ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section A Chapter 8 – 10

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ML Aggarwal Class 12 Maths Solutions Case Study Based Questions Section A Chapter 8 – 10

Chapter 8 Integrals
Case study Based Questions (Solved)

Case-Study 1:

If the derivative of a function is given, the function itself is called antiderivative or integral of the given function. If
\(\frac{d}{d x}\) (F(x)) = f(x),
then ∫ f(x) dx = F (x) + C
The function F (x) is called antiderivative or Integral of the function f(x) and C is called constant of integration.

On the basis of above information, answer the following questions:

(i) If f'(x) = 4x³, then antiderivative of 4x³ is
(a) x⁵ + C
(b) x⁴ + C
(c) x³ + C
(d) x² + C
Answer:
(b) x⁴ + C

Given f'(x) = 4x³
⇒ ∫ f'(x) dx = ∫ 4x³ dx + C
⇒ f(x) = x⁴ + C

(ii) If f'(x) = 4x³, the value of k is
(a) 3
(b) 2
(c) 4
(d) – 3
Solution:
(a) 3

Let ∫ tan² kx dx = \(\frac{\tan 3 x}{x}\) – x + C
⇒ sec² (kx – 1) dx = \(\frac{\tan 3 x}{x}\) – x + C
⇒ \(\frac{\tan k x}{k}\) – x + C = \(\frac{\tan 3 x}{x}\) – x + C
∴ k = 3

(iii) If ∫ e^x (tan x + 1) sec x dx = e^x f(x) + C, then the value of f(x) is
(a) tan x
(b) sin x
(c) sec x
(d) none of these
Solution:
(c) sec x

Let I = ∫ e^x (tan x +1) sec x dx
= ∫ e^x tan x sec x dx + ∫ e^x sec x dx
= e^x sec x – ∫ e^x sec x dx + ∫ e^x sec x dx + C
= e^x sec x + C
given I= e^x f(x) + C
∴ f(x) = sec x

(iv) If ∫ (e^ax + bx) dx = \(\frac{e^{4 x}}{4}+\frac{3 x^2}{2}\) + C, then the values of a and b are
(a) a = 4, b = 3
(b) a = 3, b = 4
(c) a = b = 1
(d) a = 0, b = 1
Solution:
(a) a = 4, b = 3

Let I = ∫ (e^ax + bx) dx
= \(\frac{e^{a x}}{a}+\frac{b x^2}{2}\)
Given I = \(\frac{e^{4 x}}{4}+\frac{3 x^2}{2}\) + C
∴ a = 4 ; b = 3

Case – Study 2 :

For any function f(x), \(\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^d f(x) d x+\int_d^b f(x) d x\). where a < c < d < b.

(i) The value of \(\) f(x) dx, where f(x) = \(\left\{\begin{array}{cc} 2 x+3, & \text { if } x \leq 1 \\ x-4, & \text { if } x>1
\end{array}\right.\) is
(a) \(\frac{7}{2}\)
(b) \(\frac{13}{2}\)
(c) \(\frac{7}{4}\)
(d) \(\frac{9}{4}\)
Solution:
(a) \(\frac{7}{2}\)

\(\int_{-1}^2 f(x) d x=\int_{-1}^1(2 x+3) d x+\int_1^2(x-4) d x\)
= \(\left.\left.\frac{(2 x+3)^2}{4}\right]_{-1}^1+\frac{(x-4)^2}{2}\right]_1^2\)
= \(\frac{1}{4}\) [25 – 1] + \(\frac{1}{2}\) [4 – 9]
= 6 – \(\frac{5}{2}\)
= \(\frac{7}{2}\)

(ii) \(\int_0^2\) |x – 1| dx =
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1

When x ≤ 1
⇒ x – 1 ≤ 0
⇒|x – 1| = – (x – 1)

(iii) \(\int_0^\pi\) |cos x| dx =
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

