Clearly
(i) \(f(x) \geq 0\)
(ii) To prove \(\int_{-\infty}^{x} f(x) d x=1\)
\( \text { LHS } =\int_{-x}^{x} f(x) d x=\frac{1}{\pi} \int_{-x}^{x} \frac{1}{1+x^{2}} d x \)
\( =\frac{1}{\pi} \left[\tan ^{-1} x\right]_{-\infty}^{\pi} \)
\( =\frac{1}{\pi}\left[\tan ^{-1} \infty-\tan ^{-1}(-\infty)\right] \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \)
\( =\frac{1}{\pi}\left(\frac{2 \pi}{2}\right)=1=\text { RHS }\)

Clearly
(i) \(f(x) \geq 0\)
(ii) To prove \(\int_{-x}^{x} f(x) d x=1\)
\( \text {LHS } =\int_{-\infty}^{x} f(x) d x=\int_{0}^{3} \frac{2 x}{9} d x \)
\( =\frac{2}{9}\left[\frac{x^{2}}{2}\right]_{0}^{3}=\frac{1}{9}(9-0) \)
\( =1=\text { RHS } \)
f is p.d.f

\( \mathrm{F}(2) =P(X \leq 2) \)
\( =\int_{-x}^{2} f(x) d x \)
\( =\int_{0}^{2} 2 e^{-2 x} d x \)
\( =2\left[\frac{e^{-2 x}}{-2}\right]_{0}^{2} \ \)
\( =-\left[e^{-4}-1\right] \)
\( =1-e^{-4} \)
\( =1-\frac{1}{e^{4}}=\frac{e^{4}-1}{e^{4}}\)

\(\mathrm{F}(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\)
\( P(0 \leq x \leq 1) =\mathrm{F}(1)-\mathrm{F}(0) \)
\( =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right) \)
\( =\frac{1}{2}+\frac{1}{\pi}\left(\frac{\pi}{4}\right)-\frac{1}{2}-\frac{1}{\pi}(0)=\frac{1}{4} \)

f is a p.d.f
\( \int_{-\pi}^{x} f(x) d x=1 \)
\( k \int_{0}^{1} x(1-x)^{10} d x=1 \)
By properties of definite integral
\( \int_{0}^{0} f(x) d x^{2} =\int_{0}^{1} f(a-x) d x \)
\(k \int_{0}^{1}(1-x)(1-(1-x))^{10} d x =1 \)
\(k \int_{0}^{1}(1-x) x^{10} d x =1 \)
\(k \int_{0}^{1}\left(x^{10}-x^{11}\right) d x =1 \)
\(k\left[\frac{x^{11}}{11}-\frac{x^{12}}{12}\right]_{0}^{1} =1 \)
\(k\left[\frac{1}{11}-\frac{1}{12}\right] =1 \)
\(k\left[\frac{12-11}{132}\right] =1 \)
k  = 132

Mean = E(X)=\(\int _{ -\infty }^{ \infty }{ x.f(x)dx=\int _{ -12 }^{ 12 }{ x.\left( \frac { 1 }{ 24 } \right) } dx } \)
\(=\frac { 1 }{ 24 } \int _{ -12 }^{ 12 }{ x.dx } \)
\(=0[\because \int _{ -a }^{ a }{ f(x)dx=0 } when\ f(x)\ is\ an\ odd\ function]\)
\(\therefore E(X)=0\)

Let X denote the amount
∴ X is a random variable. taking the values 10 and -5 when 3 coins are tossed, sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH,TTT}
∴ P(X = 10) = P (getting 3 heads or 3 tails)
\(=\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)
P(X = -5) = p(getting 1head or 2heads)
\(=\frac { 6 }{ 8 } =\frac { 3 }{ 4 } \)
∴ Probability distribution function is

∴ Expected gain E(X) = Σx_ip_i = 10(\(\frac{1}{4}\))-5(\(\frac{3}{4}\))
\(=\frac { 10 }{ 4 } -\frac { 15 }{ 4 } =-\frac { 5 }{ 4 } =-1.25\)
∴E(X) = -1.25 [A loss of Rs. 1.25]

