Students must start practicing the questions from RBSE 12th Maths Model Papers Set 5 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 5 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) Consider a binary operation * on N defined as a * b = a^{3} + b^{3}. Choose the correct Answer. (1)

(a) * is both associative and commutative.

(b) * is commutative but not associative.

(c) * is associative but not commutative.

(d) * is neither commutative nor associative.

Answer:

(b) * is commutative but not associative.

(ii) The value of sin^{-1}\(\left(\cos \frac{3 \pi}{5}\right)\) is:

(a) \(\frac{\pi}{10}\)

(b) \(\frac{3 \pi}{5}\)

(c) \(\frac{- \pi}{5}\)

(d) \(\frac{- 3 \pi}{5}\)

Answer:

(c) \(\frac{- \pi}{5}\)

(iii) If A = [2, – 3, 4], B = \(\left[\begin{array}{l}

3 \\

2 \\

2

\end{array}\right]\), X = [1, 2, 3] and Y = \(\left[\begin{array}{l}

2 \\

3 \\

4

\end{array}\right]\), then AB + XY equals:

(a) [28]

(b) [24]

(c) 28

(d) 24

Answer:

(a) [28]

(iv) If A_{ij} is the cofactor of the element a_{ij} of the determinant \(\left|\begin{array}{ccc}

2 & -3 & 5 \\

6 & 0 & 4 \\

1 & 5 & -7

\end{array}\right|\), then the value of a_{32},A_{32} is

(a) 5

(b) 100

(c) – 110

(d) 110

Answer:

(d) 110

(v) If x = a cos θ and y = o sin θ, then \(\) is:

(a) cot θ

(b) tan θ

(c) cosec^{2} θ

(d) – cot θ

Answer:

(d) – cot θ

(vi) \(\int \frac{d x}{\sin ^{2} x \cos ^{2} x}\) equals:

(a) tan x + cot x + C

(b) tan x – cot x + C

(c) tan x cot x + C

(d) tan x – cot 2x + C

Answer:

(b) tan x – cot x + C

(vii) A homogeneous differential equation of the form \(\) = h\(\) can be solved by making the substitution:

(a) y = vx

(b) v – yx

(c) x = vy

(d) x = v

Answer:

(c) x = vy

(viii) The position vector of a point, which divides the join of points with position vectors \(\vec{a}-2 \vec{b}\) and \(2 \vec{a}+\vec{b}\) externally in the ratio 2 : 1 is.

(a) \(3 \vec{a}-4 \vec{b}\)

(b) \(3 \vec{a}+4 \vec{b}\)

(c) \(4 \vec{a}+3 \vec{b}\)

(d) \(4 \vec{a}-3 \vec{b}\)

Answer:

(b) \(3 \vec{a}+4 \vec{b}\)

(ix) From a pack of 52 cards two cards are drawn randomly one by one without replacement. The probability that both of them are red is: (1)

(a) \(\frac{1}{2}\)

(b) \(\frac{25}{51}\)

(c) \(\frac{25}{102}\)

(d) \(\frac{2}{51}\)

Answer:

(c) \(\frac{25}{102}\)

(x) If A = \(\left[\begin{array}{rr}

\cos \alpha & -\sin \alpha \\

\sin \alpha & \cos \alpha

\end{array}\right]\), then for what value of α, A is an identity matrix?

(a) 0

(b) \(\frac{\pi}{2}\)

(c) π

(d) \(\frac{\pi}{4}\)

Answer:

(a) 0

(xi) The value of the constant k, so that the given function is continuous at the indicated point, is:

(a) 5

(b) 0

(c) 10

(d) 25

Answer:

(c) 10

(xii) The projection of the vector î + 3 ĵ + 7 k̂ on the vector 7 î – ĵ + 8k̂ is :

(a) \(\frac{60}{\sqrt{114}}\)

(b) \(\frac{30}{\sqrt{115}}\)

(c) \(\frac{8}{\sqrt{111}}\)

(d) \(\frac{7}{\sqrt{114}}\)

Answer:

(a) \(\frac{60}{\sqrt{114}}\)

Question 2.

