(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
\(P(0.2\le X\le 0.7)=\int _{ 0.2 }^{ 0.7 }{ f(x) } dx=\int _{ 0.2 }^{ 0.7 }{ 1dx } =0.5\)

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(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

n = 5, X B{n, p)
P(X = 1) = 0.4096
P(X = 2) 0.2048
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, .., n
ஃnC₁,p¹q₄ 0.4096
⁵C₁, p²q⁴ = 0.4096
⁵C² p²q³ = 0.2048
5pq⁴ = 0.4096  .....(1)
10 p² q³ = 0.2048  ....(2) 
Dividing (2) by (1) we get 
\(\cfrac { 5{ pq }^{ 4 } }{ 10{ p }^{ 2 }{ q }^{ 3 } } =2\) 
q = 4p
q = 4(1- q)
q = 4 - 4q
5q = 4
q = 4/5
\(p=1-q= p=\frac { 1 }{ 5 } \) 
\(Mean=np=5\times \frac { 1 }{ 5 } =1\)
\( Variance =n p q=\not 5 \times \frac{1}{\not 5} \times \frac{4}{5}=\frac{4}{5}\) 
Distribution
(i) \(P(X=x)= ^5C_{ x }\left( \frac { 1 }{ 5 } \right) ^{ x }\left( \frac { 4 }{ 5 } \right) ^{ 5-x }\) , x = 0,1,2..n

X B(n, p)
Given 4P(X = 4) = P(X = 2) and n = 6.
4. [⁶C₄p⁴ (1 - p)²] = ⁶C₂p² q⁴
⇒ 4p² = q² 
⇒ 4(1-q²) = q²  
4(1-2q+q²) = q²  
⇒ 3q² - 8q +4 = 0
⇒ (q - 2)(3q - 2) = 0
\( -q=\frac { 2 }{ 3 } \ \ \ (q\neq2)\) 
\(p = 1- q=\cfrac { 1 }{ 3 } \)  
Distribution
P(X = x) = nC_xp^x (1- p)^n-x, x = 0, 1,2, ... n
(i) \(P(X=x)= ^6C_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 6-x }\) , x = 0,1,2..n
(ii) \(mean=np=6\times \frac { 1 }{ 3 } =2\) 
(iii) standard deviation = \(\sqrt { npq } =\sqrt { 2 \times \frac {2}{ 3 } } \) 
= \( { \frac { 2 }{ \sqrt 3 } } \)

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

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)
\( =1-{ }^{10} \mathrm{C}_{0} \times\left(\frac{1}{4}\right)^{0} \times\left(\frac{3}{4}\right)^{10} \\ =1-\frac{3^{10}}{4^{10}} \)

\(f(x)= \begin{cases}\frac{1}{30} & 0
Mean =\(E(X)=\int _{ 0 }^{ 30 }{ x3f(x)dx } \) 
= \(\int _{ 0 }^{ 30 }{ x.\frac { 1 }{ 30 } dx } \) 
\(E(X)=\frac { 1 }{ 30 } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 30 }\) 
= \(\frac { 1 }{ 30 } [ \frac{30\times 30}{2}-0]\)  
E(X) = 15 minutes
The average waiting time for the student is 15| minutes.

Given E(X + 3) = 10
E(aX + b) = aE(X) + b
⇒ E(X) + 3 = 10
E(X) + 3 = 10
⇒ E(X) = 7
⇒μ = 7 ...(1)
E(X + 3)² = 116
E(X² + 6x + 9) 116
E(X²) + 6E(X) + 9 = 116 [ஃ E(9) = 9]
E(X²) + 6(7) + 9 = 116
E(X²) + 116 - 42 - 9 116 - 51
E(X²) = 65 ...(2)
Var(X) = E(X²) - [E(X)²]
65 - 7² = 65 - 49 = 16
ஃμ = 7 and σ² = 16.

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 getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

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

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 \(1-\left( \cfrac { 3 }{ 4 } \right) ^{ 10 }\)  

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 \(V(x)=\frac { 4000 }{ 11 } \)

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 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

(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
 
= \(\cfrac { 1 }{ 21 } \int _{ 3 }^{ 4 }{ { x }^{ 2 }dx+0=\cfrac { 1 }{ 21 } \left( \cfrac { { x }^{ 3 } }{ 3 } \right) } _{ 3 }^{ 4 }\) 
= \(\cfrac { 1 }{ 21 } \left( \cfrac { { 4 }^{ 3 }-{ 3 }^{ 3 } }{ 3 } \right) =\cfrac { 37 }{ 63 } \)