When 0 ≤ x ≤ \(\frac{\pi}{2}\)
⇒ cos x ≥ 0
⇒ |cos x| = + cos x
When \(\frac{\pi}{2}\) ≤ x ≤ π
⇒ cos x ≤ 0
⇒ |cos x| = – cos x
∴ \(\int_0^\pi\) |cos x| dx = \(\int_0^{\frac{\pi}{2}}\) |cos x| dx + \(\int_{\pi / 2}^\pi\) |cos x| dx
= \(\int_0^{\frac{\pi}{2}}\) cos x dx + \(\int_{\pi / 2}^\pi\) – cos x dx
= \(\left.\sin x]_0^{\pi / 2}-\sin x\right]_{\pi / 2}^\pi\)
= [1 – 0] – [0 – 1] = 2

(iv) \(\int_{-1}^1\) e^|x| dx =
(a) e – 1
(b) e + 1
(c) 2 (e – 1)
(d) 2 (e + 1)
Solution:

When – 1 ≤ x – 0
⇒ |x| = – x
When 0 ≤ x ≤ 1
⇒ |x| = x
∴ \(\int_{-1}^1\) e dx = \(\int_{-1}^0\) e^|x| dx + \(\int_0^1\) e^|x| dx
= \(\int_{-1}^0\) e^{– x} dx + \(\int_0^1\) e^x dx
= \(\int_{-1}^0 e^{-x} d x+\int_0^1 e^x d x\)
= – (1 – e) + (e – 1)
= 2 (e – 1)

Case – Study 3 :

For any function f(x), \(\int_a^b \frac{f(a+b-x)}{f(x)+f(a+b-x)} d x=\frac{b-a}{2}\).

Based on the above information, answer the following questions:

(i) \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\) dx =
(a) \(\frac{\pi}{12}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Solution:
(a) \(\frac{\pi}{12}\)

(ii) \(\int_0^{\frac{3}{2}} \frac{1}{1+\tan ^3 x}\) dx =
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(c) \(\frac{\pi}{4}\)

(iii) \(\int_0^{\frac{\pi}{2}} \frac{\sec x}{\sec x+\ {cosec} x}\) dx =
(a) 0
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

Let f(x) = cosec x;
a = 0 ; b = \(\frac{\pi}{2}\)
Here f (a + b – x) = f (0 + \(\frac{\pi}{2}\) – x)
= cosec (\(\frac{\pi}{2}\) – x)
= sec x
We know that
\(\int_a^b \frac{f(a+b-x) d x}{f(x)+f(a+b-x)}=\frac{b-a}{2}\)
∴ \(\int_0^{\pi / 2} \frac{\sec x d x}{\ {cosec} x+\sec x}=\frac{\frac{\pi}{2}-0}{2}\)
= \(\frac{\pi}{4}\)

(iv) \(\int_2^8 \frac{\sqrt[3]{10-x}}{\sqrt[3]{x}+\sqrt[3]{10-x}}\) dx =
(a) 10
(b) 5
(c) 3
(d) \(\frac{3}{2}\)
Solution:
(c) 3

Let f(x) = \(\sqrt[3]{x}\) ;
a = 2 ;
b = 8
∴ f (a + b – x) = f (2 + 8 – x)
= f (10 – x)
= \(\sqrt[3]{10-x}\)
We know that,
\(\int_a^b \frac{f(a+b-x) d x}{f(x)+f(a+b-x)}=\frac{b-a}{2}\)
⇒ \(\int_2^8 \frac{\sqrt[3]{10-x} d x}{\sqrt[3]{x}+\sqrt[3]{10-x}}=\frac{8-2}{2}\)
= 3

Chapter 9 Differential Equations
Case study Based Questions (Solved)

Consider the differential equation of the form \(\frac{d y}{d x}\) + Py = Q,
where P and Q are constants or function of x. It is known as linear differential equation in y.
To solve this type of differential equation
we calculate Integrating Factor (I.F.) = e^{∫ P dx}
and solution is given by
y × (I.F.) = ∫ Q × (T.F) dx + C
Consider the equation
x \(\frac{d y}{d x}\) – y – 2x³ = 0.