Fill in the blanks :

(i) If A = {1, 2, 3}, then number of equivalence relations containing (1, 2) is ____________ . (1)

Answer:

2

(ii) The principal value of cosec^{-1} (- √2) is ____________ . (1)

Answer:

\(\left(-\frac{\pi}{4}\right)\)

(iii) If A = \(\left[\begin{array}{ll}

6 & 9 \\

2 & 3

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

2 & 6 & 0 \\

7 & 9 & 8

\end{array}\right]\), then AB = ____________ . (1)

Answer:

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

75 & 117 & 72 \\

25 & 39 & 24

\end{array}\right]\)

(iv) The value of \(\frac{d}{d x}\) (sin x^{2}) is ____________ . (1)

Answer:

2x cosx^{2}

(v) The value of

is ____________ . (1)

Answer:

2

(vi) If \(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂, then |\(\vec{a}\) . \(\vec{b}\)| is ____________ . (1)

Answer:

√507

Question 3.

Very Short Answer Type Questions

(i) Show that the function f: R → R given by f(x) = x^{3} is injective. (1)

Answer:

Here f : R → R

f(x) = x^{3}

then x_{1}, x_{2} ∈ R (domain)

f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} = x_{2}^{3}

⇒ x_{1} = x_{2}

Thus, f is injective.

(ii) Write the following function in the simplest form :

tan^{-1} \(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), \(\frac{-\pi}{4}\) < x < \(\frac{3 \pi}{4}\)

Answer:

(iii) If A = \(\left[\begin{array}{c}

-2 \\

4 \\

5

\end{array}\right]\), B = [1 3 – 6], verify that (AB)’ = B A’.

Answer:

(iv) Find the value of x if \(\left|\begin{array}{cc}

x+1 & x-1 \\

x-3 & x+2

\end{array}\right|\) = \(\left|\begin{array}{cc}

4 & -1 \\

1 & 3

\end{array}\right|\)

Answer:

Given

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

x+1 & x-1 \\

x-3 & x+2

\end{array}\right|=\left|\begin{array}{cc}

4 & -1 \\

1 & 3

\end{array}\right|\)

On expanding, we get

⇒ (x + 1) (x + 2) – (x – 1) (x – 3) = 4 × 3 – (- 1) × 1

⇒ (x^{2} + x + 2x + 2) – (x^{2} – x – 3x + 3) = 12 + 1

⇒ (x^{2} + 3x + 2) – (x^{2} – 4x + 3) = 13

⇒ x^{2} + 3x + 2 – x^{2} + 4x – 3 = 13

⇒ 7x – 1 = 13

⇒ 7x – 1 + 13

⇒ 7x = 14

Thus, x = 2

(v) Find the derivative of the function sin^{-1}(e^{-x}) with repec1 to x. (1)

Answer:

Let y = sin^{-1}(e^{-x})

Differentiating both sides w.r.t.’x’,

we get

(vi) Integrate the function ∫\(\frac{x^{3}}{\sqrt{1-x^{8}}}\) dx (1)

Answer:

Let I = ∫\(\frac{x^{3}}{\sqrt{1-x^{8}}}\) dx

Putting x^{4} = t

⇒ 4x^{3} dx = dt

or x^{3} dx = \(\frac{1}{4}\) dt

∴ I = \(\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}\) = \(\frac{1}{4}\) sin^{-1}t + C

= \(\frac{1}{4}\) sin^{-1} (x^{4}) + C

(vii) Show that the function y = ax + 2a^{2} is a solution of the differential equation

2\(\left(\frac{d y}{d x}\right)^{2}\) + x\(\left(\frac{d y}{d x}\right)\) – y = 0

Answer:

The given function is

y = ax + 2a^{2} ……. (1)

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

\(\frac{d y}{d x}\) = a

Substituting the value of ‘a’ in equation, we get

(viii) Show that: \((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)

Answer:

(ix) A couple has two children. Find the probability that both children are males, if it is known that at least one of the children is male. (1)

Answer:

If a couple has two children then the sample space is

S = { (b, b),(b, g),(g, b),(g, g)}

Let E and F respectively denote the events that both children are males and at least one of the children is a male.