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

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

Since X is a discrete random variable, from the given data, X takes on the values
−2, −1, 0, and 1.
For discrete random variable X, by definition, we have f (x) = P(X = x)
Therefore left hand limit of f(x) at x = -2 is F(− 2⁻ ) 
f (−2) = P(X =-2 ) = F(-2 ) - F(- 2^- )= 0.25-0 = 0.25
Similarly for other jump points, we have
f (−1) = P(X = -1) = F(-1) - F(-2) = 0.60 - 0.25 = 0.35.
f (0) = P(X ) 0) = F(0) - F(-1) = 0.90 - 0.60 = 0.30 ,
f (1) = P(X =1) = F(1) - F(0) 1- 0.90 = 0.10 .
Therefore the probability mass function is

The distribution function F(x) has jumps at x = -2, -1, 0, and 1. The jumps are respectively 0.25, 0.35, 0.30, and 0.1 is shown in the figure given below.
These jumps determine the probability mass function

(i) \(P(X<0)=\sum _{ -\infty }^{ -1 }{ P(X=x)=P(X=-1)=0.25+0.35 } =0.60\)
(ii) \(P(X\ge -1)=\sum _{ -1 }^{ 1 }{ P(X=x)=P(x=-1) } +P(X=0)+P(X=1)=0.35+030+0.10=0.75\)

Since X denotes the total score in two throws, it takes on the values 2, 3, 4, 5 and 6. From the Sample space S, we have

\(P(X=2)=\frac { 1 }{ 36 } \), \(P(X=3)=\frac { 4 }{ 36 } \) 
\(P\left( X=4 \right) =\frac { 10 }{ 36 } \) , \(P(X=5)=\frac { 12 }{ 36 } \) and 
\(P(X=6)=\frac { 9 }{ 36 } \)
(i) Probability mass function is

(ii) Cumulative distribution function By definition of the cumulative distribution function for discrete random variable we have
\(f(x)=P(X\le x)=\underset { x_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\)
\(P(X for -∞
\(F(2)=P(X\le 2)=\sum _{ -\infty }^{ 2 }{ P(X=x)=P\left( X \right) <2)+P(X=2) } =0+\frac { 1 }{ 36 } =\frac { 1 }{ 36 } \)
\(F(3)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)=0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(4)=P(X\le 3)=\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( x<2 \right) +P(X=2)+P(X=3)+P(X=4)\)
\(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(5)=P\left( X\le 5 \right) =\sum _{ -\infty }^{ 5 }{ P(X=x) } =P\left( X<2 \right) +P(X=3)+P\left( X=4 \right) +P\left( X=5 \right) \)
= \(0+\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 } \)
\(F(6)=P(X\le 6)=\sum _{ -\infty }^{ 6 }{ P(X=x) } \) 
= \(P(X<2)+P(X=2)+P(X=3)+P(x=4)+P(x=5)P(X=6)\)
 \(0+\frac { 1 }{ 36 } +{ \frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =1 }\) 
Therefore the cumulative distribution function is


(iii) \(P(3\le X\le 6)=\sum _{ x=3 }^{ 5 }{ P(X={ { x }_{ 1 })=P(X=3) }+P(X=4) } +P(X=5)\)
\(=\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } \)
(iv) \(X\ge 4)=\sum _{ x=4 }^{ 5 }{ P(X={ x }_{ 1 }) } \)
 = \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)

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

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)]
= \(1-\left[ ^{18}{ C }_{ 0 }\left( \frac { 1 }{ 3 } \right) ^{ 0 }\left( \frac { 2 }{ 3 } \right) ^{ 18 }+^{ 18}{C }_{ 1 }\left( \frac { 1 }{ 3 } \right) ^{ 1 }\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \)
=  \(1-\left[ \left( \frac { 2 }{ 3 } \right) ^{ 18 }+6\left( \frac { 2 }{ 3 } \right) ^{ 17 } \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left[ \frac { 2 }{ 3 } +6 \right] \) 
= \(1-\left( \frac { 2 }{ 3 } \right) ^{ 17 }\left( \frac { 20 }{ 3 } \right) \)
= \(1-\frac { 20 }{ 3 } \left( \frac { 2 }{ 3 } \right) ^{ 17 }\)

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 \(f(x)=\begin{cases} \begin{matrix} 16{ xe }^{ -4x } & foex>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}\) 
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) \(=\frac{1}{8} \)

Let X b the random variable denotes number of heads when four coins are tossed once.
Then X take the values 0,1,2,3,4.
n(S) = 16