Based on the above information, answer the following questions :

(i) The values of P and Q in the given equation are
(a) P = – \(\frac{1}{x}\), Q = 2x²
(b) P = \(\frac{1}{x}\), Q = – 2x²
(c) P = – \(\frac{1}{x}\), Q = – 2x²
(d) P = \(\frac{1}{x}\), Q = 2x²
Solution:
(a) P = – \(\frac{1}{x}\), Q = 2x²

Given differential eqn. can be written as
\(\frac{d y}{d x}-\frac{y}{x}\) = 2x²
which is inear differential eqn. in y and on
comparing with \(\frac{d y}{d x}\) + Py = Q
we have,
P = – \(\frac{1}{x}\) ;
Q = 2x²

(ii) I.F. in the given equation is
(a) – \(\frac{1}{x}\)
(b) e^-x+
(c) \(\frac{1}{x}\)
(d) e^x
Solution:
(c) \(\frac{1}{x}\)

∴ I.F. = e^{∫ P dx}
= e^{∫ – \(\frac{1}{x}\) dx}
= x^-1
= \(\frac{1}{x}\)

(iii) Solution of the given differential equation is
(a) y = x³ + C
(b) y = x³ + Cx
(c) y = – x³ + Cx
(d) y = x² + Cx
Solution:
(b) y = x³ + Cx

∴ general solution be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . \(\frac{1}{x}\) = ∫ 2x² . \(\frac{1}{x}\) dx + C
⇒ \(\frac{y}{x}\) = x² + C
⇒ y = x³ + Cx ……………(1)

(iv) If y(1) = 2,then y =
(a) x³ + 1
(b) – x³ + x
(c) x³ + x
(d) x² + C
Solution:

Given y(1) = 2
i.e. when x = 1
⇒ y = 2
∴ from (1) ;
2 = 1³ + C
⇒ C = 1
Thus eqn. (1) becomes;
y = x³ + x

Case – Study 2:

An equation which involves unknown functions and their derivatives with respect to one or more independent variables is called a differential equation.
Now, consider ifa curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

On the basis of above information, answer the following questions:

(i) The differential equation for the curve is
(a) \(\frac{d y}{d x}\) . x = y
(b) \(\frac{d y}{d x}\) . y = x
(c) \(\frac{d y}{d x}\) (x + y) = 1
(d) \(\frac{d y}{d x}\) = xy
Solution:
(b) \(\frac{d y}{d x}\) . y = x

We know that,
slope of tangent to given curve at any point = \(\frac{d y}{d x}\)
Then according to given condition,
y \(\frac{d y}{d x}\) = x

(ii) The order of differential equation is
(a) 1
(b) 2
(c) 0
(d) – 1
Solution:
(a) 1

The highest ordered derivative existing in given diff. eqn. be \(\frac{d y}{d x}\)
so order of given diff. eqn. be 1.

(iii) The degree of differential equation is
(a) 0
(b) 2
(c) 1
(d) not defined
Solution:
(c) 1

Further the given differential equation can be expressible as polynomial in \(\frac{d y}{d x}\).
Thus degree of diff. eqn. be the highest exponent of \(\frac{d y}{d x}\) which is 1.

(iv) The equation of the curve when x = 0 and y = – 2 is
(a) y = x² + 4
(b) x² = y² + 4
(c) y = x² + 4
(d) none of these
Solution:
(a) y = x² + 4

After variable separation, we have
y dy = x dx ;
on integrating
\(\frac{y^2}{2}=\frac{x^2}{2}+\frac{\mathrm{C}}{2}\)
⇒ y² = x² + C …………..(1)
Given x = 0 when y = – 2
∴ from (1) ;
4 = 0 + C
⇒ C = 4
Thus eqn. (1) reduces to ;
y² = x² + 4

10 Probability
Case-Study Based Questions (Solved)

Case – Study 1 :

Competitive exams are considered an egalitarian way to select worthy applicants without risking influence peddling, bias or other concerns. Self motivated and hard worker candidates find these exams very simple. To clear a prestigious competitive exam to get admission in a world reputed college, a student has 75% chances of passing in atleast one subject out of three subjects A, B and C, 50% chances of passing in atleast two subjects and 40% chances of passing in exactly two subjects.