∴ E ∩ F = {(b, b)} ⇒ P(E ∩ F) = \(\frac{1}{4}\)

P (E) = \(\frac{1}{4}\); P (F) = \(\frac{3}{4}\)

⇒ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{\frac{4}{3}}{4}\) = \(\frac{1}{3}\)

(x) Show that point A(a, b + c), B(b, c + a) and C(c, a + b) are collinear. (1)

Answer:

Three joints will be collinear if area of ∆ABC is zero.

Area of triangle

(Since C_{1} and C_{3} are identical)

Thus, given points are collinear.

Hence Proved

(xi) Verify that the given function is a solution of the corresponding differential

equation x + y = tan^{-1} y : y^{2}y’ + y^{2} + 1 = 0 (1)

Answer:

The given function is:

x + y = tan^{-1}y …….. (1)

Differentiating equation (1) w.r.t.

‘x’, we get

1 + y’ = \(\frac{1}{1+y^{2}}\) . y’

⇒ (1 + y^{2}) + (1 + y^{2})y’ = y’

⇒ 1 + y^{2} + y’ + y^{2}y’ = y’

⇒ y^{2}y’ + y^{2} + 1 = 0

verified

(xii) Find the angle between \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\), where \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\).

Answer:

Section – B

Short Answer Type Questions

Question 4.

Show the function f: R → R defined by f(x) = \(\frac{x}{x^{2}+1}\), as injection, surjection or bijection. (2)

Answer:

(i) Injection test : Let x_{1}, x_{2} ∈ R

f(x_{1}) = f(x_{2}) = \(\frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}\)

⇒ x_{1}(x_{2}^{2} + 1) = x_{2}(x_{1}^{2} + 1)

⇒ x_{1} x_{2}^{2} + x_{1} = x_{2} x_{1}^{2} + x_{2}

⇒ x_{2} x_{2} (x_{2} – x_{1}) = (x_{2} – x_{1})

⇒ (x_{2} – x_{1}) (x_{1}x_{2} – 1) = 0

⇒ x_{2} = x_{1} or x_{1}x_{2} = 1

⇒ x_{1} = x_{2} or x_{1} = \(\frac{1}{x_{2}}\)

Here, f is not an injection.

(ii) Surjection test: Let y ∈ R (co-domain) be any arbitrary element.

Consider, y = f(x)

∴ y = \(\frac{x}{x^{2}+1}\) ⇒ x^{2}y + y = x

⇒ x^{2}y – x + y = 0

⇒ x = \(\frac{1 \pm \sqrt{1-4 y^{2}}\frac{-1}{2}}{2 y}\), which does not exist for 1 – y^{2} < 0, i.e., for y > \(\frac{1}{2}\) and y < \(\).

In particular for y = 1 ∈ R

(co-domain), there does not exist any x ∈ R (domain) such that f(x) = y.

∴ f is not surjection.

Thus, f is neither injection nor surjection.

Question 5.

Express the matrix \(\left[\begin{array}{ccc}

3 & -2 & -4 \\

3 & -2 & -5 \\

-1 & 1 & 2

\end{array}\right]\) as the sum of a symmetric and skew-symmetric matrix and verify your result.

Answer:

So, P is a symmetric matrix and Q is a skew symmetric matrix.

Now, P + Q = \(\) (A + A’) + \(\frac{1}{2}\) (A – A’)

Thus, matrix A is expressed as the sum of symmetric matrix and skew-symmetric matrix.

