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

\(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.

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) \(\cfrac { 1 }{ 8 } \) \(\cfrac { 3 }{ 8 } \) \(\cfrac { 3 }{ 8 } \)  \(\cfrac { 1 }{ 8 } \) \(\therefore f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & \begin{matrix} for & x=0,3 \end{matrix} \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } & \begin{matrix} for & x=1,2 \end{matrix} \end{matrix} \end{cases}\)

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

Values of the Random Variable	2	3	4	5	6	Total
Number of elements in inverse images	1	4	10	12	9	36

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

x	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12}{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

(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) \(X\ge 4)=\sum _{ x=4 }^{ 5 }{ P(X={ x }_{ 1 }) } \)
 = \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)

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

Values of random variable	0	1	2	Total
Number of elements in inverseimage	3	12	6	21

The probability mass function is

x	0	1	2
f(x)	\(\cfrac { 1 }{ 7 } \)	\(\cfrac { 4 }{ 7 } \)	\(\cfrac { 2 }{ 7 } \)

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
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(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 probability mass function is

x	0	1	2	3	4
f(x)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)

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

x	0	1	2
f(x)	\(\cfrac { 1 }{ k } \)	\(\cfrac { 2 }{ k } \)	\(\cfrac {5 }{ k } \)

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

x	0	1	2
f(x)	\(\cfrac { 1 }{ 8 } \)	\(\cfrac { 2 }{ 8 } \)	1

\(F(0)=P(X<0)\\P(x=0)\\ =\frac { 1 }{ 8 } \) 
\(F(1)=P(X = 0)+ P(X = 1)\\ \frac { 1 }{ 8 } +\frac { 2 }{ 8 } =\frac { 3 }{ 8 } \)
\(F(2)=P(X= 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } +\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =1\) 
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 thrown random variable denotes the total in two the thrown a die.
Sample space S

I/II	1	3	3	5	5	5
1	2	4	4	6	6	6
3	4	6	6	8	8	8
3	4	6	6	8	8	8
5	6	8	8	10	10	10
5	6	8	8	10	10	10
5	6	8	8	10	10	10

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

Values of the random variable	2	4	6	8	10	Total
No. of elements in inverse images	1	4	10	12	9	36

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

x	2	4	6	8	10
f(x)	\(\\ \cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12 }{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

(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}