Based on the above information, answer the following questions :

(i) What is the value of P (A ∩ B ∩ C)?
(a) 0.50
(b) 0.30
(c) 0.70
(d) 0.10
Solution:
(d) 0.10

Let us define the events are as follows:
A : studcnt passed in subject A
B : student passed in subject B
C : student passed in subject C
Given P (passing atleast onc subject) = 75 %
= \(\frac{75}{100}\) ……………………..(1)
P (passing atleast two subjects) = 50%
= \(\frac{50}{100}\) ………………………(2)
P (passing in exactly two subjects) = 40%
= \(\frac{40}{100}\) …………………….(3)

from (1) ;
1 – P (Not passing in any of subjects) = \(\frac{75}{100}=\frac{3}{4}\)
⇒ 1 – P \((\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\frac{3}{4}\)
P \((\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})\) = 1 – \(\frac{3}{4}=\frac{1}{4}\) ……………………(4)
from (2) ;
P (\(\bar{A}\) ∩ B ∩ C) + P (A ∩ \(\bar{B}\) ∩ C) + P (A ∩ B ∩ \(\bar{C}\)) = \(\frac{40}{100}=\frac{2}{5}\) ……………….(6)
From (5) and (6) ; we have
P(A ∩ B ∩ C) = \(\frac{1}{2}-\frac{2}{5}\)
= \(\frac{1}{10}\) = 0.1

(ii) What is the probability that student is not able to clear any of three subjects A, B and C?
(a) 0.20
(b) 0.25
(c) 0.35
(d) 0.40
Solution:
(b) 0.25

required Prob. = P \((\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})\)
= \(\frac{1}{4}\) [from (4)]

(iii) What is the value of P (A ∩ B) + P (A ∩ C) + P (B ∩ C)?
(a) 0.75
(b) 0.30
(c) 0.70
(d) 0.60
Solution:
(c) 0.70

from (6) ; we have
[P (B ∩ C) – P (A ∩ B ∩ C)] + [P (A ∩ C) – P (A ∩ B ∩ C)] + [P (A ∩ B) – P (A ∩ B ∩ C)] = \(\frac{2}{5}\)
[∵ P(A ∩ \(\overline{\mathbf{B}}\)) = P (A) – p (A ∩B)]
⇒ P(B ∩ C) + P(C ∩ A) + P (A ∩ B) -3 P (A ∩ B ∩C) = \(\frac{2}{5}\)
P(B ∩ C) + P(C ∩ A) + P(A ∩ B) – 3 \(\times \frac{1}{10}=\frac{2}{5}\)
P(B ∩ C) + P(C ∩ A) + P(A ∩ B) = \(\frac{2}{5}+\frac{3}{10}\)
= \(\frac{7}{10}\) = 0.7

(iv) What is the value of P (A) + P (B) + P (C)?
(a) \(\frac{3}{4}\)
(b) \(\frac{27}{20}\)
(c) \(\frac{1}{4}\)
(d) 1
Solution:
(b) \(\frac{27}{20}\)

from (4);
P \((\overline{\mathrm{A}} \cap \overline{\mathrm{B}} \cap \overline{\mathrm{C}})=\frac{1}{4}\)
⇒ P \([(\overline{\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}})]=\frac{1}{4}\)
⇒ 1 – P (A ∪ B ∪ C) = \(\frac{1}{4}\)
⇒ P (A ∪ B ∪ C) = \(\frac{3}{4}\)
⇒ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) = \(\frac{3}{4}\)
⇒ P(A) + P(B) + P(C) – \(\frac{7}{10}+\frac{1}{10}=\frac{3}{4}\)
⇒ P(A) + P(B) + P(C) = \(\frac{3}{4}+\frac{6}{10}=\frac{27}{20}\)
= \(\frac{27}{20}\)

Case-Study 2 :

Let X be the random variable which count the number of hours a student of class XII studies.
X has a probability distribution P (X) of the following form:
P(X) = \(\left\{\begin{array}{rc}
k, & \text { if } X=0 \\
2 k, & \text { if } X=1 \\
3 k, & \text { if } X=2 \\
0, & \text { otherwise }
\end{array}\right.\), where k Is constant.