Question 6.

The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result, we get 7. By adding second and third numbers to three times the first number, we get 12. Using matrices find the number.

Answer:

Let the three numbers be x, y and z respectively. Then,

x + y + z = 6

x + 2z = 7

3x + y + z =12

We obtain the following system of simultaneous linear equations:

x + y + z = 6

x + 0y + 2z = 7

3x + y + z = 12

The above system of equation can be written in matrix form as:

= 1 (0 – 2) – (1 – 6) + 1 (1 – 0)

= – 2 + 5 + 1 = 4 ≠ 0

So, the above system of equations has a unique solutions given by X = A^{-1}B

Cofactors of | A | are :

a_{11} = -2, a_{12} = 5, a_{13} = 1, a_{21} = 0, a_{22} = – 2, a_{23} = 2, a_{31} = 2, a_{32} = – 1 and a_{33} = – 1

Matrix formed by the cofactors of |A| is:

x = 3, y = 1, z = 2

Hence, the number are 3, 1 and 2 respectively.

Question 7.

Find the relationship between a and b so that the function f defined by

is continuous at x = 3. (2)

Answer:

∴ 3a + 1 = 3b + 3

⇒ 3a = 3b + 3 – 1

⇒ 3a = 3b + 2

⇒ a = b + \(\frac{2}{3}\)

If b = k an arbitrary real number, then a = k + \(\frac{2}{3}\)

Thus, a = k + \(\frac{2}{3}\) and b = k

Question 8.

Evaluate ∫\(\frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}\) dx

Answer:

⇒ 2t + 1 = A(t + 4) + Bt

⇒ 2t + i = (A + B)t + 4A

On comparing the coefficients of t and constant terms, we get

A + B = 2

and 4A = 1

On solving these two equations, we get

A = \(\frac{1}{4}\) and B = \(\frac{7}{4}\)

Question 9.

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces. (2)

Answer:

The number of aces is a random variable.

Let it be denoted by X. Clearly, X can take the values 0,1, or 2.

Now, since the draws are done with replacement, therefore, the two draws form independent experiments.

Therefore, P(X = 0) = P(non-ace and non-ace)

= P(non-ace) × P(non-ace)

= \(\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)

P(X = 1) = P(ace and non-ace or non-ace ana ace)

= P(ace and non-ace) + P(non-ace and ace)

= P(ace). P(non-ace) + P(non-ace).P(ace)

= \(\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}=\frac{24}{169}\)

and P(X = 2) = P(ace and ace)

\(\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)

Thus, the required probability distribition is

Question 10.

Find x, if [x – 5 – 1] \(\left[\begin{array}{lll}

1 & 0 & 2 \\

0 & 2 & 1 \\

2 & 0 & 3

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

x \\

4 \\

1

\end{array}\right]\) = 0 (2)

Answer:

We have

[(x – 2) × x – 10 × 4 + (2x – 8) × 1] = [0]

[x^{2} – 2x – 40 + 2x – 8] = [0]

⇒ x^{2} – 48 = 0

⇒ x^{2} = 16 × 3

⇒ x = ±4√3

Question 11.

If y = sin^{-1}\(\left\{x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^{2}}\right\}\) and 0 < x < 1, then find \(\frac{d y}{d x}\). (2)

Answer:

Given

Question 12.

Using properties of determinants, prove that:

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

(b+c)^{2} & a^{2} & b c \\

(c+a)^{2} & b^{2} & c a \\

(a+b)^{2} & c^{2} & a b

\end{array}\right|\) = (a – b) (b – c) (c – a) (a + b + c) (a^{2} + b^{2} + c^{2})

Answer:

Expanding along C1we get

= (b – a) (c – a) (a^{2} + b^{2} + c^{2}) [- b(b + a) + c(c + a)]

= (b – a) (c – a) (a^{2} + b^{2} + c^{2}) [- b^{2} – ab + c^{2} + ac]

= (b – a) (c – a) (a^{2} + b^{2} + c^{2}) [(c^{2} – b^{2}) + (ca – ab)]

= (b – a) (c – a) (a^{2} + b^{2} + c^{2}) [(c – b) (c + b) + a(c – b)]

= (b – a) (c – a) (a^{2} + b^{2} + c^{2}) (c – b) (a + b + c)

= (a – b) (b – c) (c – a) (a + b + c) (a^{2} + b^{2} + c^{2})

= RHS.

