\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
Let X be a random variable whose values x are the number of fours.
The sample space S is given in the table.
It can also be written as
S = {(i, j)} , where i = 1, 2, 3, 6 and j = 1, 2, 3, 6
Therefore X takes on the values of 0, 1 and 2.
We observe that
(i) X = 0, if (i,  j) for i ≠ 4, j≠ 4,
(ii) X = 1, if (1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(iii) X = 2, if (4, 4) ,
Therefore,

The probabilities are
\(f(0)=P(X=0)\cfrac { 25 }{ 36 } \) 
\(f(1)=P(X=1)=\cfrac { 10 }{ 36 } \) 
and \(f(20=P(X=2)=\cfrac { 1 }{ 36 } \)
Clearly the function f(x) satisfies the conditions
(i) f (x) ≥ 0, for x = 0, 1, 2 and
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=-2 }{ f(x) } =f(0)+f(1)+f(2)=1\) 
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=1\)
The probability mass function is presented as

(or) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

(i) Differentiating F(x) with respect to x at continuity points of f(x), we get
\(f(x)={ F }^{ 1 }(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & x\ge 1 \end{matrix} \end{cases}\) 
The pdf f(x) is not continuous at x = 0, or at x = 1. We can define f(0) and f(1) in any manner. Choosing f(0) = 1, and f(1) = 0 .
Therefore the probability density function f(x) is
\(f(x)=\begin{cases} \begin{matrix} 1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(ii) P(0.2 ≤ X ≤ 0.7) = F(0.7) − F(0.2)
= 0.7-0.2 = 0.5
\(P(0.2\le X\le 0.7)=\int _{ 0.2 }^{ 0.7 }{ f(x) } dx=\int _{ 0.2 }^{ 0.7 }{ 1dx } =0.5\)

The sample space S = {H,T} \(\times\) {H,T}
That is S = {TT, TH, HT, HH}
Let X be the random variable denoting the number of heads.
Therefore
X (TT ) = 0 , X (TH ) = 1,
X (HT) = 1, and X (HH) = 2 .
Then the random variable X takes on the values 0, 1 and 2

The probabilities are given by

\(f(0)=P(X=0)=\cfrac { 1 }{ 4 } \) 
\(f(1)=P(X=1)=\cfrac { 1 }{ 2 } \) 
and \(f(2)=P(X=2)=\cfrac { 1 }{ 4 } \) 
The function f (x) satisfies the conditions
(i) f (x) ≥ 0 , for x = 0, 1, 2
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=2 }{ f(x) } =f(0)+f(1)+f(2)\) 
= \(\cfrac { 1 }{ 4 } +{ \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } =1 }\) 
Therefore f (x) is a probability mass function.
The probability mass function is given by

(or)
\(f(x)\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & forx=0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } & forx=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 4 } & forx=2 \end{matrix} \end{cases}\)

The possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

Let us denote white and red balls as w₁, w₂, r₁, r₂ and r₃
The sample space consists of 5C₃ = 10 different samples of size 3.
That is S = \(\left\{w_{1} w_{2} r_{1}, w_{1} w_{2} r_{2}, w_{1} w_{2} r_{3}, w_{1} r_{1} r_{2}, w_{1} r_{2} r_{3}, w_{1} r_{1} r_{3}, w_{2} r_{1} r_{2}, w_{2} r_{2} r_{3}, w_{2} r_{1} r_{3}, r_{1} r_{2} r_{3}\right\} .\)
The random variable X takes on the values 1, 2, and 3.

