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

(i) The sample space S = {H,T}\(\times\){H,T}
That is S = {TT,TH,HT,HH}
(ii)  Let X : S ⟶R be the number of tails
Then X (TT) = 2 (2 Tails)
X (TH ) = 1 (1 Tail)
X (HT) = 1 (1 Tail)
and X (HH) = 0 (0 Tails).
Then X is a random variable that takes on the values 0, 1 and 2.
Let X (ω) denotes the number of tails, this gives
\(\\ \\ \\ \\ \\ \\ X\left( \omega \right) =\begin{cases} \begin{matrix} 2 & if\omega =TT \end{matrix} \\ \begin{matrix} 1 & if\omega =HT,TH \end{matrix} \\ \begin{matrix} 0 & if\omega =HH \end{matrix} \end{cases}\) 
The inverse images of 1 {TH, HT} . That is X^-1{1} = {TH, HT}.
(iii) Number of elements in inverse images are shown in the table.

Given P (hitting the target) = \(\frac { 1 }{ 4 } \Rightarrow P=\frac { 1 }{ 4 } \) 
n = 10,
(i) P(X = 4)
\(P(X+4)=\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ x }(1-p)^{ n-x },x\) 
= 0,1,2,...n
(i) Probability of hitting the target exactly 4 times
P(X = 4) = \(^{10}{ C }_{ 4 } \times\left( \begin{matrix} 1 \\ 4 \end{matrix} \right) ^{ 4 }\times \left( \cfrac { 3 }{ 4 } \right) ^{ 6 }\)
\( =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2} \times \frac{1}{4^{4}} \times \frac{3^{6}}{4^{6}} \\ =210 \times \frac{3^{6}}{4^{10}} \) 
(ii) Probability of hitting atleast one time
= P(X≥1) = 1-P(x<1)
= 1-P(X = 0)
\( =1-{ }^{10} \mathrm{C}_{0} \times\left(\frac{1}{4}\right)^{0} \times\left(\frac{3}{4}\right)^{10} \\ =1-\frac{3^{10}}{4^{10}} \)

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

n (S) = 36
X = {4,5,6,7,8}
From the sample space

Let X be the random variable denotes the Winning amount.
X (Both are black balls) = Rs. 2 (-10) = Rs. -20
X (one red and oneblack ball) = Rs.15-Rs. 10 = Rs. 5
X (both are red ball) = Rs. 2 (15) = Rs. 30
= {-20, 5, 30}
The sample space consists of ¹⁴C₂ = 91
X = -20, Both are black balls= ⁸C₁ = 28
X = 5, One black, one redball = ⁸C₁ x ⁶C₁ = 8 x 6 = 48
X = 30, Both are white balls = ⁶C₁ = 15

Let X be the random variable of number of black cards occur.
X = {0,1,2}
Sample space ⁵²C₂ = 1326
Let X denote the number of black cards drawn.
X = 0, X (both are red cards) = ²⁶C₂ = 325  
X = 1 (1 black card and 1 red card) = ²⁶C₁ \(\times\) ²⁶C₁ = 676  
X = 2 (both are black cards) = ²⁶C₂ = 325  
∴ X takes the values 0, 1, 2

\(f(x)= \begin{cases}2(x-1) & 1
\(Mean=E(X)=\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ 2((x-1)dx } } \) 
\( =2\left[\frac{8}{3}-\frac{4}{2}-\frac{1}{3}+\frac{1}{2}\right] \)
\( =2\left(\frac{7}{3}-\frac{3}{2}\right) \)
\( =2 \times \frac{5}{6} \)
\( =\frac{5}{3} \)
\(E({ x }^{ 2 })=\int _{ 1 }^{ 2 }{ { x }^{ 2 }f(x)dx } \) 
= \(\int _{ 1 }^{ 2 }{ { x }^{ 2 }.2\left( x-1 \right) } dx\) 
= \(2\int _{ 1 }^{ 2 }{ ({ x }^{ 3 }-{ x }^{ 2 })dx } \) 
= \(2\left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\) 
= \(2\left[ \left( 4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right) \right] \) 
= \(2\left[ \frac { 4 }{ 3 } +\frac { 1 }{ 12 } \right] =2\left[ \frac { 16+1 }{ 12 } \right] \) 
= \(\frac { 17 }{ 6 } \) 
ஃ Var(X) = E(X²) - [E(X)]²
= \(\frac { 17 }{ 6 } -(\frac{5}{ 3 }^{ 2 })=\frac { 17 }{ 6 } -\frac { 25 }{ 9 } \)
= \(\frac{51-50}{18}\)
= \(\frac{1}{18}\)

P(0.5≤X<1.5)
= \(\int _{ 0.5 }^{ 1.5 }{ f(x)dx } =\int _{ 0.5 }^{ 1 }{ f(x)dx+\int _{ 1 }^{ 1.5 }{ f(x)dx } } \) 
= \(\int _{ 0.5 }^{ 1 }{ xdx+\int _{ 1 }^{ 1.5 }{ (2-x)dx } } \) 
= \(\left[ \cfrac { { x }^{ 2 } }{ 2 } \right] _{ 0.5 }^{ 1 }+\left[ 2x-\cfrac { { x }^{ 2 } }{ 2 } \right] _{ 1 }^{ \\ 1.5 }\) 
= \(\frac { 1 }{ 2 } -\frac { (0.5)^{ 2 } }{ 2 } +\left[ 2(1.5)-\frac { (1.5)^{ 2 } }{ 2 } \right] -\left( 2-\frac { 1 }{ 2 } \right) \)
\(= 1+3-2-\frac { 2.50 }{ 2 } \)
\( 2-\frac { 2.5 }{ 2 } =\frac { 1.5 }{ 2 } \)
= 0.75

 P(0.2≤X < 0.6)
= \(\int _{ 0.2 }^{ 0.6 }{ f(x)dx } =\int _{ 0.2 }^{ 0.6 }{ x.dx } =\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0.2 }^{ 0.6 }\)  
= 0.18 - 0.02
= 0.16

Given \(f(x)=\begin{cases} \begin{matrix} 16{ xe }^{ -4x } & foex>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\) 
Mean :
= \(E(x)=\int _{ 0 }^{ \infty }{ x.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ x.16.xe^{ -4x }dx } \) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 2 } } .{ e }^{ -4x }dx=16\times \frac { 2! }{ { 4 }^{ 3 } } \)  \(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\frac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(16\times \frac { 2 }{ 64 } \)
\(=\frac { 1 }{ 2 } \) 
Variance :
\(E({ x }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x)dx } \) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.{ e }^{ -4x } } dx\) 
= \(16\int _{ 0 }^{ \infty }{ { x }^{ 3 }{ e }^{ -4x }dx } \) 

ஃ Var(X) = E(X²) - [E(x)]²
\( =\frac{3}{8}-\frac{1}{4} \)
\(=\frac{1}{8} \)
∴ Var(X) \(=\frac{1}{8} \)