Question 13.

If

then , find the value of a. (2)

Answer:

Question 14.

For the following differential equation, find the general solution: (2)

sec^{2} x tan y dx + sec^{2}y tan x dy = 0

Answer:

Given differential equation is

sec^{2}x tan y dx + sec^{2}y tan x dy = 0

Dividing equation (1) by tan x tany

\(\frac{\sec ^{2} x}{\tan x}\)dx + \(\frac{\sec ^{2} y}{\tan y}\)dy = 0 …. (2)

Integrating equation (2), we get

∫ \(\frac{\sec ^{2} x}{\tan x}\)dx + ∫ \(\frac{\sec ^{2} y}{\tan y}\)dy = 0

log |tan x| + log |tan y| = log C

or log |tan x tan y| = log C

or tan x tan y = C…(3)

(x, y ∈ R and x, y is not the odd multiple of π/2)

Equation (3) is the required solution of given differential equation.

Question 15.

If î + ĵ + k̂, 2î + 5ĵ, 3î + 2ĵ – 3k̂ and î – 6ĵ – k̂ are the position vectors of points A, B, C and D respectively, then find the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\). Deduce that \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear. (2)

Answer:

Note that if θ is the angle between AB and CD, then θ is also the angle between \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\).

Now, \(\overrightarrow{A B}\) = Position vector of B – Position vector of A

Since 0 ≤ θ ≤ π, it follows that θ = π. This shows that \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear.

Question 16.

A die is thrown. If E is the event “the number appearing is a multiple of 3” and F be the event “the number appearing is even”, then find whether E and F are independent? (2)

Answer:

Sample space,

S= (1, 2, 3, 4, 5, 6)

Now E = (3, 6), F = {2,4,6}

and E ∩ F = {6}

then P(E) = \(\frac{2}{6}=\frac{1}{3}\)

P(F) = \(\frac{3}{6}=\frac{1}{2}\)

and P(E ∩ F) = \(\frac{1}{6}\)

Clearly P(E ∩ F) = P(E).P(F)

= \(\frac{1}{3} \times \frac{1}{2}=\frac{1}{6}\)

Thus, events E and F are independent events.

Section – C

Long Answer Type Questions

Question 17.

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

Prove that: sin^{-1}: sin^{-1}\(\left(2 x \sqrt{1-x^{2}}\right)\) = 2 cos^{-1}x, \(\frac{1}{\sqrt{2}}\) ≤ x ≤ 1.

Answer:

(tan^{-1} x)^{2} + (cot^{-1} x)^{2} = \(\frac{5 \pi^{2}}{8}\)

⇒ (tan^{-1}x + cot^{-1})^{2} – 2 tan^{-1}x cot^{-1}x = \(\frac{5 \pi^{2}}{8}\)

⇒ 16(tan^{-1} x)^{2} – 8π(tan^{-1} x) – 3π^{2} = 0

(on solving by tan^{-1} x + cot^{-1} x = \(\frac{\pi}{2}\))

⇒ 16(tan^{-1}x – \(\frac{3 \pi}{4}\)) (tan^{-1}x + \(\frac{\pi}{4}\)) = 0

⇒ tan^{-1}x = – \(\frac{\pi}{4}\)

⇒ x = – 1 [∵ – \(\frac{\pi}{2}\) < tan^{-1} x < \(\frac{\pi}{2}\)]

Question 18.

