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

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
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) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

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 \(\frac { 1 }{ { \lambda }^{ 2 } } \)

Let X denote the winning amount. The possible events of selection are 
(i) both balls are black, or
(ii) one white and one black or
(iii) both are white 
Therefore X is a random variable that can be defined as
X (both are black balls) = Rs. 2(20) = Rs. 40
X (one black and one white ball) = Rs. 20 − Rs. 10 = Rs. 10
X (both are white balls) = (Rs. 20) = - Rs. 20
Therefore X takes on the values 40,10 and −20
Total number of balls n = 12
Total number of ways of selecting 2 balls = \(\left( \begin{matrix} 12 \\ 2 \end{matrix} \right) =\frac { 12\times 11 }{ 1\times 2 } =66\)
Number of ways of selecting 2 black balls =  \(\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) =6\) 
Number of ways of selecting one black ball and one white ball = \(\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) =32\) 
Number of ways of selecting 2 white balls = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Probability mass function is

Mean :
\(E(X)\Sigma xf(x)=40.\left( \frac { 6 }{ 66 } \right) +10.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) .\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
That is expected winning amount is 0
 Variance :
\(\Sigma x^{ 2 }=\Sigma { x }^{ 2 }f(x)=40^{ 2 }.\left( \frac { 6 }{ 66 } \right) +10^{ 2 }.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) ^{ 2 }.\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
 (E(X )² = 0² = 0
This gives  \(V(X)=E({ X }^{ 2 })-\left( E(X))^{ 2 } \right) =\frac { 4000 }{ 11 } -0=\frac { 4000 }{ 11 } \)
Therefore E(X ) = 0 and \(V(x)=\frac { 4000 }{ 11 } \)

Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 5 }{ kdx } +\int _{ 5 }^{ \infty }{ 0dx } =1\) 
\(0+k\left( x \right) _{ 1 }^{ 5 }+0=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 0 & Otherwise \end{matrix} \end{cases}\)
(i) Distribution function
The distribution function 
\(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1, \(F(x)=\int _{ -\infty }^{ x }{ f(u)du } =\int _{ -\infty }^{ x }{ oldu } =0\) 
When 1 ≤ x ≤ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u)du=\int _{ -\infty }^{ x }{ 0du } +\int _{ 1 }^{ x }{ odu } +\int _{ 1 }^{ x }{ \frac { 1 }{ 4 } du } =\frac { 1 }{ 4 } (x-1) } \) 
When x ≥ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du=\int _{ -\infty }^{ x }{ odu } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 5 }^{ 5 }{ odu } =1\) 
Thus \(F(x)=\begin{cases} \begin{matrix} 0 & x<1 \end{matrix} \\ \begin{matrix} \frac { x-1 }{ 1 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 1 & x>5 \end{matrix} \end{cases}\) 
(ii) P(X < 3) = P(X ≤ 3) = F(3) = \(\frac { 3-1 }{ 2 } =\frac { 1 }{ 2 } \) (Since F(x) is continuous)
(iii) P(2 < X < 4) = P(2 ≤ X ≤ 4) F(4) - F(2) = \(\frac { 3 }{ 4 } -\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \) 
(iv) P(3 ≤ X ) = P(X ≥ 3) = 1− P(X < 3) = 1 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

Since the given function is a probability mass function, the total probability is one. That is \(\underset { x }{ \Sigma } f(x)=1\)
 From the given data k + 2k + 6k + 5k + 6k +10k+1
\(30k=1\Rightarrow k=\frac { 1 }{ 30 } \) 
Therefore the probability mass function is

(i) P(2 < X < 6) = f(3)+ f(4)+ f(5) = \(\frac { 6 }{ 30 } +\frac { 5 }{ 30 } +\frac { 6 }{ 30 } =\frac { 17 }{ 30 } \) 
(ii) P(2≤X≤5) = f(2)+f(3)+f(4) = \(\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 13 }{ 30 } \) 
(iii) P(2≤4) = f(1)+f(2)+f(3)+f(4) = \(\frac { 1 }{ 30 } +\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 14 }{ 30 } \) 
(iv) P(3>X) = f(4)+f(5)+f(6) = \(\frac { 5 }{ 30 } +\frac { 6 }{ 30 } +\frac { 10 }{ 30 } =\frac { 21 }{ 30 } \)   

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) \(P(X\ge -1)=\sum _{ -1 }^{ 1 }{ P(X=x)=P(x=-1) } +P(X=0)+P(X=1)=0.35+030+0.10=0.75\)

Since X denotes the total score in two throws, it takes on the values 2, 3, 4, 5 and 6. From the Sample space S, we have

\(P(X=2)=\frac { 1 }{ 36 } \), \(P(X=3)=\frac { 4 }{ 36 } \) 
\(P\left( X=4 \right) =\frac { 10 }{ 36 } \) , \(P(X=5)=\frac { 12 }{ 36 } \) and 
\(P(X=6)=\frac { 9 }{ 36 } \)
(i) Probability mass function is

(ii) Cumulative distribution function By definition of the cumulative distribution function for discrete random variable we have
\(f(x)=P(X\le x)=\underset { x_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\)
\(P(X for -∞
\(F(2)=P(X\le 2)=\sum _{ -\infty }^{ 2 }{ P(X=x)=P\left( X \right) <2)+P(X=2) } =0+\frac { 1 }{ 36 } =\frac { 1 }{ 36 } \)
\(F(3)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)=0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(4)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)\)
\(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(5)=P\left( X\le 5 \right) =\sum _{ -\infty }^{ 5 }{ P(X=x) } =P\left( X<2 \right) +P(X=3)+P\left( X=4 \right) +P\left( X=5 \right) \)
= \(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 } \)
\(F(6)=P(X\le 6)=\sum _{ -\infty }^{ 6 }{ P(X=x) } \) 
= \(P(X<2)+P(X=2)+P(X=3)+P(x=4)+P(x=5)P(X=6)\)
 \(0+\frac { 1 }{ 36 } +{ \frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =1 }\) 
Therefore the cumulative distribution function is


(iii) \(P(3\le X\le 6)=\sum _{ x=3 }^{ 5 }{ P(X={ { x }_{ 1 })=P(X=3) }+P(X=4) } +P(X=5)\)
\(=\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } \)
(iv) \(X\ge 4)=\sum _{ x=4 }^{ 5 }{ P(X={ x }_{ 1 }) } \)
 = \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)

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