Students must start practicing the questions from RBSE 12th Maths Model Papers Set 6 with Answers in English Medium provided here.
RBSE Class 12 Maths Board Model Paper Set 6 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 f: R’ → R be given by f(x) = (3 – x3)\(\frac{1}{3}\), then fof(x) is : (1)
(a) x\(\frac{1}{3}\)
(b) x3
(c) x
(d) (3- x3)
Answer:
(c) x
(ii) The simplest form of cot-1\(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\), x >1 is: (1)
(a) tan-1x
(b) sec-1x
(c) cosec-1x
(d) sin-1x
Answer:
(b) sec-1x
(iii) The simplest form of the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3 : (1)
(a) 81
(b) 27
(c) 64
(d) 8
Answer:
(a) 81
(iv) Minor of an element of a determinant of order (n ≥ 2) is a determinant of order : (1)
(a) n
(b) n – 1
(c) n + 1
(d) n + 2
Answer:
(b) n – 1
(v) If x = a(θ – sin θ) and y = a(1 + cos θ), then \(\frac{dy}{dx}\) at θ = \(\frac{\pi}{3}\) is : (1)
(a) √2
(b) √3
(c) – √3
(d) 1
Answer:
(c) – √3
(vi) ∫\(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) is equal to: (1)
(a) tanx + cotx + C
(b) tanx + cosecx + C
(c) -tanx + cotx + C
(d) tanx + secx + C
Answer:
(a) tanx + cotx + C
(vii) The sum of the order and degree of the differential equation
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{3}\) + x4 = 0. (1)
(a) 2
(b) 3
(c) 5
(d) 4
Answer:
(d) 4
(viii) If \(\vec{a}\) is a non-zero vector of magnitude ‘a’ and λ a non-zero scalar, then λ \(\vec{a}\) is unit vector if : (1)
(a) λ = 1
(b) λ = – 1
(c) a = |λ|
(d) a = 1/|λ|
Answer:
(d) a = 1/|λ|
(ix) If P(Ā) > 0.7, P(B) = 0.7 and P(\(\frac{B}{A}\)) = 0.5, then P(A ∪ B) is : (1)
(a) 0.15
(b) 0.85
(c) 3/14
(d) 5/12
Answer:
(b) 0.85
(x) The simplest form of cos θ\(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sin θ \(\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\) (1)
(a) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
(c) 1
(d) 2
Answer:
(a) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
(xi) If v = 5 cos x – 3 sin x, then \(\frac{d^{2} y}{d x^{2}}\) is: (1)
(a) – 5 sin x – 3 cos x
(b) – 5 cos x + 3 sin x
(c) 3 sin x – 5 cos x
(d) 3 cos x + 5 sin x
Answer:
(b) – 5 cos x + 3 sin x
(xii) The unit vector perpendicular to each of the vertors (1)
(a) \(\frac{7 \hat{i}-6 \hat{j}+10 \hat{k}}{\sqrt{185}}\)
(b) \(\frac{7 \hat{i}+6 \hat{j}+10 \hat{k}}{\sqrt{185}}\)
(c) \(\pm\left(\frac{7 \hat{i}-6 \hat{j}-10 \hat{k}}{\sqrt{185}}\right)\)
(d) \(\pm\left(\frac{6 \hat{i}-7 \hat{j}+10 \hat{k}}{\sqrt{185}}\right)\)
Answer:
(c) \(\pm\left(\frac{7 \hat{i}-6 \hat{j}-10 \hat{k}}{\sqrt{185}}\right)\)
Question 2.
Fill in the blanks :
(i) If * is a binary operation on N given by a * b = LCM (a, b) for all a, b e N, then 5 * 7 is ____________. (1)
Answer:
35
(ii) The value of tan-1(tan\(\frac{7 \pi}{6}\)) is ____________. (1)
Answer:
\(\frac{\pi}{6}\)
(iii) If A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), then (A + A’) is ____________. (1)
Answer:
\(\left[\begin{array}{cc}
2 & 11 \\
11 & 14
\end{array}\right]\)
(iv) If y = tan-1\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\) then \(\frac{d y}{d x}\) is ____________. (1)
Answer:
\(\frac{3}{1+x^{2}}\)
(v) The value of ∫32\(\frac{d x}{x^{2}-1}\) is ____________. (1)
Answer:
\(\frac{1}{2}\)log\(\frac{3}{2}\)
(vi) If \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors, \(|\vec{a}+\vec{b}|\) = 13 and |\(\vec{a}\)| = 5, then the value ____________ (1)
Answer:
12
Question 3.