Based on the above information, answer the following questions:

(i) What is the value of k?
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{6}\)
(d) none of these
Solution:
(c) \(\frac{1}{6}\)

Given P(X) = \(\left\{\begin{array}{rc}
k, & \text { if } X=0 \\
2 k, & \text { if } X=1 \\
3 k, & \text { if } X=2 \\
0, & \text { otherwise }
\end{array}\right.\)

Since Σ P(X) = 1
⇒ P (X = 0) + P (X = 1) + P (X = 2) + P (X > 2) = 1
⇒ k + 2k + 3k + 0 = 1
⇒ 6k = 1
⇒ k = \(\frac{1}{6}\)

(ii) Find P (X < 2).
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{3}\)
Solution:
(c) \(\frac{1}{2}\)

P (X < 2) = P (X = 0) + P (X = 1)
= k + 2k = 3k
= 3 × \(\frac{1}{6}\)
= \(\frac{1}{6}\)

(iii) Find P (0 ≤ X < 4)
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Solution:
(a) 1

P (0 ≤ X < 4) = P (X = 0) + P (X = 1) + P (x = 2) + P (X = 3) = k + 2k + 3k + 0 = 6k = 6 × \(\frac{1}{6}\) = 1 (iv) Find P (X ≥ 2)
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{3}{4}\)
(d) none of these
Solution:
(a) \(\frac{1}{2}\)

P (X ≥ 2) = P (X = 2) + P (X > 2)
= 3k + 0
= 3 × \(\frac{1}{6}\)
= \(\frac{1}{2}\)

Case – Study 3 :

The binomial distribution with parameters n and p is the discrete probability distribution.
The probability distribution function is given as
P (X = r) = \({ }^n \mathbf{C}_r p^r q^{n-r}\) r = 0, 1, 2, 3, …………………, n,
p, q > 0 such that (p + q)ⁿ = 1,
where n = number of triais,
p = probability of success
and q = probability of failure.
A fair coin is tossed 6 times.

Based on the above information, answer the following questions :

(i) The values of parameters of this binomial experiment are
(a) n = 6, p = \(\frac{1}{3}\)
(b) n = 6, p = \(\frac{1}{2}\)
(c) n = 6, p = \(\frac{2}{3}\)
(d) n = 5, p = \(\frac{1}{2}\)
Solution:
(b) n = 6, p = \(\frac{1}{2}\)

Given p = prob. of getting head in a single toss = \(\frac{1}{2}\)
and q = prob. of getting tail 1 – p
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)
Since a coin is tossed so all its outcomes are equally likely.
∴ it is a problem of binomial distribution with
n = 6 ;
p = q = \(\frac{1}{2}\)

(ii) The probability of getting exactly 4 heads
(a) \(\frac{15}{64}\)
(b) \(\frac{13}{}\)
(c) \(\frac{3}{32}\)
(d) \(\frac{2}{23}\)
Solution:
(a) \(\frac{15}{64}\)

Then by binomial distribution, we have
P (X = r) = \({ }^n \mathrm{C}_r\) p^r q^n-r
= \({ }^6 \mathrm{C}_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{6-r}\)
= \({ }^6 C_r\left(\frac{1}{2}\right)^6\)
∴ P (X = 4) = \({ }^6 C_4\left(\frac{1}{2}\right)^6\)
= \(\frac{6 \times 5}{2} \times \frac{1}{2^6}\)
= \(\frac{15}{64}\)

