Students must start practicing the questions from RBSE 12th Maths Model Papers Set 2 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 2 with Answers in English

Time : 2 Hours 45 Min.

Maximum Marks : 80

General Instructions to the Examinees:

- Candidate must write first his/her Roll. No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.
- Write down the serial number of the question before attempting it.

Section – A

Question 1.

Multiple Choice Questions

(i) If the binary operation * on the set of integers Z, is defined by a * b = a + 3b^{2}, then find the value of 8 * 3 is : (1)

(a) 11

(b) 24

(c) 17

(d) 35

Answer:

(d) 35

(ii) If sin^{-1} x = y, then : (1)

(a) 0 ≤ y ≤ π

(b) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)

(c) 0 < y < π

(d) \(-\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)

Answer:

(b) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)

(iii) If \(\left[\begin{array}{cc}

x+y & 4 \\

-5 & 3 y

\end{array}\right]=\left[\begin{array}{rr}

3 & 4 \\

-5 & 6

\end{array}\right]\), then the values of x and y are: (1)

(a) x = 1, y = 2

(b) x – 2, y = 3

(c) x = 2, y = 1

(d) x = 3, y = 2

Answer:

(a) x = 1, y = 2

(iv) If A = \(\left[\begin{array}{lll}

a & 0 & 0 \\

0 & a & 0 \\

0 & 0 & a

\end{array}\right]\), then det (adj A) equals : (1)

(a) a^{27}

(b) a^{9}

(c) a^{6}

(d) a^{2}

Answer:

(c) a^{6}

(v) If y = cos(√3x), then find \(\frac{dy}{dx}\) is: (1)

(a) \(\frac{\sin \sqrt{3 x}}{2 \sqrt{3 x}}\)

(b) \(\frac{-3 \sin \sqrt{3 x}}{2 \sqrt{3 x}}\)

(c) \(\frac{-3 \sin \sqrt{x}}{2}\)

(d) \(\frac{\sin \sqrt{3 x}}{2 x}\)

Answer:

(b) \(\frac{-3 \sin \sqrt{3 x}}{2 \sqrt{3 x}}\)

(vi) ∫\(\frac{(x-1)(x-\log x)^{3}}{x}\)dx is equal to: (1)

(a) \(\frac{(1-\log x)^{4}}{4}\) + C

(b) \(\frac{(3 x-\log x)^{4}}{3}\) + C

(c) \(\frac{(x-\log x)^{3}}{4}\) + C

(d) \(\frac{(x-\log x)^{4}}{4}\) + C

Answer:

(d) \(\frac{(x-\log x)^{4}}{4}\) + C

(vii) The sum of the order and degree of the differential equation \(\frac{d}{d x}\left\{\left(\frac{d y}{d x}\right)^{3}\right\}\) = 0 is: (1)

(a) 1

(b) 2

(c) 3

(d) 4

Answer:

(c) 3

(viii) If λ(3î + 2ĵ – 6k̂) is a unit vector, then the value of λ is : (1)

(a) ± 7

(b) ± \(\sqrt{43}\)

(c) ± \(\frac{1}{\sqrt{43}}\)

(d) ± \(\frac{1}{7}\)

Answer:

(c) ± \(\frac{1}{\sqrt{43}}\)

(ix) Three dice are thrown simultaneously. The probability of obtaining a total score of 5 is: (1)

(a) \(\frac{5}{216}\)

(b) \(\frac{1}{6}\)

(c) \(\frac{1}{36}\)

(d) \(\frac{1}{49}\)

Answer:

(d) \(\frac{1}{49}\)

(x) If A is square matrix such that A^{2} = A, then(I + A)^{3} – 7A is equal to: (1)

(a) A

(b) I – A

(c) I

(d) 3A

Answer:

(c) I

(xi) Find \(\frac{d^{2} y}{d x^{2}}\) if y = x^{3} + tan x. (1)

(a) 3x^{2} + sec^{2} x

(b) 6x + 2 sec^{2} x tanx

(c) 6x + sec^{2} x

(d) sec^{2} x tan x

Answer:

(b) 6x + 2 sec^{2} x tanx

(xii) If \(\vec{a} \cdot \vec{b}=-|\vec{a}||\vec{b}|\) then the angle between a and b is : (1)

(a) 180°

(b) 60°

(c) 45°

(d) 90°

Answer:

(a) 180°

Question 2.