Very Short Answer Type Questions
(i) Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor translitive. (1)
Answer:
R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive as (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.
(ii) Find the value of tan-1[2 cos(2 sin-1\(\frac{1}{2}\))] (1)
Answer:
(iii) Construct a 2 × 2 matrix, A = [aij] whose elements are given by: \(\frac{(i+2 j)^{2}}{2}\). (1)
Answer:
Given, aij = \(\frac{(i+2 j)^{2}}{2}\)
Then
(iv) If ∆ = \(\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\), write the cofactor of element a32. (1)
Answer:
Cofactor of elements a32
= (-1)3+2\(\left|\begin{array}{ll}
5 & 8 \\
2 & 1
\end{array}\right|\) = (-1)(5 – 16) = 11
(v) Differentiate tan-1\(\left(\frac{1+\cos x}{\sin x}\right)\) with respect to x. (1)
Answer:
Let y = tan-1\(\left(\frac{1+\cos x}{\sin x}\right)\)
(vi) Evaluate : ∫cos 4x cos x dx. (1)
Answer:
Let I = ∫cos 4x cos x dx
[cos(A + B) + cos(A – B) – 2cos A cos B]
∴ ∫cos 4x.cos x dx
= \(\frac{1}{2}\)∫ 2cos4xcosx dx
= \(\frac{1}{2}\)∫ cos (4x + x) + cos (4x – x) dx
= \(\frac{1}{2}\)∫(cos5x + cos3x) dx
= \(\frac{1}{2}\)∫ cos 5x dx + \(\frac{1}{2}\)∫ cos 3x dx
= \(\frac{1}{2} \times \frac{1}{5}\)sin5x + \(\frac{1}{2} \times \frac{1}{3}\)sin3x + C
= \(\frac{\sin 5 x}{10}+\frac{\sin 3 x}{6}\) + C
(vii) Find the general solution of the differential equation \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\) (1)
Answer:
The given differential equation is:
\(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\) …..(i)
Since, 1 + y2 ≠ 0, therefore, separating the variables, the given differential equation can be written as
\(\frac{d y}{1+y^{2}}=\frac{d x}{1+x^{2}}\) ……(ii)
Integrating both sides of equation (ii), we get:
∫\(\frac{d y}{1+y^{2}}\) = ∫\(\frac{d y}{1+x^{2}}\)
⇒ tan-1y = tan-1x + C
which is the general solution of equation (i).
(viii) Find the position vector of the mid-point of. the vector joining the points P(2, 3, 4) and Q(4, 1, – 2). (1)
Answer:
Let origin is O, then position vector of P and Q with respect to O are \(\overrightarrow{O P}\) and \(\overrightarrow{O Q}\).
Now \(\overrightarrow{O P}\) = 2î + 3ĵ + 4k̂ and \(\overrightarrow{O Q}\) = 4î + ĵ – 2k̂
Let R is the mid point of PQ, then
Hence, the required mid-point is 3î + 2ĵ + k̂.
(ix) Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black. (1)
Answer:
There are 26 black cards in a deck of 52 cards.
Let P(A) be the probability of getting a black card in the first draw.
∴ P(A) = \(\frac{26}{52}=\frac{1}{2}\)
Let P(B) be the probability of get¬ting a black card on the second draw. Since the card is not replaced,
P(B) = \(\frac{25}{51}\)
Thus, probability of getting both the cards black
= \(\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\)
(x) For what value of x, the matrix \(\left[\begin{array}{cc} (1)
2 x+4 & 4 \\
x+5 & 3
\end{array}\right]\) is a singular matrix?
Answer:
Let A = \(\left[\begin{array}{cc}
2 x+4 & 4 \\
x+5 & 3
\end{array}\right]\)
If matrix A is singular, then |A| = 0
⇒ \(\left[\begin{array}{cc}
2 x+4 & 4 \\
x+5 & 3
\end{array}\right]\) = 0
⇒ (2x + 4) × 3 – (x + 5) × 40
⇒6x + 12 – 4x – 20 = 0 ⇒ 2x = 8
Thus, x = 4
(xi) Form a differential equation representing the family of curve \(\frac{x}{a}+\frac{y}{b}\) = 1 by eliminating arbitrary constants a and b. (1)
Answer:
The given equation is :
\(\frac{x}{a}+\frac{y}{b}\) = 1 ……..(i)
Differentiating equation (i) w.r.t. ‘x’, we get
Hence, equation (iii) is the required differential equation.