(iii) The probability of getting atleast 4 heads is
(a) \(\frac{15}{64}\)
(b) \(\frac{13}{}\)
(c) \(\frac{3}{32}\)
(d) \(\frac{2}{23}\)
Solution:
(c) \(\frac{3}{32}\)

P (getting atleast 4 heads) = P (X ≥ 4) = P (X = 4) + P (X = 5) + p (X = 6)
= \({ }^6 \mathrm{C}_4\left(\frac{1}{2}\right)^6+{ }^6 \mathrm{C}_5\left(\frac{1}{2}\right)^6+{ }^6 \mathrm{C}_6\left(\frac{1}{2}\right)^6\)
= \(\left(\frac{1}{2}\right)^6\left[{ }^6 \mathrm{C}_4+{ }^6 \mathrm{C}_5+{ }^6 \mathrm{C}_6\right]\)
= \(\frac{1}{2^6}\left[\frac{6 \times 5}{2}+6+1\right]\)
= \(\frac{22}{64}=\frac{11}{32}\)

(iv) In a binomial epcrimcnt, if n = 5 and P (X = 0) = \(\frac{32}{243}\), then the value of P (X = 3) is
(a) \(\frac{80}{243}\)
(b) \(\frac{40}{243}\)
(c) \(\frac{20}{243}\)
(d) \(\frac{1}{243}\)
Solution:
(b) \(\frac{40}{243}\)

We know that
P(X = r) = \({ }^n \mathrm{C}_r\) p^r q^n-r
= \({ }^5 \mathrm{C}_r\) (p)^r (q)^5-r
∴ P (X = 0) = \(\frac{32}{243}\)
⇒ \({ }^5 C_0 p^0 q^5=\frac{32}{243}\)
⇒ q⁵ = \(\left(\frac{2}{3}\right)^5\)
⇒ q = \(\frac{2}{3}\)
∴ p = 1 – q
= 1 – \(\frac{2}{3}\)
= \(\frac{1}{3}\)
Thus P (X = 3) = \({ }^5 C_3\) p³ q²
= \({ }^5 C_3\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2\)
= \(\frac{5 \times 4}{2} \times \frac{4}{3^5}\)
= \(\frac{40}{243}\)

Case-Study 4 :

In a game, a man wins Rs 8 for getting a number greater than 3 and loses ₹ 3 otherwise, when a fair die is thrown. The man decided to throw a die 4 times but to quit as and when he gets a number greater than 3.

Based on the above information, answer the following questions:

(i) If X denotes the amount which the man wins or loses, then all possible values of X are
(a) 8, 5, 2, – 1, – 12
(b) 8, 5, 2, – 1
(c) 8, 5, 3, 0, – 3
(d) 8, 5, 2, 0, – 3
Solution:
(a) 8, 5, 2, – 1, – 12

Given X be the random variable denotes the amount which the man wins or loses.
Since a man decided to throw a die 4 times and to quit as when he gets a number > 3.
Let p = prob. of getting a number> 3 ¡n single throw of die
= \(\frac{3}{6}=\frac{1}{2}\)
and q = prob.of failure = 1 – p
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)
When a man gets a number > 3 in first throw and wins 8
Then X=8
When a man gets a number > 3 in 2nd throw then he not getting a number > 3 in first throw.
Then x = – 3 + 8 = 5
When a man gets a number > 3 in 3rd throw.
∴ X = – 3 – 3 + 8 = 2
When a man gets a number > 3 in 4th throw.
Then X = – 3 – 3 – 3 + 8 – 1
When a man does not get a number> 4 in all four throws.
∴ X = – 3 – 3 – 3 – 3
= – 12
Thus X can take values 8, 5, 2, – 1, – 12.

