X~ B(n, p)
Given mean np = 6
\(S.D=\sqrt { npq } =2\) 
\(\Rightarrow npq=4\) 
\( \rightarrow \frac { npq }{ np } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \) 
\(\Rightarrow q=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-P=\frac { 2 }{ 3 } \) 
\(\Rightarrow 1-\frac { 2 }{ 3 } =P\) 
\(\therefore P=\frac { 1 }{ 3 } \)  
\(n\times \frac { 1 }{ 3 } =6\Rightarrow n=18\)
(i) The probability mass function
P(X = x) nC_x p^x (1 - p )^n-x,
X = 0,1,2, ... , n
\(\therefore P(X=x)=\ ^{18}{ C }_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 18-x }\) 
x=0,1,2...,8
(ii) \(P(X=3)=\ ^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 18-3 }\) 
= \(^{ 18}{C }_{ 3 }\left( \frac { 1 }{ 3 } \right) ^{ 3 }\left( \frac { 2 }{ 3 } \right) ^{ 15 }\)
(iii) P(X ≥ 2)  
 P(X ≥ 2) 1 -P(X < 2)
= 1 - [P(X = 0) + P(X = 1)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

Given 

(i) The probability density function. Differentiating F(x) with respect to 'x' at continuity points  of F(x), we get
\(F(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } ({ 2x }+1) & 0\le x<1 \end{matrix} \\ \begin{matrix} 1 & x\ge 1 \end{matrix} \end{cases}\) 
(ii) \(p(0.3\le X\le 0.6)=\int _{ 0.3 }^{ 0.6 }{ f(x)dx } \) 
= \(\int _{ 0.3 }^{ 0.6 }{ \frac { 1 }{ 2 } \left( 2x+1 \right) dx } =\frac { 1 }{ 2 } \left[ \frac { { 2x }^{ 2 } }{ 2 } +x \right] _{ 0.3 }^{ 0.6 }\) 
= \(\frac { 1 }{ 2 } \left( { x }^{ 2 }+x \right) _{ 0.3 }^{ 0.6 }=\frac { 1 }{ 2 } \left[ \left( { 0.6 }^{ 2 }+0.6 \right) -\left( { 0.3 }^{ 2 }+0.3 \right) \right] \) 
= \(\frac { 1 }{ 2 } \left[ \left( .36.6 \right) \right] -\left( .09+0.3 \right) ]\) 
= \(\frac { 1 }{ 2 } \left[ 0.96-.39 \right] =\frac { 0.57 }{ 2 } =0.285\) 
= 0.285

\(f(x)=\begin{cases} \begin{matrix} x+1 & -1\le x<0 \end{matrix} \\ \begin{matrix} -x+1 & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
(i) Distribution function
Case 1 : x < -1
F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \) = 0
Case 2 : -1 ≤ x < 0
\(\int _{ -\infty }^{ x }{ f(u)du } \) 
= \(\int _{ -\infty }^{ x }{ f(x) } dx=\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] _{ -1 }\) 
= \(\left( \frac { {u }^{ 2 } }{ 2 } +u \right)=\frac{x^2}{2}+x -\left( \frac { 1 }{ 2 } +1 \right) \) 
= \(\frac { { x }^{ 2 } }{ 2 } +x+\frac { 1 }{ 2 } \) 
Case 3 : 0 ≤ x < 1,
 \(F(X)=\int _{ 0 }^{ x }{ (-x+1)dx } =\left[ -\frac { { x }^{ 2 } }{ 2 } +x \right] _{ 0 }^{ x }\) 
= \(\left( -\frac { { x }^{ 2 } }{ 2 } +x \right) -\left( 0 \right) =\frac { { x }^{ 2 } }{ 2 } +x\) 
When 1 ≤ x,
\(F(x)=\int _{ 1 }^{ x }{ f(x)dx } =\int _{ 1 }^{ x }{ 0dx } \)
= \(\therefore F(X)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 } }{ 2 } +x+\frac { 1 }{ 2 } & -1\le x<0 \end{matrix} \\ \begin{matrix} -\frac { { x }^{ 2 } }{ 2 } +x & 0\le x<1 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(ii) p(0.5 ≤ X ≤ 0.5)
= \(\int _{ -0.5 }^{ 0.5 }{ f(x)dx } =\int _{ 0.5 }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 0.5 }{ f(x)dx } } \)
= \(\int _{ -0.5 }^{ 0 }{ (x+1) } dx+\int _{ 0 }^{ 0.5 }{ \left( -x+1 \right) } dx\) 
= \(\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] _{ -0.5 }^{ 0 }+\left[ \frac { -{ x }^{ 2 } }{ 2 } +x \right] _{ 0 }^{ 0.5 }\) 
= \(0-\left( \frac { { 0.5 }^{ 2 } }{ 2 } -0.5 \right) +\left( -\frac { \left( 0.5 \right) ^{ 2 } }{ 2 } +0.5 \right) -0\) 
= \(-\left( \frac { .25 }{ 2 } -0.5 \right) +\left( \frac { -0.25 }{ 2 } +0.5 \right) \) 
= \(\frac { .25 }{ 2 } +0.5-\frac { 0.25 }{ 2 } +0.5=0.25+1\)
= 0.75

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

Let p be the probability that indicates the defective rate of an electronic device
n = 10
\(P=5\%=0.05 \) 
q = 1 - p
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1,2, .., n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 - P(X < 1)
= 1-P(X = 0)
= 1-¹⁰C₀ (0.05)⁰ (0.95)¹⁰
P(X ≥1) = 1 - (0.95)¹⁰
(ii) Exactly two defective items  
P(X = 2) =¹⁰C₂(0.05)² (0.95)⁸