Fill in the blanks :

(i) A relation in a set A is called ____________ relation, if each element of A is related to itself. (1)

Answer:

reflexive

(ii) If tan^{-1}x + tan^{-1}y = \(\frac{\pi}{4}\) xy < 1, then the value of x + y + xy is ____________ (1)

Answer:

1

(iii) If [2 1 3]\(\left[\begin{array}{ccc}

-1 & 0 & -1 \\

-1 & 1 & 0 \\

0 & 1 & 1

\end{array}\right]\left[\begin{array}{c}

1 \\

0 \\

-1

\end{array}\right]\) = A, then the order of matrix A is ____________ (1)

Answer:

|x|

(iv) The number of points of discontinuity of f defined by f(x) = |x| – |x + 1| is ____________ (1)

Answer:

zero

(v) The value of ∫^{π/2} _{0} cos 2x dx is ____________. (1)

Answer:

0

(vi) The magnitude of \(\left|\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\right|\) is ____________. (1)

Answer:

1

Question 3.

Very Short Answer Type Questions

(i) Find gof and fog, if f(x) = 8x^{3} and g(x) = x^{1/3}. (1)

Answer:

Given, f(x) = 8x^{3} and g(x) = x^{1/3}

then gofix) = g(f(x)) = g(8x^{3})

= (8x^{3})^{1/3} = [(2x)^{3}]^{1/3}

= 2x

Now ,fog{x) =f(g(x)) = f(x^{1/3})

= 8(x^{1/3})^{3} = 8x

(ii) Find the domain of sec^{-1}(3x – 1). (1)

Answer:

We know that the range of sec x is (- ∞, – 1) ∪ (1, ∞)

3x -1 < -1 and 3x – 1 ≥ 1

⇒ 3x < 0 and 3x ≥ 2

⇒ x < 0 and x ≥ \(\frac{2}{3}\)

∴ x ∈ (- ∞, 0) x ∈ (\(\frac{2}{3}\), ∞)

∴ x ∈ (- ∞, 0) ∪ (\(\frac{2}{3}\), ∞)

Thus, domain is (- ∞, 0) ∪ (\(\frac{2}{3}\), ∞)

(iii) Construct a 2 × 2 matrix, A = [a_{ij}], whose elements are given by: a_{ij} = \(\frac{(i+j)^{2}}{2}\) (1)

Answer:

Given a_{ij} = \(\frac{(i+j)^{2}}{2}\)

(iv) Let A be a square matrix of order 3 × 3. Write the value of |2A|, where | A | =4. (1)

Answer:

Given (A) = 4

|2A| = 2^{3}. | A |

(Using | kA | = k^{n}. | A |)

= 2^{3} × 4 = 8 × 4 = 32 (here k = 2, n = 3)

(v) If y – x|x|, find \(\frac{dy}{dx}\) for x < 0; (1)

Answer:

We have, y = x|x|

When x < 0, then | x | = – x

y = x(- x) = – x^{2}

Thus, \(\frac{dy}{dx}\) = – 2x

(vi) Evaluate: ∫\(\frac{\sin ^{6} x}{\cos ^{8} x}\)dx (1)

Answer:

(vii) Find the differential equation representing the family of curves y = ae^{2x} + 5, where a is an arbitrary constant. (1)

Answer:

Given y = ae^{2x} + 5 ……..(i)

(viii) Find |\(\vec{a}\)| and |\(\vec{b}\)|, if \(|\vec{a}|=2|\vec{b}|\) and \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 12. (1)

Answer:

(ix) If A and B are independent events and P(A) = 0.2, P(B) – 0.5, then find P(A ∪ B). (1)

Answer:

Given,

P(A) = 0.2 and P(B) = 0.5

Now, P(A u B)

= P(A) + P(B) – P(A)P(B)

= 0.2 + 0.5 – 0.2 × 0.5

= 0.70 – 0.10 = 0.60

(x) Find the value of λ, so that the points (1, -5), (4, – 5) and (λ, 7) are collinear. (1)

Answer:

Given, the points (1, – 5), (-4, 5) and (λ, 7) are collinear, then

\(\left|\begin{array}{ccc}

1 & -5 & 1 \\

-4 & 5 & 1 \\

\lambda & 7 & 1

\end{array}\right|\) = 0

On applying R_{2} → R_{2} – R_{1} and R_{2} → R_{3} – R_{1}

⇒ \(\left|\begin{array}{ccc}

1 & -5 & 1 \\

-5 & 10 & 0 \\

\lambda-1 & 12 & 0

\end{array}\right|\) = 0

Expanding along C_{3}, we get

1[- 60 – 10(λ -1)] = 0

⇒ -6o – 10λ + 10 = o ⇒ – 10λ = 5o

Thus, λ = – 5

(xi) Verify that the given functions (explicit or implict) is a solution of the corresponding differential equation : y = cos x + C : y’ + sin x = 0. (1)

Answer:

The given function is :

y = cos x + C …(1)

Differentiating equation (1) w.r.t ‘x’, we get

y’ = – sin x

or y’ + sin x = 0

Hence, the given function y = cos x + C is the solution of the differential equation y’ + sin x = 0.