(xii) Find the area of a triangle having the points A (1, 1, 1), B (1, 2, 3) and C (2, 3,1) as its vertices. (1)
Answer:
We have
\(\overrightarrow{A B}\)= ĵ + 2k̂ and \(\overrightarrow{A C}\) = î + 2ĵ
The area of the given triangle is \(\frac{1}{2}\)||
Thus, the required area is \(\frac{1}{2}\)\(\sqrt{21}\)
Section – B
Short Answer Type Questions
Question 4.
Consider a bijective function f: R+ → (7, ∞) given by f(x) = 16x2 + 24x + 7, where R+ is the set of all positive real numbers. Find the inverse function of f. (2)
Answer:
The function is given by f(x) = 16x2 + 24x + 7
Let y = f(x)
⇒ y = 16x2 + 24x + 7
⇒ y = 16x2 + 24x + 7 + 2 – 2
⇒ y = 16x2 + 24x + 9 – 2
⇒ y = (4x + 3)2 – 2
⇒ (4x + 3)2 = y + 2
⇒ 4x + 3 = \(\sqrt{(y+2)}\)
⇒ x = \(\frac{\sqrt{(y+2)}-3}{4}\)
Now, y = f(x)
Thus, f-1 = x = \(\frac{\sqrt{(y+2)}-3}{4}\)
Question 5.
If F(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 1 & 1
\end{array}\right]\), show that : F(x) F(y) = F(x + y). (2)
Answer:
Question 6.
Find A-1, where A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\). Hence, solve the system of equations,
x + 2y – 3z = 4, 2x + 3y + 2z = 2 and 3x – 3y – 4z = 11. (2)
Answer:
We have A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
2 & 3 & 2 \\
3 & -3 & -4
\end{array}\right]\)
= 1(-12 + 6) – 2(-8 – 6) – 3(-6 – 9)
= -6 + 28 + 45 = 67 ≠ 0
So, A is invertible.
Cofactors of |A| are
Matrix formed by the cofactors of |A| is
The given system of equation is
x + 2y – 3z = – 4
2x + 3y + 2z = – 2
3x – 3y – 4z = 11
We know that AX = B
Thus, x = 3, y = – 2 and z = 1 is the required solution.
Question 7.
If f(x) = \(\sqrt{x^{2}+1}\), g(x) = \(\frac{x+1}{x^{2}+1}\) and h(x) = 2x – 3, then find f'(h(g'(x)). (2)
Answer:
Given, f(x) = \(\sqrt{x^{2}+1}\), g(x) = \(\frac{x+1}{x^{2}+1}\) and h(x) = 2x – 3
On differentiating above functions w.r.t. x, we get
Question 8.
Evaluate : ∫x\(\sqrt{1+2 x^{2}}\) dx (2)
Answer:
Let I = ∫x\(\sqrt{1+2 x^{2}}\) dx
Puffing 1 + 2x2 = t
⇒ 4xdx = dt
⇒ xdx = \(\frac{dt}{4}\)
∴ ∫x\(\sqrt{1+2 x^{2}}\) dx = ∫\(\frac{t^{1 / 2} d t}{4}\)
= \(\frac{1}{4}\)∫t1/2 = \(\frac{1}{4} \frac{t^{3 / 2}}{3 / 2}\) + C
= \(\frac{2}{4 \times 3}\) (1 + 2x2)3/2 + C
= \(\frac{1}{6}\)(1 + 2x2)3/2 +C
Question 9.
Find the probability distribution of X, the number of heads in a simultaneous toss of two coins. (2)
Answer:
When two coins are tossed, there may be 1 head, 2 heads or no head at all. Thus, the possible values of X are 0, 1, 2.
Now, P(X = 0)
= P (Getting no head) = P(TT) = \(\frac{1}{4}\)
P(X = 1) = P (Getting one head)
= P(HT or TH)
= \(=\frac{2}{4}=\frac{1}{2}\)
P(X = 2) = P (Getting two heads) = P(HH) = \(\frac{1}{4}\)
Thus, the required probability distribution of X is
Question 10.
Show that :
Answer:
Question 11.
Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π? Verify. (2)
Answer:
Given function,
f(x) = x2 – sin x + 5
For continuity at x = π
Question 12.