The probability mass function is

Mean
= \(\\ E(x)=E(X)=\Sigma xf(x)\) 
= \(0\left( \frac {4 }{ 16 } \right) +1\left( \frac {24 }{ 16 } \right) +2\left( \frac { 36 }{ 16 } \right) +3\left( \frac { 16}{ 16 } \right) = \frac { 80 }{ 16 } \) 
= \(\frac { 1 }{ 4 } +\frac { 3 }{ 4 } +\frac { 3 }{ 4 } +\frac { 1 }{ 4 } =\frac { 8 }{ 4 } =2\) 
Variance
\(E({ X }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \(0^{ 2 }\left( \frac { 1 }{ 16 } \right) +1^{ 2 }\left( \frac { 1 }{ 4 } \right) +2^{ 2 }\left( \frac { 3 }{ 8 } \right) +3^{ 2 }\left( \frac { 1 }{ 4 } \right) +4^{ 2 }\left( \frac { 1 }{ 16 } \right) \)
\(0\left( \frac {4 }{ 16 } \right) +1\left( \frac {24 }{ 16 } \right) +2\left( \frac { 36 }{ 16 } \right) +3\left( \frac { 16}{ 16 } \right) = \frac { 80 }{ 16 } \) 
\( =5\) 
(X) = E(X²) - [E(X)]²
= 5 - 2² = 5 - 4 = 1

Let X be the random variablc denotes number of red balls.
Then X take the values 0, 1, 2
Sample space = ⁷C₂ = 21
Let X denote the drawing the red ball.
Then X take the values 0, 1, 2
P(X = 0), X^-1 (BB) = ³C₂ = 3
P(X = 1), X^-1 (BR) = ³C₁ x ⁴C₁ = 12
P(X = 2), X^-1 (BR) = ³C₂ = 6

The probability mass function is

Mean :
\(E(x)=\Sigma x.f\left( x \right) \) 
= \(0(\frac { 1 }{ 7 } )+1\left( \frac { 4 }{ 7 } \right) +2\left( \frac { 2 }{ 7 } \right) \) 
= \(\frac { 4 }{ 7 } +\frac { 4 }{ 7 } =\frac { 8 }{ 7 }\)

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
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(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 \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(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)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

(i) Probability mass function
For a discrete random variable we have
f(x) = p(X = x)
\(\therefore f(0)=F(0)=\frac { 1 }{ 2 } \) 
f(1) = F(1) - F(0)
= \(\frac { 3 }{ 5 } -\frac { 1 }{ 2 } =\frac { 6-5 }{ 10 } =\frac { 1 }{ 10 } \) 
f(2) = F(2)-F(1)
= \(\frac { 4 }{ 5 } -\frac { 3 }{ 5 } =\frac { 1 }{ 5 } \) 
f(3) = F(3) - F(2)
\(\frac { 9 }{ 10 } -\frac { 4 }{ 5 } =\frac { 9-8 }{ 10 } =\frac { 1 }{ 10 } \)
f(4) = F(4)-F(3)
= \(1-\frac { 9 }{ 10 } =\frac { 1 }{ 10 } \) 
ஃThe probability mass function is

(ii) p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)
= \(\frac { 1 }{ 2 } +\frac { 1 }{ 10 } +\frac { 1 }{ 5 } =\frac { 5+1+2 }{ 10 } =\frac { 8 }{ 10 } \) 
= \(\frac { 4 }{ 5 } \)
(iii) p(x≥2) = p(x = 2) + p(x = 3) + p(x = 4)
= \(\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2+1+1 }{ 10 } =\frac { 4 }{ 10 } \)
= \(\frac { 2 }{ 5 } \)

Given probability mass function is

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

Given
\(\\ \\ \\ \\ \\ f(x)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 }+1 }{ k } & forx=0,1,2 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\\ \\ \\ \\ \\ \\ \)
The random variable X take the values 0, 1, 2.
Probability mass function.

\(\sum _{ i=0 }^{ 2 }{ f(x_{ i })=1\Rightarrow f(0)+f(1)+f(2)=1 } \)
\(\Rightarrow \frac { 0+1 }{ k } +\frac { 1+1 }{ k } +\frac { 4+1 }{ k } \) 
\(\Rightarrow \frac { 1 }{ k } +\frac { 2 }{ k } +\frac { 5 }{ k } =1\) 
\(\Rightarrow \frac { 8 }{ k } =1\) 
\(\Rightarrow k=8\) 
(ii) Cumulative distribution function

\(F(0)=P(X<0)\\P(x=0)\\ =\frac { 1 }{ 8 } \) 
\(F(1)=P(X = 0)+ P(X = 1)\\ \frac { 1 }{ 8 } +\frac { 2 }{ 8 } =\frac { 3 }{ 8 } \)
\(F(2)=P(X= 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } +\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =1\) 
Cumulative distribution function is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & for & x\le 0 \end{matrix} \\ \begin{matrix} \frac { 2 }{ 8 } +\frac { 1 }{ 8 } & for & \frac { 3 }{ 8 } forx\le 1 \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } +\frac { 5 }{ 8 } =1 & for & x\le 2 \end{matrix} \end{cases}\)
(iii) p(x ≥ 1) = p(x = 1) + p(x = 2)
= \(\frac { 2 }{ 8 } +\frac { 5 }{ 8 } \)
\(p(x\ge 1)=\frac { 7 }{ 8 } \)

Let X be the random variable denotes number of| girl child among 4 children
X = {0, 1, 2, 3, 4}
X =2) (0) {BBBB}
X(1) = {GBBB, BGBB, BBGB, BBBG}
X(2) = {GGBB, BBGG, GBGB, BGBG, BGGB, GBBG}
X(3) = {BGGG, GGGB, GBGG, GGBG}
X(4) = {GGGG}

(i) Probability mass function

(ii) Cumulative distribution function
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)P(X = x_i)
P(X<0) = 0 for -\(\infty\) < x < 0
\(F(0)=\frac { 1 }{ 16 } \) 
\(F(1)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } =\frac { 5 }{ 16 } \) 
\(F(2)=\frac { 5 }{ 16 } +\frac { 3 }{ 8 } =\frac { 5 }{ 16 } +\frac { 6 }{ 16 } =\frac { 11 }{ 16 } \)
\(F(3)=\frac { 11 }{ 6 } +\frac { 1 }{ 4 } =\frac { 11 }{ 16 } +\frac { 4 }{ 16 } =\frac { 15 }{ 16 } \) 
\(F(4)=\frac { 15 }{ 16 } +\frac { 1 }{ 16 } =\frac { 16 }{ 16 } =1\)
\(F(x)=\left\{\begin{array}{lll} \frac{0}{16} & \text { for } & x<0 \\ \frac{1}{16} & \text { for } & x \leq 0 \\ \frac{5}{16} & \text { for } & x \leq 1 \\ \frac{11}{16} & \text { for } & x \leq 2 \\ \frac{15}{16} & \text { for } & x \leq 3 \\ 1 & \text { for } & x \leq 4 \end{array}\right.\)

Let X be the thrown random variable denotes the total in two the thrown a die.
Sample space S

n (S) = 36
X = {2, 4, 6, 8, 10}

\(p(x=2)=\cfrac { 1 }{ 36 } \)
\(p(x=4)=\cfrac { 4 }{ 36 } \)
\(p(x=6)=\cfrac { 10 }{ 36 } \)
\(p(x=8)=\cfrac { 12 }{ 36 } \)
\(p(x=10)=\cfrac { 9 }{ 36 } \)
(i) Probability mass function is

(ii) Cumulative distribution function .
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)(X = x_i)
P(X<2) = 0 for \(\infty\) < x < 2
\(F(2)=\frac { 1 }{ 36 } \)
\(F(4)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 36 } \)
\(F(6)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(8)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 }\)
\(F(10)=\frac { 27 }{ 36 } +\frac { 9 }{ 36 } =\frac { 36 }{ 36 } =1\)
∵ The cumulative distribution function n
\(F(x)=\left\{\begin{array}{lll} 0 & \text { for } & x<2 \\ \frac{1}{36} & \text { for } & x \leq 2 \\ \frac{5}{36} & \text { for } & x \leq 6 \\ \frac{15}{36} & \text { for } & x \leq 8 \\ 1 & \text { for } & x \leq 10 \end{array}\right.\)
(iii) p(4≤ X < 10) = p(x = 4) + p(x = 6) + p(x = 8)
= \(\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } =\frac { 13 }{ 18 } \)  
(iv) p(x ≥ 6) = p(x = 6) + p(x = 8) + p(x = 10)
= \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)
Sample space = {4 childrens}

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

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
Let X be a random variable whose values x are the number of fours.
The sample space S is given in the table.
It can also be written as
S = {(i, j)} , where i = 1, 2, 3, 6 and j = 1, 2, 3, 6
Therefore X takes on the values of 0, 1 and 2.
We observe that
(i) X = 0, if (i,  j) for i ≠ 4, j≠ 4,
(ii) X = 1, if (1, 4), (2, 4), (3, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(iii) X = 2, if (4, 4) ,
Therefore,

The probabilities are
\(f(0)=P(X=0)\cfrac { 25 }{ 36 } \) 
\(f(1)=P(X=1)=\cfrac { 10 }{ 36 } \) 
and \(f(20=P(X=2)=\cfrac { 1 }{ 36 } \)
Clearly the function f(x) satisfies the conditions
(i) f (x) ≥ 0, for x = 0, 1, 2 and
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=-2 }{ f(x) } =f(0)+f(1)+f(2)=1\) 
\(=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=1\)
The probability mass function is presented as

(or) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

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]

= \(\frac { 135 }{ 512 } \)

Let p be the probability of getting heads
q = 1-p
 \(p=\frac { 1 }{ 2 } \)
\(Mean=np=100\times \frac { 1 }{ 2 } =50\) 
\(Variance=npq=100\times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =25\) 
(ii)Let p be the probability of getting 4 when a die is thrown
n = 240 
\(p=\frac { 1 }{ 6 } \) [ஃ 4 appears only one]
\(\therefore Mean=np =240\times \frac { 1 }{ 6 } =40\)
\(Variance=npq= 40\times \frac { 5 }{ 6 } \)  
\(Variance=\frac { 100 }{ 3 } \)

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

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

Given \(f(x)=\begin{cases} \begin{matrix} { 3e }^{ -3x } & x>0 \end{matrix} \\ \begin{matrix} 0 & elsewhere \end{matrix} \end{cases}\)
\(E(X)=\int _{ 0 }^{ \infty }{ x.f(x)dx } =\int _{ 0 }^{ \infty }{ x.3.{ e }^{ -3x }dx } \) 
= \(3\int _{ 0 }^{ \infty }{ x.{ e }^{ -3x }dx } \left[ \therefore \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(3\times \frac { 1! }{ { 3 }^{ 2 } } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
ஃ Expected life of the electronic equipment is = \(\frac { 1 }{ 3 } \) 

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

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

Given
\(f(x)=\cfrac { 4-x }{ 6 } \) 
\(f(x)=\cfrac { 4-1 }{ 6 } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 } \) 
\(f(2)=\cfrac { 4-2 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \) 
\(f(3)=\cfrac { 4-3 }{ 6 } =\cfrac { 1 }{ 6 } \) 
ஃ The probability mass function is

Mean \(E(X)=\Sigma xf(x)\) 
= \(1\left( \cfrac { 1 }{ 2 } \right) +2\left( \cfrac { 1 }{ 3 } \right) +3\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 6 } =\cfrac { 3+4+6 }{ 6 } \) 
= \(\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } =1.67\) 
\(E({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \({ 1 }^{ 2 }\left( \cfrac { 1 }{ 2 } \right) +{ 2 }^{ 2 }\left( \cfrac { 1 }{ 3 } \right) +3^{ 2 }\left( \cfrac { 1 }{ 6 } \right) \) 
= \(\cfrac { 1 }{ 2 } +\cfrac { 4 }{ 9 } +\cfrac { 9 }{ 6 } =\cfrac { 3+8+9 }{ 6 } \) 
= \(\cfrac { 20 }{ 6 } =\cfrac { 10 }{ 3 } =3.33\)
var(X) E(X²) - [E(X)]²
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\) 

\(\therefore Mean=E(x)=\Sigma xf(x)=0\left( \frac { 1 }{ 5 } \right) +1\left( \frac { 1 }{ 5 } \right) +2\left( \frac { 1 }{ 10 } \right) +3\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 1 }{ 5 } \right) +5\left( \frac { 1 }{ 10 } \right) \) 
\(\frac { 1 }{ 5 } +\frac { 1 }{ 5 } +\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { 1 }{ 2 } \) 
= \(\frac { 2+2+6+8+5+ }{ 10 } =\frac { 23 }{ 10 } =2.3\) 
= \(f({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\)
= \({ 0 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 1 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +2^{ 2 }\left( \frac { 1 }{ 10 } \right) +{ 3 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 4 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 5 }^{ 2 }\left( \frac { 1 }{ 10 } \right) \) 
= \(0+\frac { 1 }{ 5 } +\frac { 4 }{ 10 } +\frac { 9 }{ 5 } +\frac { 16 }{ 5 } +\frac { 25 }{ 10 } \) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\)
Variance = E(X²) - [E(X)]²
= 8.1- (2.3)2
= 8.1- 5.29
= 2.81

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(1.2≤X<1.8)
= \(\int _{ 1.2 }^{ 1.8 }{ f(x)dx } =\int _{ 1.2 }^{ 1.8 }{ \left( 2-x \right) } dx\) 
\(=\left[ 2x-\frac { { x }^{ 2 } }{ 2 } \right] _{ 1.2 }^{ 1.8 }\) 
= 3.6- 1.62- 2.4 +  0.72 = 0.3

 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