(i) Since f (x) is a probability mass function, f (x) ≥ 0 for all x , and d \(\sum_{x} f(x)=1\)
Thus, \(\sum_{x} f(x)=1\)
\(c^{2}+2 c^{2}+3 c^{2}+4 c^{2}+c+2 c=0\)
\(c=\frac{1}{5} \text { or }-\frac{1}{2}\)
Since f x( ) ≥ 0 for all x , the possible value of c is \(\frac{1}{5}\)
Hence, the probability mass function is

 (ii) To find mean and variance, let us use the following table

Mean :  \(E(X)=\Sigma xf(x)=\frac { 115 }{ 25 } =4.6\) 
Variance : \(V(x)=E\left( x \right) ^{ 2 }=\Sigma { x }^{ 2 }f(x)-\left( \Sigma xf(x) \right) ^{ 2 }\) 
= \(\frac { 585 }{ 25 } -\left( \frac { 115 }{ 25 } \right) ^{ 2 }=23.40-21.16=2.24\)
Therefore the mean and variance are 4.6 and 2.24 respectively.

(i) Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ - }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 0 }{ 0dx } +\int _{ 0 }^{ \infty }{ k{ e }^{ -2x }dx } =1\) 
\(0+k\left( \frac { { e }^{ -2x } }{ -2 } \right) =1\Rightarrow k\left( \frac { { e }^{ -\infty }-{ e }^{ 0 } }{ -2 } \right) =1\Rightarrow k=2\) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} 2{ e }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx \end{matrix}\le 0 \end{cases}\)
 (ii) Distribution function
By definition the distribution function \(F(x)=P\left( x\le x \right) =\int _{ -\infty }^{ x }{ f(u) } du\) 
When x≤0 \(F(x)=\int _{ -\infty }^{ x }{ F(u) } du=\int _{ -\infty }^{ x }{ odu=0 } \) 
When x > 0 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du\int _{ -\infty }^{ x }{ 0du } +\int _{ 0 }^{ x }{ { 2e }^{ -2x }du\left( \frac { { e }^{ -2x } }{ -2 } \right) } =1-{ e }^{ 2x }\) 
This gives \(F(x)=\begin{cases} \begin{matrix} 0 & forx\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ 2x } & forx>0 \end{matrix} \end{cases}\) 
(iii) P(X < ) = P(X ≤2 ) = F(2 ) = 1-e^2\(\times\)2 (since F(x) is continuous)
(iv) The probability that X is at least equal to four unit of time is
P(X ≥ 4 ) = 1 - P(X < 4 ) = 1- F( 4) = 1 - ( 1-e^{-2\(\times\)4}) = e⁸
(v) In the continuous case, f (x) at x = a is not the probability that X takes the value a, that is f (x) at x = a is not equal to P( X ) a.  If X is continuous type, P(X = a) = 0 for a ∈ R. Therefore P(x = 3) = 0.

Given that n = 6
Probability for selecting a defective product is \(\frac { 20 }{ 100 } \) that is \(p=\frac { 1 }{ 5 } \) 
Since X denotes the number defective products, X can take on the values 0,1,2,...,6
The probability for defective (success) is \(p-\frac { 1 }{ 5 } \) and for failure \(q=1-p=\frac { 4 }{ 5 } \), and n = 6 
Therefore X follows a binomial distribution denoted by \(X\sim B\left( 6,\frac { 1 }{ 5 } \right) \) 
This gives \(f(x)=\left( \begin{matrix} 6 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ x }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-x }\), x = 0,1,2,...,6,
(i) Probability for two defective products is
\(P(X=2)=f(2)=\left( \begin{matrix} 6 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ x }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-x }=15\left( \cfrac { { 4 }^{ 4 } }{ { 5 }^{ 6 } } \right) \)
(ii) Probability for at most one defective products is 
P(X ≤1) = P(X = 0) + P(X = 1)
\(-\left( \begin{matrix} 6 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 5 } \right) ^{ 0 }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-0 }+\left( \begin{matrix} 6 \\ 1 \end{matrix} \right) \left( \frac { 1 }{ 5 } \right) ^{ 1 }\left( \cfrac { 4 }{ 5 } \right) ^{ 6-1 }\) 
\(-\left( \cfrac { 4 }{ 5 } \right) ^{ 6 }+\left( 6 \right) \left( \cfrac { { 4 }^{ 2 } }{ { 5 }^{ 2 } } \right) =2\left( \cfrac { 4 }{ 5 } \right) ^{ 2 }\)
Probability for at most one defective products is \(2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\) 
(iii) Probability for at least two defective products is 
P(X≥2)-1-P(X<2) = 1-P(X≤1) = \(1-2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\) 
Probability for at least two defective products is \(1-2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\)

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 \(V(x)=\frac { 4000 }{ 11 } \)

To find the probabilities, the values of the parameters n and p must be known.
Given that
Mean = np = 2 and variance = npq = 1.5
This gives \(\frac { npq }{ np } =\frac { 1.5 }{ 2 } =\frac { 3 }{ 4 } \) 
\(q=\frac { 3 }{ 4 } \) and \(p=1-q=1-3\frac { 4 }{ 4 } =\frac { 1 }{ 4 } \)
np = 2  gives \(n=\frac { 2 }{ p } =8\) . Therefore \(X\sim B\left( 8,\frac { 1 }{ 4 } \right) \) 
Therefore probability distribution is
\(P(X=x)=f(x)=\left( \begin{matrix} 8 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ x }\left( \cfrac { 3 }{ 4 } \right) ^{ 8-x }\) 
(i) \(P(X=0)=f(0)=\left( \begin{matrix} 8 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 0 }\left( \cfrac { 3 }{ 4 } \right) ^{ 8-0 }=\left( \cfrac { 3 }{ 4 } \right) ^{ 8 }\) 
(ii) \(P(X=1)=f(1)=\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) \left( \cfrac { 3 }{ 4 } \right) ^{ 8-1 }=2\left( \cfrac { 3 }{ 4 } \right) ^{ 2 }\) 
(iii)  P(X≥1) = 1-P(X<1) = 1-P(X = 0) = \(1-\left( \frac { 3 }{ 4 } \right) ^{ 8 }\)