By using properties of determinants show that: \(\left|\begin{array}{ccc}
x+4 & 2 x & 2 x \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\) = (5x + 4)(4 – x)2
Answer:
Question 13.
Evaluate:
Answer:
Question 14.
Find the general solution of the differential equation x5 \(\frac{d y}{d x}\) = – y5. (2)
Answer:
Given differential equation is
x5 \(\frac{d y}{d x}\) = – y5 or \(\frac{d y}{y^{5}}=-\frac{d x}{x^{5}}\) ………(i)
Integrating both sides of equation (i), we get
Equation (ii) is the required solution of given differential equation.
Question 15.
Show that the points A (2î + 3ĵ + 5k̂), B(î + 2ĵ + 3k̂) and C(7î – k̂) are collinear. (2)
Answer:
Here,
AB = (1 + 2)î + (2-3)ĵ + (3 – 5)k̂
= 3î – ĵ – 2k̂
BC = (7 – 1)î + (0 – 2)ĵ + (-1 – 3)k̂
= 6î – 2ĵ – 4k̂
AC =(7 + 2)î + (0-3)ĵ + (-1 – 5)k̂
= 9î – 3ĵ – 6k̂
|\(\overrightarrow{A B}\)| = \(\sqrt{14}\), |\(\overrightarrow{B C}\)| = 2\(\sqrt{14}\)
and |\(\overrightarrow{A C}\)| = 3\(\sqrt{14}\)
∴ \(|\overrightarrow{A C}|=|\overrightarrow{A B}|+|\overrightarrow{B C}|\)
Thus, the points A, B and C are collinear
Question 16.
A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is ‘1’. Find the probability that it is actually 1. (2)
Answer:
Let E1 = Event that 1 occurs in a die
E2 = Event that 1 does not occur in a die
A = Event that the man reports that 1 occur in a die
Then, P(E1) = \(\frac{1}{6}\) and P(E2) = \(\frac{5}{6}\)
∴ P(man reports that 1 occurs when 1 occur) = P\(\left(\frac{A}{E_{1}}\right)=\frac{3}{5}\)
and P (man reports that 1 occur but 1 does not occur) = P\(\left(\frac{A}{E_{2}}\right)=\frac{2}{5}\)
Thus, by Baye’s theorem, we get P(get actually 1 wfien he reports that 1 occur)
Section – C
Long Answer Type Questions
Question 17.
Show that : sin-1 \(\frac{12}{13}\) + cos-1 \(\frac{4}{5}\) + tan-1 \(\frac{63}{16}\) = π.
Or
Express tan-1 \(\frac{\cos x}{1-\sin x}\).where \(\frac{-3 \pi}{2}\) < x < \(\frac{\pi}{2}\) in the simplest form. (3)
Answer:
Let sin-1 \(\frac{12}{13}\) = x, cos-1 \(\frac{4}{5}\) = y,
tan-1 \(\frac{63}{16}\) = z
⇒ sin x = \(\frac{12}{13}\)
cos y = \(\frac{4}{5}\)
tan z = \(\frac{63}{16}\)
Thus, tan(x + y) = – tan z
⇒ tan(x + y) = tan (- z)
⇒ tan(x + y) = tan (π – z)
Therefore, x + y = – z or
x + y = π – z
Since, x, y and z are positive, x + y ≠ – z
Thus, x + y + z = π
sin-1\(\frac{12}{13}\) + cos-1\(\frac{4}{5}\) + tan-1\(\frac{63}{16}\)
Question 18.
Examine the applicability of Mean Value Theorem for the function f(x) = x2 – 1, x ∈ [1, 2]
Or
Differentiate the following function w.r.t. x : xx + xa + ax + aa, ∀ a > 0 and x > 0. (3)
Answer:
It is polynomial function which is continuous in interval [1, 2] and is differentiable in interval (1, 2).
Again f(1)= 12 – 1 = 0 and
f(2)= 22 – 1 = 4 – 1 = 3
f'(x) = 2x exists in interval (1, 2)
f(x) = x2 – 1 is differentiable in interval (1, 2).
Then, in interval (1,2) a point c exists such that
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
i.e., f'(c) = \(\frac{f(2)-f(1)}{2-1}=\frac{3-0}{1}\) = 3
f'(c) = 3 ⇒ 2c = 3
⇒ c = \(\frac{3}{2}\) ∈ (1, 2)
Thus c = \(\frac{3}{2}\) ∈ (1, 2) such that
f'(c) = \(\frac{f(2)-f(1)}{2-1}\)
Hence, mean value theorem is verified.