Hence Proved.

(xii) Find the projection of the vector \(\vec{a}\) = 2î + 3ĵ + 2k̂ on the vector \(\vec{b}\) = î + 2ĵ + k̂ . (1)

Answer:

The projection of vector \(\vec{a}\) on the vector \(\vec{b}\) is

Section – B

Short Answer Type Questions

Question 4.

Check if the relation R on the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y) : y is divisible by x} is (i) symmetric (ii) transitive. (2)

Answer:

Given, A = {1, 2, 3, 4, 5, 6}

R = {(x, y) :y is divisible by x}

(i) Symmetric: Now, (2,4) ∈ R [as 4 is divisible by 2]

But (4, 2)jfR [as 2 is not divisible by 4]

∴ R is not symmetric.

(ii) Transitive ; Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y-

∴ z is divisible by x.

⇒ (x, y) ∈ R

∴ R is transitive

Thus, R is reflexive but not symmetric.

Question 5.

Find the matrix A such that \(\left[\begin{array}{rr}

2 & -1 \\

1 & 0 \\

-3 & 4

\end{array}\right]\)A = \(\left[\begin{array}{rr}

-1 & -8 \\

1 & -2 \\

9 & 22

\end{array}\right]\) (2)

Answer:

On equating the corresponding elements, we get

⇒ 2a – c = — 1 …(i)

2b -d = – 8 …(ii)

a = 1 …(iii)

b = -2 …(iv)

-3a + 4c = 9 …(v)

– 3b+ M = 22 …(vi)

On solving these equations, we get a = 1,b = -2, c = 3, d = 4

Thus, A = \(\left[\begin{array}{cc}

1 & -2 \\

3^{-} & 4

\end{array}\right]\)

Question 6.

Solve the following system of linear equations by matrix method : (2)

2x – y = – 2 and 3x + 4y = 3

Answer:

Given system of equations is:

2x – y = -2

3x + 4y = 3

Writing the given system of equations in matrix form,

AX = B…….(i)

= 2 x 4 – 3 x (-1) = 8 + 3 = 11

∴ | A | = 11 ≠ 0

Since, matrix A is non-singular so A^{-1} exists, and system is consistent.

If A_{ij} is cofactor of a., in A

A_{11} = (-1)^{1+1} 4 = 4

A_{12} = (- 1)^{1+2} = 3 = – 3

A_{21} = (-1)^{2+1} (-1)

= (-1)(-1) = 1

A_{22} = (-1)^{2+2} 2 = 2

Matrix formed by the cofactor of |A| is:

Question 7.

If y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \infty}}}\), then find \(\frac{dy}{dx}\). (2)

Answer:

Given y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \infty}}}\)

⇒ y = \(\sqrt{\log x+y}\)

Squaring both sides, we get

y^{2} = logx + y

Differentiating both sides by x, we get

Question 8.

Evaluate : ∫ sin x. log cos x dx. (2)

Answer:

Let

I = ∫ sin x. log cos x dx

Put cos x = t ⇒ – sin x dx = dt

I = – ∫ log tdt = -∫ (log t).1 dt

= -[log t∫1 dt – ∫{\(\frac{d}{dx}\)(log t)∫1 dt}dt]

= – t log t + ∫1.dt = – t log t + t + C

= – cos x log cos x + cos x + c

Question 9.

From well shuffled pack of 52 cards, 4 cards are drawn one by one and drawn cards are replaced. Find the probability that : (2)

(i) all cards are of heart (ii) 3 cards are of heart

Answer:

Probability to get card of heart

p = \(\frac{13}{52}=\frac{1}{4}\)

∴ Probability not to get card of heart

q = 1 – p = 1 – \(\frac{1}{4}=\frac{3}{4}\)

If random variable X denotes num¬ber of cards of heart, then by formula P(X = r) = ^{n}C. p^{r} q^{n-r} where n = 4

(i) P(X = 4) = ^{4}C_{4}p^{4}q^{4-4}

= 1 × \(\left(\frac{1}{4}\right)^{4} \times\left(\frac{3}{4}\right)^{0}=\left(\frac{1}{4}\right)^{4}\)

(ii) P(X = 3) = ^{4}C_{3}p^{3}q^{4-3}

= 4 × \(\left(\frac{1}{4}\right)^{3} \times\left(\frac{3}{4}\right)^{1}=12\left(\frac{1}{4}\right)^{4}\)

Question 10.