Let X be a random variable. That denote the mark obtained by a student for answering a question.
∴ X can take values 1 and -\(\frac{1}{2}\)
∴ P(X = 1) = P (answering a question correctly) 
= \(\frac{1}{4}\)
P(X = -\(\frac{1}{2}\)) = P(answering a question wrongly)
=\(1-\frac{1}{4}=\frac{3}{4}\)
∴ Probability distribution function is

\(\therefore E(x)=\sum { xp(x)=1(\frac { 1 }{ 4 } )-\frac { 1 }{ 2 } \left( \frac { 3 }{ 4 } \right) =\frac { 1 }{ 4 } -\frac { 3 }{ 8 } } \)
\(=\frac { 2-3 }{ 8 } =-\frac { 1 }{ 8 } \)
∴ Expectation of mark for answering a single question is -\(\frac{1}{8}\)
∴ Expectation of mark for answering 120 questions = 120(-\(\frac{1}{8}\)) = -15.

Since f(x) is a probability density function,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 0 }^{ \infty }{ { ce }^{ -ax }dx=1 } \Rightarrow C\int _{ 0 }^{ \infty }{ { e }^{ -ax }dx=1 } \)
\(\Rightarrow c{ \left[ \frac { { e }^{ -ax } }{ -a } \right] }_{ 0 }^{ \infty }=1\Rightarrow \frac { -c }{ a } [{ e }^{ -\infty }-{ e }^{ 0 }]\)
\(\Rightarrow \frac { -c }{ a } [0-1]=1\quad [\because { e }^{ -\infty }=0,{ e }^{ 0 }=1]\)
\(\Rightarrow \frac { c }{ a } =1\Rightarrow C=a\quad \therefore C=a\)

Clearly f(x) ≥0 for all real values of x
\(\therefore \int _{ -\infty }^{ \infty }{ f(x)dx } =\int _{ 0 }^{ 3 }{ \frac { 2x }{ 9 } dx=\frac { 2 }{ 9 } \int _{ 0 }^{ 3 }{ xdx } } \)
\(=\frac { 2 }{ 9 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }=\frac { 2 }{ 9 } \left[ \frac { 9 }{ 2 } -0 \right] =\frac { 2 }{ 9 } \times \frac { 9 }{ 2 } =1\)
∴ f(x) is a probability density function.

For the probability distribution, ∑p_i = 1
⇒ a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1
⇒81a = 1 ⇒ a = \(\frac{1}{81}\)
Also P(X< 3) = P(X = 0)+P(X = 1) + P(X = 2)
= a + 3a + 5a = 9a = 9(\(\frac{1}{81}\))
= \(\frac{1}{9}\)

Since the random variable. X is the probability mass function, Σp_i = 1
\(\Rightarrow \frac { k }{ 6 } +\frac { k }{ 4 } +\frac { k }{ 12 } =1\Rightarrow \frac { 2k+3k+k }{ 12 } =1\)
\(\Rightarrow \frac { 6k }{ 12 } =1\Rightarrow k=\frac { 12 }{ 6 } =2\quad \therefore k=2\)

A bag contains 2 bad eggs and 8 good eggs
∴ Total number of eggs = 10
We are going to select 3 eggs, out of that 2 must be bad eggs.
∴ Required probability \(=\frac { { 2C }_{ 2 }\times { 8C }_{ 1 } }{ 10{ C }_{ 3 } } =\frac { 1\times 8 }{ \frac { 10\times 9\times 8 }{ 3\times 2\times 1 } } \)
\(=\frac { 1\times 8\times 3\times 2\times 1 }{ 10\times 9\times 8 } =\frac { 1 }{ 15 } \)
∴ Probability of getting two bad eggs = \(\frac{1}{15}.\)

When a die is rolled, sample space 
S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
∴P(X = 0) = Probability of getting an odd number = \(\frac{3}{6}\)[∵ Their are 3 favourable events]
= \(\frac{1}{2}\)
Thus, the probability distribution of the random variable X is given by

P(X = 0) + P(X = 1) + P(X = 2)
= 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
Hence the given distribution of probabilities is not a probability distribution.