(i) Given that five fair coins are tossed once. Since the coins are fair coins the probability of getting an head in a single coin is
 \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 1 }{ 2 } \) 
Let X denote the number of heads that appear in five coins. X is binomial random variable that takes on the values 0, 1, 2,  3, 4 and 5 and \(p=\frac { 1 }{ 2 } \) That is \(X\sim B\left( 5,\cfrac { 1 }{ 2 } \right) \) 
Therefore the binomial distribution is
\(f(x)=\left( \begin{matrix} n \\ x \end{matrix} \right) p*\left( 1-p \right) ^{ n-x }\), x = 0, 1, 2,..,n
becomes
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ x }\left( \cfrac { 1 }{ 2 } \right) ^{ n-x }\), x = 0, 1, 2,..,5
That is 
\(f(x)=\left( \begin{matrix} 5 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 2 } \right) ^{ n }\), x = 0, 1, 2...,n
(ii) A fair die is rolled ten times and X denotes the number of times 4 appeared. X is binomial
random variable that takes on the values 0, 1, 2, 3,...10 , with n = 10 and \(p=\cfrac { 1 }{ 6 } \). That is \(X\sim B\left( 10,\cfrac { 1 }{ 6 } \right) \) 
Probability of getting a four in a die is \(p=\frac { 1 }{ 2 } \) and \(q=1-p=\frac { 5 }{ 6 } \)
Therefore the binomial distribution is 
\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 6 } \right) ^{ x }\left( \cfrac { 5 }{ 6 } \right) ^{ 10-x }\) x = 0, 1, 2,...,10

(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

Values of the Random Variable	0	1	2	Total
Number of elements in inverse images	1	2	1	4

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

x	0	1	2
f(x)	\(\cfrac { 1 }{ 4 } \)	\(\cfrac { 1 }{ 2 } \)	\(\cfrac { 1 }{ 4 } \)

(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.

Values of the Random Variable	0	1	2	Total
Number of elements in inverse image	1	2	1	4

Sample space s = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
Let X be the random variable denoting the number of heads
X = {0, 1, 2, 3}
Values of randomn variabl 0 1 2 3 Total Number of elements in inverse image 1 3 3 1 8 Probability mass function 
x 0 1 2 3 f(x) \(\cfrac { 1 }{ 8 } \) \(\cfrac { 3 }{ 8 } \) \(\cfrac { 3 }{ 8 } \)  \(\cfrac { 1 }{ 8 } \) \(\therefore f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & \begin{matrix} for & x=0,3 \end{matrix} \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } & \begin{matrix} for & x=1,2 \end{matrix} \end{matrix} \end{cases}\)

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

II	2	3	3	4	4	4
I
2	4	5	5	6	6	6
3	5	6	6	7	7	7
3	5	6	6	7	7	7
4	6	7	7	8	8	8
4	6	7	7	8	8	8
4	6	7	7	8	8	8

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

Values of random variable	4	5	6	7	8	Total
No of points in inverse image	1	4	10	12	9	36

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

Values of random variable X	0	1	2	Total
Number of elements in inverse images	325	676	325	1326

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 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

x	1	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 30 } \)	\(\cfrac { 2 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 5 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 10 }{ 30 } \)

(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) = \(\frac { 5 }{ 30 } +\frac { 6 }{ 30 } +\frac { 10 }{ 30 } =\frac { 21 }{ 30 } \)   

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

Let p be the probability of the useful life hours of a fluorescent light.
n = 12
P = 0.9
q = 1-p = 0.1
P(X= x)= ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Exactly 10
P(X = 10) = ¹²C₁₀(0.9)¹⁰(1 - 0.9)²
= ¹²C₁₀(0.9)¹⁰ (0.1)²
(ii) Atleast 11
P(X≥11) = R(X = 11) + P(X = 12)
= ¹²C₁₁(0.9)¹¹ (0.1)¹ + ¹²C₁₂(0.9)¹²(0.1)⁶
= ¹²C₁ (0.9)¹¹ (0.1) + (0.9)¹²
= ¹²(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (0.9)¹¹ (2.1)
(ii) Atleast 2 will not have a useful
P(X,10) = 1 -P(X > 10)
= 1 - [P(X = 11) + P(X = 12)]
= 1-(2.1) (0.9)¹¹

Given n = 12
P = 0.9
(i) P(X = 10) = 12C₁₀(0.9)¹⁰(1 - 0.9)²
= 12C₁₀(0.9)¹⁰ (0.1)²
(ii) P(X 2: 11) = R(X = 11) + P(X = 12)
= 12C₁₁(0.9)¹¹ (0.1) + 12C₁₂(0.9)¹²(0.1)⁶
= 12C (0.9)¹¹ (0.1) + (0.9)¹²
= 12(0.9)¹¹ (0.1) + (0.9)¹²
= (0.9)¹¹ ((12)(0.1) + 0.9)
= (0.9)11 (1.2 + 0.9) = (0.9)11 (2.1)
(iii) P( at least 2 will not have a useful life of atleast 600 hours) = 1 - P( atleast 11 will have a useful life of atleast 600 hours)
= 1 - P(X ≥11)
= 1-2.1 (0.9)¹¹

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Given \(p=\frac { 3 }{ 4 } \) 
n = 5
P(X = x) = nC_xp^x (1-p)^n-x
\(P(X=3)={ 5C }_{ 3 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( 1-\frac { 3 }{ 4 } \right) ^{ 2 }\) 
\(P(X=3)={ 5C }_{ 2 }\left( \frac { 3 }{ 4 } \right) ^{ 3 }\left( \frac { 1 }{ 4 } \right) ^{ 2 }\) [∵nCr = nC_n-r]

= \(\frac { 135 }{ 512 } \)

 

\(\mathrm{n}=9, \mathrm{p}=\frac{1}{2}, \mathrm{k}=7 \)
\( \mathrm{P}(X=x)={ }^{n} C_{x} p^{x} q^{n-x}, x=0,1,2, \ldots, n \)
\(p =\frac{1}{2} \)
\(q =1-p=\frac{1}{2} \)
\(P(X=7) ={ }^{9} C_{7}\left(\frac{1}{2}\right)^{7}\left(\frac{1}{2}\right)^{2} \)
\( =\frac{9 \times 8}{2} \times \frac{1}{2^{9}} \)
\( =36 \times \frac{1}{512}=\frac{9}{128} \)

\(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
= \(\frac { 17 }{ 6 } -(\frac{5}{ 3 }^{ 2 })=\frac { 17 }{ 6 } -\frac { 25 }{ 9 } \)
= \(\frac{51-50}{18}\)
= \(\frac{1}{18}\)