If A = \(\left[\begin{array}{ccc}

0 & 6 & 7 \\

-6 & 0 & 8 \\

7 & -8 & 0

\end{array}\right]\) B = \(\left[\begin{array}{lll}

0 & 1 & 1 \\

1 & 0 & 2 \\

1 & 2 & 0

\end{array}\right]\) and C = \(\left[\begin{array}{c}

2 \\

-2 \\

3

\end{array}\right]\), then calculate AC, BC and (A + B) Also, verify that (A + B)C = AC + BC (2)

Answer:

Question 11.

If sin y = x cos (a + y), then show that \(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos a}\). Also, show that \(\frac{d y}{d x}\) = cos a,when x = 0.

Answer:

Given, sin y = x cos (a + y) ……….(i)

Question 12.

By using properties of determination, prove that: \(\left|\begin{array}{lll}

a+b & b+c & c+a \\

b+c & c+a & a+b \\

c+a & a+b & b+c

\end{array}\right|=2\left|\begin{array}{lll}

a & b & c \\

b & c & a \\

c & a & b

\end{array}\right|\) (2)

Answer:

Question 13.

Evaluate : ∫^{1} _{-1}\(\frac{d x}{x^{2}+2 x+5}\) (2)

Answer:

Question 14.

Find the particular solution of the differential equation (1 – y^{2})(1 + log | x |) dx + 2xy dy = 0 given that ij = 0, when x = 1. (2)

Answer:

Given differential equation is :

(1 – y^{2})(1 + log | x |) dx + 2xy dy = 0

⇒ \(\frac{(1+\log |x|)}{x}\)dx + \(\frac{2 y}{1-y^{2}}\)dy = 0

On integrating, we get

Question 15.

Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) , having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{1}{2}\). (2)

Answer:

Question 16.

P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact ? Do you think, when they agree, mean both are speaking truth ? (2)

Answer:

Let P_{T}: Event that P speaks truth and Q_{T}: Event that Q speaks truth.

Given, P(P_{T}) = \(\frac{70}{100}\) ,then

P(P̄_{T}) = 1 – \(\frac{70}{100}=\frac{30}{100}\)

and P(Q_{T}) = \(\frac{80}{100}\), then

P̄(Q_{T}) = 1 – \(\frac{80}{100}=\frac{20}{100}\)

P(A and B are agree to each other)

Percentage of P(A and B are agree to each other)

= \(\frac{62}{100}\) × 100 = 62%

No, agree does not mean that they are speaking truth.

Section – C

Long Answer Type Questions

Question 17.

Prove that: \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^{-1} = \(\left(\frac{1}{3}\right)=\frac{9}{4}\)sin^{-1}\(\left(\frac{2 \sqrt{2}}{3}\right)\) (3)

Or

If (tan^{-1}x)^{2} + (cot^{-1}x)^{2} = \(\frac{5 \pi^{2}}{8}\), then find x. (3)

Answer:

Question 18.

If x\(\sqrt{1+y}\) + y\(\sqrt{1+x}\) = 0, then prove that: (1 + x)^{2}\(\frac{d y}{d x}\) + 1 = 0. (3)

Or

If y = sin^{-1}\(\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\), then show that: \(\frac{d y}{d x}=\frac{-1}{2 \sqrt{1-x^{2}}}\) (3)

Answer:

Given, x\(\sqrt{1+y}\) + y\(\sqrt{1+x}\) = 0

⇒ x\(\sqrt{1+y}\) = – y\(\sqrt{1+x}\)

⇒ x^{2}\(\sqrt{1+y}\) = – y^{2}\(\sqrt{1+x}\)

⇒ x^{2} + x^{2}y = y^{2} + xy^{2}

⇒x^{2} – y^{2} = xy^{2} – x^{2}y

⇒ (x – y) (x + y) = – xy(x – y)

⇒ x + y = -xy ⇒ x = -xy – y

⇒ x = -y(x + 1) ⇒ y = \(\frac{-x}{1+x}\)

Differentiating both sides w.r.t.x, we get

Question 19.

Evaluate : ∫\(\frac{x^{2}}{(x \sin x+\cos x)^{2}}\) dx. (3)

Or

Evaluate : ∫\(\frac{\sqrt{1-\sin x}}{1+\cos x}\)e^{-x/2} dx. (3)

Answer:

Let I = ∫\(\frac{x^{2}}{(x \sin x+\cos x)^{2}}\) dx

Multiplying N^{r} and D^{r} by cos x, we get

Putting x sin x + cos x = t

⇒ (x cos x + sin x – sin x)dx = dt

⇒ xcos xdx=dt

Let I_{1} = ∫\(\frac{x \cos x}{(x \sin x+\cos x)^{2}}\) dx

= ∫\(\frac{d t}{t^{2}}=\frac{-1}{t}=\frac{-1}{x \sin x+\cos x}\)

[Putting t = x sin x + cos x]

Now, integrating Eq. (i) by parts, we get

Question 20.

Let \(\vec{a}\) = î + ĵ + k̂,\(\vec{b}\) = 4î – 2ĵ + 3k̂ and \(\vec{c}\) = î – 2ĵ + k̂.Find a vector of magnitude 6 units, which is parallel to the vector 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\). (3)

Or

Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î -2j + 7k̂ and \(\vec{c}\) =2î – j + 4k̂.Find a vector \(\vec{p}\), which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{p} \cdot \vec{c}\) = 18. (3)

Answer:

Given,

\(\vec{a}\) = î + ĵ + k̂,\(\vec{b}\) = 4î – 2ĵ + 3k̂ and \(\vec{c}\) = î – 2ĵ + k̂

∴ 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\)

2(î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3(î – 2ĵ + k̂)

2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂+ 3î – 6ĵ + 3k̂

= î – 2ĵ + 2k̂

Now, a unit vector in the direction of vector

Hence, vector of magnitude 6 units parallel to the vector

2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\) = 6(\(\frac{1}{3}\)î – \(\frac{2}{3}\)ĵ + \(\frac{2}{3}\)k̂)

= 2î – 4ĵ + 4k̂

Section – D

Essay Type Questions

Question 21.

Evaluate: ∫^{5} _{2}[|x – 2| + |x – 3| + |x – 5|]dx. (4)

Or

Prove that : ∫^{π/4} _{0}(\(\sqrt{\tan x}+\sqrt{\cot x}\)) dx = √2.\(\frac{\pi}{2}\). (4)

Answer:

First, we redefined the integrand of the integral between the given limits (2, 5). After that integrate and simplify it.

For, 2 ≤ x < 5, | x – 2| =(x-2), 2 ≤ x < 3, | x – 3 | = – (x – 3)

3 ≤ x < 5, | x – 3| = (x – 3) and 2 ≤ x < 5, | x – 51 = (5 – x)

Now, I = ∫^{5} _{2}[|x – 2| + |x – 3| + |x – 5|]dx

Question 22.

Find the particular solution of the differential equation e^{x} \(\sqrt{1-y^{2}}\) dx + \(\frac{y}{x}\)dy = o, given that y = 1,when x = 0. (4)

Or

Form the differential equation of the family of circles touching the x-axis at origin. (4)

Answer:

Given, differential equation is

e^{x} \(\sqrt{1-y^{2}}\) dx + \(\frac{y}{x}\)dy = o

⇒ e^{x} \(\sqrt{1-y^{2}}\) dx = – \(\frac{y}{x}\)dy

On separating the variables, we get

\(\frac{-y}{\sqrt{1-y^{2}}}\) dy = x e^{x} dx

On integrating both sides, we get

∫\(\frac{-y}{\sqrt{1-y^{2}}}\) dy = ∫x e^{x} dx

On putting 1 – y^{2} = t ⇒ -ydy = \(\frac{d t}{2}\) in LHS, we get

using integration by parts]

⇒ \(\sqrt{1-y^{2}}\) = xe^{x} – ∫e^{x} dx [put t = 1 – y^{2}]

⇒ \(\sqrt{1-y^{2}}\) = x e^{x} – e^{x} + C …(i)

Also, given that y = 1, when x = 0

On putting y = 1 and x = 0 in Eq.(i), we get

\(\sqrt{1-1}\) = 0 – e°+ C

⇒ C = 1 [∵ e° = 1]

On substituting the value of C in Eq. (i), we get

\(\sqrt{1-y^{2}}\) = x e^{x} – e^{x} + 1

which is the required particular solution of given differential equation.

Question 23.

An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question? (4)

Or

Consider the experiment of throwing a dice, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’. (4)

Answer:

The given data can be tabulated as :

Let us denote: E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions

Total number of questions = 1400

Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is

P(E ∩ M) = \(\frac{500}{1400}=\frac{5}{14}\)

Probability of selecting a multiple choice questions, P(M), is

\(\frac{900}{1400}=\frac{9}{14}\)

P(E/M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.

∴ P(E/M) = \(\frac{P(E \cap M)}{P(M)}=\frac{\frac{5}{14}}{\frac{9}{14}}=\frac{5}{9}\)

Therefore, the required probability is \(\frac{5}{9}\)

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