(ii) The probability distribution of X is
(a)

(b)

(c)

(d)
Solution:
(c)

P (X = 8) = P (getting a number > 3 in Ist throw)
= p
= \(\frac{1}{2}\)
P (X = 5) = (qp)
= \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\)
P (X = 2) = qqp
= \(\left(\frac{1}{2}\right)^2 \times \frac{1}{2}=\frac{1}{8}\)
P (X = – 1) = (qqqp)
= \(\left(\frac{1}{2}\right)^4=\frac{1}{16}\)
P (X = – 12) = qqqq
= \(\left(\frac{1}{2}\right)^4=\frac{1}{16}\)

Thus probability distribution of X is given below:

(iii) The expected value of X i.e. E (X) is
(a) \(\frac{67}{8}\)
(b) \(\frac{75}{16}\)
(c) \(\frac{75}{8}\)
(d) \(\frac{67}{16}\)
Solution:
(b) \(\frac{75}{16}\)

E(X) = Σ X P(X)
= \(8 \times \frac{1}{2}+5 \times \frac{1}{4}+2 \times \frac{1}{8}-1 \times \frac{1}{16}-12 \times \frac{1}{16}\)
= \(4+\frac{5}{4}+\frac{1}{4}-\frac{1}{16}-\frac{3}{4}\)
= \(\frac{64+20+4-1-12}{16}=\frac{75}{16}\)

(iv) The standard deviation of X is
(a) \(\frac{21 \sqrt{15}}{8}\)
(b) \(\frac{21 \sqrt{15}}{4}\)
(c) \(\frac{21 \sqrt{15}}{6}\)
(d) \(\frac{21 \sqrt{15}}{16}\)
Solution:
(d) \(\frac{21 \sqrt{15}}{16}\)

σ² = E (X²) – [E(X)]²

Case – Study 5 :

A biased die is such that probability of getting 4 is and other scores being equally likely. The die is tossed twice. X denotes the number of 4 obtained.

Based on the above information, answer the following questions:

(i) All possible values of random variable X are
(a) 1, 2
(b) 0, 1, 2
(c) 1, 2, 3
(d) 0, 1, 2, 3
Solution:
(b) 0, 1, 2

Given P (prob. of getting 4) = \(\frac{1}{10}\) = P
P(prob. of not getting 4) = 1 – \(\frac{1}{10}\)
= \(\frac{9}{10}\) = q
Since other scores being equally likely.
∴ P (1) = P (2) = P (3) = P (5) = P (6) = \(\frac{9}{50}\)
Since die is tossed thrice and X be the random variable denotes the number 4 is obtained then X can take values 0, 1, 2,

(ii) The probability of getting atleast one 4 is
(a) \(\frac{19}{100}\)
(b) \(\frac{81}{100}\)
(c) \(\frac{18}{100}\)
(d) 1
Solution:
(a) \(\frac{19}{100}\)

P (getting atleast one 4) = 1 – P (getting no. 4)
= 1 – q . q
= 1 – \(\frac{9}{10} \times \frac{9}{10}\)
= \(\frac{19}{100}\)

(iii) The probability distribution of X is
(a)

(b)

(c)

(d)

Solution:
(c)

P(X = 0) = P(getting no. 4)
= q . q
= \(\frac{9}{10} \times \frac{9}{10}=\frac{81}{100}\)
P(X = 1) = P(getting one 4)
= pq + qp
= \(\frac{1}{10} \times \frac{9}{10}+\frac{9}{10} \times \frac{1}{10}\)
= \(\frac{18}{100}\)
P (X = 2) = P (getting exactly two 4’s)
= p . p
= \(\frac{1}{10} \times \frac{1}{10}\)
= \(\frac{1}{100}\)
The probability distribution of X is given as under:

(iv) Variance of the random variable X is
(a) 0.18
(b) 0.15
(c) 0.13
(d) 0.11
Solution:
(a) 0.18

Variance (σ) = σ²
= X² P(X) – [X P (X)]²
= \(\left[0^2 \times \frac{81}{100}+1^2 \times \frac{18}{100}+2^2 \times \frac{1}{100}\right]-\left[0 \times \frac{81}{100}+1 \times \frac{18}{100}+2 \times \frac{1}{100}\right]^2\)
= \(\frac{22}{100}-\frac{400}{10000}\)
= \(\frac{1800}{10000}\)
= \(\frac{18}{100}\)
= 0.18