X B(n, p)
Given 4P(X = 4) = P(X = 2) and n = 6.
4. [⁶C₄p⁴ (1 - p)²] = ⁶C₂p² q⁴
⇒ 4p² = q² 
⇒ 4(1-q²) = q²  
4(1-2q+q²) = q²  
⇒ 3q² - 8q +4 = 0
⇒ (q - 2)(3q - 2) = 0
\( -q=\frac { 2 }{ 3 } \ \ \ (q\neq2)\) 
\(p = 1- q=\cfrac { 1 }{ 3 } \)  
Distribution
P(X = x) = nC_xp^x (1- p)^n-x, x = 0, 1,2, ... n
(i) \(P(X=x)= ^6C_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 6-x }\) , x = 0,1,2..n
(ii) \(mean=np=6\times \frac { 1 }{ 3 } =2\) 
(iii) standard deviation = \(\sqrt { npq } =\sqrt { 2 \times \frac {2}{ 3 } } \) 
= \( { \frac { 2 }{ \sqrt 3 } } \)

Observe that the given distribution is continuous
By definition \(\mu =E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -2x } \right) dx } +\int _{ 0 }^{ \infty }{ x\left( { \lambda e }^{ -\lambda x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ x\left( { e }^{ -\lambda x } \right) dx } \) 
=  \(0+\lambda \left( \frac { 1 }{ { \lambda }^{ 2 } } \right) \) (using Gamma integral for positive integer n,\(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\cfrac { n }{ { a }^{ n+1 } } \))
= \(\frac { 1 }{ \lambda } \) 
Variance :
By definition,\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -\lambda x } \right) } dx+\int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \lambda { e }^{ -2x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( { e }^{ -2x } \right) dx } \)
 (using Gamma integral for positive integer)
Therefore Var(X ) = E(X² )- E(X )²
= \(\frac { 2 }{ { \lambda }^{ 2 } } -\left( \frac { 1 }{ \lambda } \right) ^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } } \) 
Hence the mean and variance are respectively \(\frac { 1 }{ \lambda } \)  and \(\frac { 1 }{ { \lambda }^{ 2 } } \)

Since X is a discrete random variable, from the given data, X takes on the values
−2, −1, 0, and 1.
For discrete random variable X, by definition, we have f (x) = P(X = x)
Therefore left hand limit of f(x) at x = -2 is F(− 2⁻ ) 
f (−2) = P(X =-2 ) = F(-2 ) - F(- 2^- )= 0.25-0 = 0.25
Similarly for other jump points, we have
f (−1) = P(X = -1) = F(-1) - F(-2) = 0.60 - 0.25 = 0.35.
f (0) = P(X ) 0) = F(0) - F(-1) = 0.90 - 0.60 = 0.30 ,
f (1) = P(X =1) = F(1) - F(0) 1- 0.90 = 0.10 .
Therefore the probability mass function is

The distribution function F(x) has jumps at x = -2, -1, 0, and 1. The jumps are respectively 0.25, 0.35, 0.30, and 0.1 is shown in the figure given below.
These jumps determine the probability mass function

(i) \(P(X<0)=\sum _{ -\infty }^{ -1 }{ P(X=x)=P(X=-1)=0.25+0.35 } =0.60\)
(ii) \(P(X\ge -1)=\sum _{ -1 }^{ 1 }{ P(X=x)=P(x=-1) } +P(X=0)+P(X=1)=0.35+030+0.10=0.75\)

By definition the cumulative distribution function for discrete random variable is
\(F(x)P\left( X\le x \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\) 
\(P(X<1)=0\) for -∞
\(F(1)=P\left( X\le 1 \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ i })=\sum _{ -\infty }^{ 1 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) =0+\frac { 1 }{ 12 } =\frac { 1 }{ 12 } \)
\(F(2)=P\left( X\le 2 \right) =\sum _{ -\infty }^{ 2 }{ P\left( X=x \right) } =P\left( X\le 1 \right) +P\left( X=1 \right) +P\left( X=2 \right) \) 
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } =\frac { 1 }{ 2 } \) 
\(F(3)=P\left( X\le 3 \right) =\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) \)
= \(0+\frac { 1 }{ 2 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } =\frac { 11 }{ 12 } \) 
\(F(4)=P\left( X\le 4 \right) =\sum _{ -\infty }^{ 4 }{ P\left( X=x \right) } =P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) +P(X=4)\)
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } +\frac { 1 }{ 12 } =1\)
\(F(x)= \begin{cases}0, & -\infty

(ii) \(P(X\le 3)=F(3)\frac { 11 }{ 12 } \)
(iii) \(P(X\ge 2)=1-P\left( X<2 \right) =1-P(X\le 1)=1-F(1)=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 } \)