Verify Lagrange’s mean value theorem for the following function on the indicated interval. In each case find a point ‘c’ in the indicated interval as stated by the Lagrange’s mean value theorem: f(x) = x^{2} + x – 1 on [o, 4] (3)

Or

If y = cosec^{-1}x, x > 1, then show that : x (x^{2} – 1)\(\frac{d^{2} y}{d x^{2}}\) + (2x^{2} – 1)\(\frac{d y}{d x}\) = 0. (3)

Answer:

Given,

f(x) = x^{2} + x – 1 on (0, 4) Every polynomial function is continuous everywhere on (- ∞, ∞) and differentiable for all arguments. Here, f (x) is a polynomial function. So, it is continuous in 10,4] and differentiable on (0,4). So, both the necessary conditions of Lagrange’s mean value theorem is satisfied.

Therefore, there exist a point CE (0, 4) such that:

f'(c) = \(\frac{f(4)-f(0)}{4-0}\) = \(\frac{f(4)-f(0)}{4}\)

Now, f(x) = x^{2} + x – 1

Differentiating both sides w.r.t.’x’,

we get

f(x) = 2c + 1

For f’(4), put the value of x = 4 in f(x):

f'(c) = 2c + 1

For f(4), put the value of x = 4 in f(x):

f(4) = 4^{2} + 4 – 1 = 16 + 4 – 1 = 19

For) (0), put the value of x = 0 in f(x):

f(0) = 0^{2} + 0 – 1 = 0 + 0 – 1 = – 1

Now, f’(c) = \(\frac{f(4)-f(0)}{4}\)

⇒ 2c + 1 = \(\frac{19-(-1)}{4}=\frac{20}{4}\)

⇒ 2c + 1 = 5 ⇒ 2c = 5 – 1 = 4

⇒ c = \(\frac{4}{2}\) = 2 ∈ (0,4)

Hence, Lagrange’s mean value theorem is verified.

Question 19.

Find : ∫\(\frac{1}{1+\cot x}\) dx

Or

Find : ∫\(\frac{x+3}{\sqrt{5-4 x-x^{2}}}\) dx

Answer:

Question 20.

Show that each of the given three vectors is a unit vector:

\(\frac{1}{7}\) (2î + 3ĵ + 6k̂), latex]\frac{1}{7}[/latex](3î – 6ĵ + 2k̂), \(\frac{1}{7}\)(6î + 2ĵ – 3k̂)

Or

Find the area of the triangle with vertices A(1, 1, 2), B (2, 3, 5) and C (1, 5, 5). 3

Answer:

Hence, given three vectors is a unit vector.

Section – D

Essay Type Questions

Question 21.

Evaluate:

Or

Evaluate:

Answer:

Let

Question 22.

Show that the given differential equation is homogeneous and solve it:

(x – y)dy – (x + y)dx = 0. (4)

Or

Find the particular solution of the differential equation

\(\frac{d y}{d x}\) + y cot x = 2x + x^{2} cot x (x ≠ 0); given that y = 0 when x = \(\frac{\pi}{2}\) (4)

Answer:

Given differential equation is

(x – y)dy – (x + y)dx = 0

Thus, f(x, y) is a homogeneous function of degree zero.

So, the given differential equation is homogeneous.

Now putting y = vx … (2)

which is the required solution.

Question 23.

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that

(i) all the five cards are spades?

(ii) only 3 cards are spades?

(iii) none is a spade? (4)

Or

Find the probability of throwing at most 2 sixes in 6 throws of a single die. (4)

Answer:

Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shufflied deck of 52 cards, there are 13 spade cards.

⇒ p = \(\frac{13}{52}=\frac{1}{4}\)

∴ q = 1 – \(\frac{1}{4}=\frac{3}{4}\)

X has a binomial distribution with n = 5 and p = \(\frac{1}{4}\)

P(X = x) = ^{n}C_{x} q^{n-x} p^{x}, where x = 0, 1, …….. n

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