1, 2), (1, 3) (2, 1),(2, 2) (3, 1)}
n(A) = 6
∴ P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(ii) Let B be the event of getting a sum greater than 10.
∴ The sum will be 11 or 12.
Now the numbers whose sum is 11.
= {(5, 6), (6, 5)}
The number whose sum is 12 = {(6, 6)}
n(B) = 2 + 1 = 3
∴ P(B) = \(\frac{3}{36}=\frac{1}{12}\)

(iii) Let C be the event of getting a sum 9 or 11.
Now C = {(3, 6), (4, 5)
(5, 4), (6, 3)
(5, 6), (6, 5)}
n(C) = 6
∴ P(C) = \(\frac{6}{36}=\frac{1}{6}\)

Question 3.
Three coins are tossed once. Find the probability of getting
(i) exactly two heads
(ii) at least two heads
(iii) atmost two heads.
Solution:
The sample space when three coin are tossed once is as follows:
S = {(H, H, H), (H, T, H), (T, H, H), (H, H, T), (T, T, H), (H, T, T) (T, H, T), (T, T, T)}
n(S) = 2³ = 8
(i) Let A be the event of getting exactly two heads.
∴ A = {(H, T, H) (T, H, H) (H, H, T)}
n(A) = 3
∴ n( A) = \(\frac{3}{8}\)

(ii) Let B be the event of getting at least two heads.
B = {(H, T, H), (T, H, H), (H, H, T), (H, H, H)}
n(B) = 4
∴ P(B) = \(\frac{4}{8}=\frac{1}{2}\)

(iii) Let C be the event of getting atmost two heads.
C = {(T, T, T), (H, T, T), (T, H, T), (T, T, H) (H, H, T), (T, H, H), (H, T, H)}
n( C) = 7 ∴ P(C) = \(\frac{7}{8}\)

Question 4.
A bag contains 5 white and 7 black balls. 3 balls are drawn at random. Find the probability that
(i) all are white
(ii) one white and 2 black.
Solution:
Number of white balls = 5
Number of black balls = 7
Total number of balls = 12
Selecting 3 from 12 balls can be done in

(ii) Let B be the event of selecting one white and 2 black balls.

Question 5.
In a box containing 10 bulbs, 2 are defective. What is the probability that among 5 bulbs chosen at random, none is defective?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 2
∴ Number of good bulbs = 10 – 2 = 8
Now selecting 5 from the 10 bulbs can be done in ¹⁰C₅ ways.

Question 6.
Out of 10 outstanding students in a school there are 6 girls and 4 boys. A team of 4 students is selected at random for a quiz programme. Find the probability that there are atleast two girls.
Solution:
Let A, B and C be the three possible events of selections. The number of combinations are shown below:

Question 7.
An integers is chosen at random from the first fifty positive integers. What is probability that the integer chosen is a prime or multiple of 4.
Solution:
S = {1, 2, 3, ……….. ,50} ∴ n(S) = 50
Let A be the event of getting prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
n(A) = 15, so P(A) = 15/50
Let B be the event of getting number multiple of 4
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
n(B) = 12, so P(B) = 12/50
Here A and B are mutually exclusive. (i.e.,) A ∩ B = ϕ
∴ P(A ∪ B) = P(A) + P(B) = 15/50 + 12/50 = 27/50.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

Question 1.
Find the values of
(i) sin(480°)
(ii) sin(-1110°)
(iii) cos(300°)
(iv) tan(1050°)
(v) cot(660°)
(vi) tan \(\left(\frac{19 \pi}{3}\right)\)
(vii) sin \(\left(-\frac{11 \pi}{3}\right)\)
Solution:
(i) sin(480°) = sin(360° + 120°) = sin 120°
= sin(90° + 30°) = cos 30° = \(\sqrt{3}\)/2

(ii) sin(-1110°) = -sin(1110°)
= – sin (360° × 3 + 30°)
= -sin 30° = -1/2

(iii) cos(300°) = cos(270° + 30°) = sin 30° = 1/2

(iv) tan(1050°) = tan [3(360°) – 30°]

(v) cot(660°) = cot (360° × 2 – 60°)

Question 2.
\(\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)\) is a point on the terminal side of an angle θ is standard position. Determine the trigonometric function values of angle θ .
Solution:

Question 3.
Find the values of other five trigonometric functions for the following:

(i) cos θ = \(-\frac{1}{2}\); θ lies in the III quadrant.
Solution:
Taking the Numerical values

(ii) cos θ = \(\frac{2}{3}\) ; θ lies in the I quadrant
Solution:

(iii) sin θ = –\(\frac{2}{3}\) ; θ lies in the IV quadrant
Solution:

(iv) tan θ = -2; θ lies in the II quadrant
Solution:

(v) sec θ = \(\frac{13}{5}\) ; θ lies in the IV quadrant
Solution:

Question 4.

Solution:
cot(180° + θ) = cot θ
sin (90° – θ) = cos θ
cos(-θ) = cos θ
sin (270 + θ) = – cos θ
tan(-θ) = -tan θ
cosec (360° + θ) = cosec θ

Question 5.
Find all the angles between 0° and 360° which satisfy the equation sin² θ = \(\frac{3}{4}\)
Solution:

Question 6.

Solution:
LHS = sin² 10° + sin² 20° + sin² 70° + sin² 80°
= sin² 10° + sin² (90° – 10°) + sin² 20° + sin²(90° – 20°)
= sin² 10° + (cos 10°)² + sin² 20° + (cos 20°)²
= (sin² 10+ cos² 10) + sin² 20° + cos² 20°
= 1 + 1 = 2 = RHS

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3 Additional Questions Solved

Question 1.
Prove that: sin 600°. tan (-690°) + sec 840°. cot (-945°) = \(\frac{3}{2}\)
Solution:

Question 2.
Prove that sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1 = 0
Solution:
LHS = sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1
Now, sin (270° – θ) = sin {180°+ (90°- θ)}
= – cos (90° – θ) = – sin θ
LHS = – cos θ . cos θ – (- sin θ) (- sin θ) + 1
= – cos² θ – sin² θ + 1
= – (cos² θ + sin² θ) + 1 = -1 + 1 = 0 = RHS

Question 3.
Prove that cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = \(\frac{1}{2}\)
Solution:
cos 204° = cos (180°+ 24°) = – cos 24°
cos 125° = cos (180° – 55°) = – cos 55°
LHS = cos 24° + cos 55° + (- cos 55°) + (- cos 24°) + cos 300°
= cos 24° + cos 55° – cos 55° – cos 24° + cos 300°

Question 4.

Solution:

Question 5.

Solution:

LHS = [(1 +cot α) + cosec α][(1 + cot α) – cosec α]
= (1 + cot α)² – cosec² α
= 1 + cot²α + 2 cot α – cosec² α
[∵ 1 + cot²α = cosec²α]
= cosec²α + 2 cot α – cosec²α
= 2 cot α = RHS

Question 6.

Solution:

= sec (450° – θ)
= sec[360° + (90° – θ)]
sec (90° – θ) = cosec θ

Question 7.

Solution:
cos (90° + θ) = – sin θ
sec (- θ) = sec θ
tan (180° – θ) = – tan θ
sec (360° – θ) = sec θ
sin (180° + θ) = – sin θ
cot (90° + θ) = – tan θ

Question 8.
Find x from the equation cosec (90° + A) + x cos A cot (90° + A) = sin (90° + A).
Solution: