(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) 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.

(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 possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

Let us denote white and red balls as w₁, w₂, r₁, r₂ and r₃
The sample space consists of 5C₃ = 10 different samples of size 3.
That is S = \(\left\{w_{1} w_{2} r_{1}, w_{1} w_{2} r_{2}, w_{1} w_{2} r_{3}, w_{1} r_{1} r_{2}, w_{1} r_{2} r_{3}, w_{1} r_{1} r_{3}, w_{2} r_{1} r_{2}, w_{2} r_{2} r_{3}, w_{2} r_{1} r_{3}, r_{1} r_{2} r_{3}\right\} .\)
The random variable X takes on the values 1, 2, and 3.

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(i) The sample space
S = {1, 2, 3, 4, 5, 6}\(\times\){1, 2, 3, 4, 5, 6}
consists of 36 ordered pairs (α, β) where α and β can take any integer value between 1 and 6 as shown. X is assigned to each point (α, β) the sum of the numbers on the dice .
That is X (α, β) = α + β
Therefore
X (1,1) = 1+1 = 2
X (1, 2) = X (2,1) = 3
X (1,3) = X (2,2) = X (3,1)= 4
X (1, 4) = X (2,3) = X (3, 2)  X (4,1) = 5
X (1,5) = X (2,4) = X (3,3) = X (4, 2) = X (5,1) = 6
X (1,6) = X (2,5) = X (3, 4) = X (4,3 = X (5, 2) X (6,1) = 7
X (2,6) = X (3,5) = X (4,4) = X (5,3) = X (6,2) = 8
X (3,6) = X (4,5) = X (5,4)  X (6,3) = 9
X (4,6) = X (5,5) X (6,4) = 10
X (5,6) = (6,5) = 11
X (6,6) = 12
(ii) Then the random variable X takes on the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
(iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}.
(iv) The number of inverse images are given below

(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 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 \(1-2\left( \frac { 4 }{ 5 } \right) ^{ 5 }\)

(i) Since X denotes the number of success, X can take the values 0,1, 2, ...10
The probability for success is \(p=\frac { 1 }{ 4 } \) and for failure \(q=1-p=\frac { 3 }{ 4 } \) and n = 10
Therefore X follows a binomial distribution denoted by \(X\sim B\left( 10,\frac { 1 }{ 4 } \right) \) 
This gives,\(f(x)=\left( \begin{matrix} 10 \\ x \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ x }\left( \cfrac { 3 }{ 4 } \right) ^{ 10-x }\)  x = 0, 1, 2,..,10
(ii) Probability for seven correct answers is 
\(P(X=7)=f(7)=\left( \begin{matrix} 10 \\ 7 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 7 }\left( \cfrac { 3 }{ 4 } \right) ^{ 10-7 }=120\left( \cfrac { { 3 }^{ 2 } }{ { 4 }^{ 10 } } \right) \)
Probability that the student will get seven correct answers is \(120\left( \cfrac { { 3 }^{ 2 } }{ { 4 }^{ 10 } } \right) \) 
(iii) Probability for at least one correct answer is
P(X ≥1) = 1- P(X <1) = 1- P(X = 0)
= \(1-\left( \begin{matrix} 10 \\ 0 \end{matrix} \right) \left( \cfrac { 1 }{ 4 } \right) ^{ 0 }\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }=1-\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }\)
Probability that the student will get for at least one correct answer is \(1-\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }\)  

(i) By definition \(F(x)=\le x)=\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1 \(F(x)=P\left( x\le x \right) =\int _{ 1 }^{ x }{ odu+\int _{ 1 }^{ 0 }{ (u-1) } du } \) 
When 1 ≤ x < 2 \(F(x)=P(X\le x)=\int _{ -\infty }^{ x }{ odu } =0\) 
When 1 ≤ x < 2 \(F(x)=P\left( x\le x \right) =\int _{ 1 }^{ x }{ odu+\int _{ 1 }^{ 0 }{ (u-1) } du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] =\frac { \left( x-1 \right) ^{ 2 } }{ 2 } \) 
When 2 ≤ x <3 \(F(x)=P(X\le x)=\int _{ -\infty }^{ 1 }{ du } +\int _{ 1 }^{ 2 }{ \left( u-1 \right) du } +\int _{ 2 }^{ x }{ \left( 3-u \right) du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] +\left[ \frac { (3-u)^{ 2 } }{ 2 } \right] \) 
= \(\frac { { 1 }^{ 2 }-0 }{ 2 } +\frac { 1-(3-x)^{ 2 } }{ 2 } =1\frac { \left( 3-x \right) ^{ 2 } }{ 2 } \) 
When x ≥ 3, \(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ 1 }{ odu } +\int _{ 1 }^{ 3 }{ (u-1) } +\int _{ 2 }^{ 1 }{ (3-u) } +\int _{ 3 }^{ x }{ odu } \) 
= \(\int _{ -\infty }^{ 1 }{ 0du } +\int _{ 1 }^{ 2 }{ (u-1)du } +\int _{ 2 }^{ 3 }{ (3-u) } du+\int _{ 3 }^{ x }{ 0du } \) 
= \(0+\left[ \frac { \left( u-1 \right) ^{ 2 } }{ 2 } \right] +\left[ \frac { \left( 3-u \right) ^{ 2 } }{ 2 } \right] _{ 2 }^{ 3 }+0\) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 2 } =1\) 
These give 


(ii) P(1.5 ≤ X ≤ 2.5) = F(2.5) − F(1.5)
= \(\left( 1-\frac { \left( 3-2.5 \right) ^{ 2 } }{ 2 } \right) -\left( \frac { \left( 1.5-1 \right) ^{ 2 } }{ 2 } \right) \) 
= \(\cfrac { 1.75-0.25 }{ 2 } =0.75\) 
\(P\left( 1.5\le X\le \right) =\int _{ 1.5 }^{ 2.5 }{ f(x)dx } =\int _{ 1.5 }^{ 2 }{ (x-1) } dx+\int _{ 2 }^{ 2.5 }{ (-x+3) } dx=0.75\) 

Since the given function is a probability density function

\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ f(x) } dx+\int _{ 1 }^{ 4 }{ f(x) } dx+\int _{ 4 }^{ \infty }{ f(x) } dx=1\) 
From the given information
\(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 4 }{ { Cx }^{ 2 }dx } +\int _{ 4 }^{ \infty }{ 0dx } =1\) 
\(0+C\left[ \cfrac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 4 }+0=1\Rightarrow C\left[ \cfrac { 64-1 }{ 3 } \right] =1\Rightarrow 21C\Rightarrow C=\cfrac { 1 }{ 21 } \) 
Therefore the probability density function is
\(f(x)= \begin{cases}C x^{2} & 1
Since f (x) is continuous, the probability that X is equal to any particular value is zero. Therefore when the random variable is continuous, either or both of the signs < by ≤ and > by ≥ can be interchanged. Thus
(i) P(1.5 < X < 3.5) = P(1.5 ≤ X< 3.5)= P(1.5 < X ≤3.5) = P(1.5 ≤X ≤ 3.5)
Therefore
\(P(1.5
 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) =\cfrac { 1 }{ 21 } \left( \cfrac { \left( 3.5 \right) ^{ 3 }-\left( 1.5 \right) ^{ 3 } }{ 3 } \right) \) 
= \(\cfrac { 79 }{ 126 } \) 
(ii) \(P(X\le 2)=\int _{ -\infty }^{ 2 }{ f(x) } dx=\int _{ -\infty }^{ 1 }{ f(x)dx } +\int _{ 1 }^{ 2 }{ f(x)dx } \)
 
Therefore
\(P(X\le 2)=0+\cfrac { 1 }{ 21 } \int _{ 1 }^{ 2 }{ { x }^{ 2 }dx=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } ^{ 2 }_{ 1 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 2 }^{ 3 }-{ 1 }^{ 3 } }{ 3 } \right) =\cfrac { 7 }{ 63 } \) 
(iii) \(P(3
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

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) = \(1-\left( \frac { 3 }{ 4 } \right) ^{ 8 }\)