Question 19.
Evaluate : ∫ e2x sin x dx
0r
Find : ∫ \(\sqrt{x^{2}+3 x}\) dx (3)
Answer:
Let
= e2x ∫ sin x dx
= ∫{\(\frac{d}{d x}\)e2x ∫sin x dx} dx
= e2x (- cos x) – ∫ 2e2x (- cos x) dx
= – e2x cos x + 2 ∫ e2x cos x dx
⇒ I = -e2x cos x + 2I1 .. .(i)
⇒ I1 = e2x ∫cos x dx – ∫{\(\frac{d}{d x}\)e2x∫cosx dx} dx
⇒ I1 = e2x sin x – ∫ 2e2x sin x dx + C1
Putting the value of I, in equation (i), we get
I = – e2x cos x + 2[e2x sin x – 2 ∫e2x sin x dx + C1
⇒ I = – e2x cos x + 2e2x sin x – 4 ∫e2x sin x dx + 2C1
⇒ 1 = – e2x cos x + 2e2x sin x – 4I + 2C1
⇒ I + 4I = – e2x cos x + 2e2x sin x + 2C1
⇒ 5I = – e2* cos x + 2e2x sin x + 2C1
⇒ I = –\(\frac{1}{5}\)e cosx + \(\frac{2}{3}\)e sinx + \(\frac{2}{5}\)C1
⇒ I = \(\frac{1}{5}\)e2x cosx + \(\frac{2}{5}\) e2x sinx + C1
(Where C = \(\frac{2}{5}\)C1)
I = \(\frac{e^{2 x}}{5}\)(2sin x – cos x) + C
Question 20.
Show that the points A(1, 2, 7), B(2, 6, 3) and C (3, 10, – 1) are collinear. (3)
Or
Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î – 2ĵ + 7k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and = 15. (3)
Answer:
Let O be the origin. Then w.r.t. O Position vector of point A
\(\overrightarrow{O B}\) = î + 2ĵ + 7k̂
Position vector of point B
\(\overrightarrow{O B}\) = 2î + 6ĵ + 3k̂
Position vector of point C
\(\overrightarrow{O C}\) = 3î + 10ĵ – k̂
Now, \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\)
= 2 î + 6ĵ + 3k̂- (î + 2ĵ + 7k̂)
= (2-1)î + (6-2)ĵ + (3-7)k̂
\(\overrightarrow{A B}\) = î + 4ĵ – 4k̂
Then \(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}\)
= (3î +10 ĵ- k̂) – (2î + 6ĵ + 3k̂)
= (3 – 2)î + (10 – 6)ĵ + (-1 – 3)k̂
or \(\overrightarrow{B C}\) = î + 4ĵ – 4k̂
\(\overrightarrow{A B}\) = î + 4ĵ – 4k̂ = 5C
i.e., \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) represent the same vector. So, A, B and C are the points of this vector. Thus, A, B and C are collinear.
Section – D
Essay Type Questions
Question 21.
Evaluate :
Or
Evaluate ;
Answer:
Putting tan x = t
⇒ sec2 x dx = dt
When x = 0, then t = 0,
When x = π/2, then t = ∞
Question 22.
Solve the differential equation : 2x2 \(\frac{dy}{dx}\) – 2xy + y2 = 0. (4)
Or
Shpw that the family of curves for which the slope of the tangent at any point (x, y) on it is \(\frac{x^{2}+y^{2}}{2 x y}\), is given by x2 – y2 = cx. (4)
Answer:
Given differential equation is
2x2 \(\frac{dy}{dx}\) – 2xy + y2 = 0
⇒ \(\frac{d y}{d x}=\frac{y}{x}-\frac{y^{2}}{2 x^{2}}\) ………(i)
which is a homogeneous differential equation as
\(\frac{d y}{d x}\) = F\(\left(\frac{y}{x}\right)\)
On putting y = vx ⇒ \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) in Eq. (i), we get
which is the required solution.
Question 23.
Find the mean number of heads in three tosses of a fair coin. (4)
Or
Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the values x, has the following form, where k is some unknown constant.
(i) Find the value of k.
(ii) What is the probability that you study at least two hours ? Exactly two hours ? At most two hours ? (4)
Answer:
Let X denote the success of getting heads.
Therefore, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that X can take the value of 0,1, 2, or 3.
Therefore, the required probability distribution is as follows :
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