Discrete random variable	Continuous random variable
1.	Finite number of possible values	Takes any value in the interval
2.	p(xi) ≥ 0 ∀i, \(\sum _{ i=1 }^{ n }{ p({ x }_{ i })=1 } \)	f(x) ≥ 0 ∀x and \(\\ \int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)

\(F(x)=\int _{ -\infty }^{ x }{ f(t)dt } \)
\(=\int _{ 100 }^{ x }{ \frac { 100 }{ { t }^{ 2 } } dt,\quad x\ge 100 } \)
\(={ \left[ \frac { 100 }{ -t } \right] }_{ 100 }^{ x },\quad x\ge 100\)
\(F(x)=\left[ 1-\frac { 100 }{ x } \right] ,\ge 100\)

Let X be the number of cars

X=xi	Number of Household	P(xi)
0	30	0.03
1	320	0.32
2	380	0.38
3	190	0.19
4	80	0.08
Total	1000	1.00

i) P(xi)\(\ge\)0\(\forall \) i and
ii) \(\sum _{ i=1 }^{ \infty }{ P({ x }_{ i })=p(0)+p(1)+p(3)+p(4) } \)
= 0.03+0.32+0.38+0.19+0.08 = 1
Hence p(x_i) is a probability mass function.

Given probability function is
\(p(x)=\left\{\begin{array}{l} \frac{1}{4}, \text { for } x=-2 \\ \frac{1}{4}, \text { for } x=0 \\ \frac{1}{2}, \text { for } x=10 \\ 0, \text { elsewhere } \end{array}\right.\)

X=x	-2	0	10
P(X=x)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{2}\)

P(|X|≤2)=P(-2 =P(X=-2)+(P(X=0)
\(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

We know that,
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \)
\(k\int _{ -\infty }^{ \infty }{ { ke }^{ -|x| }dx=1 } \) 
\(2k\int _{ -\infty }^{ \infty }{ { e }^{ -|x| }dx=1 } \)   (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an function)
\(2k\int _{ 0 }^{ \infty }{ { \left[ \frac { { e }^{ -x } }{ -1 } \right] }_{ 0 }^{ \infty } } =1\)
\(k=\frac { 1 }{ 2 } \)
Mean of the random variable is
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(E(X)=\int _{ -\infty }^{ \infty }{ xk{ e }^{ -|x| }dx } \)    (\(\because { xe }^{ -|x| }\) is an odd function of x)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { xe }^{ -|x| } } \)
= 0
\(E\left( { x }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } f(x)dx\)
\(=\int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { ke }^{ -|x| }dx\)
\(=\frac { 1 }{ 2 } \int _{ -\infty }^{ \infty }{ { x }^{ 2 } } { e }^{ -|x| }dx\)
\(=\int _{ 0 }^{ \infty }{ { x }^{ 2 } } { e }^{ -x }\)  (\(\because { x }^{ 2 }{ e }^{ -|x| }\) is an even function)
\(=\Gamma 3\left( \because \Gamma \left( \alpha \right) =\int _{ 0 }^{ \infty }{ { x }^{ \alpha -1 } } { e }^{ -x }dx,\alpha >0;\Gamma n=(n-1)! \right) \)
= 2
\(V(X)=E\left( { x }^{ 2 } \right) -{ \left[ E(X) \right] }^{ 2 }\)
\(=2-{ \left[ 0 \right] }^{ 2 }\)
= 2

We have F(x) = f(x) ≥ 0, where F(x) is the distribution function and f(x) is the probability density function.
Here F(x) = 0 for x ≤ 1  f(x) = 0 for x ≤ 1 
Again F(x) = 1 for x > 3 
f(x) = d/dx (1) = 0 for x > 3 
In 1 < x ≤ 3, F(x) = k(x – 1)4 
f(x) = d/dx (k(x – 1)4) = 4k(x – 1)3 
\(\therefore f(x)=4k{ (x-1) }^{ 3 }for\quad 1\le x\le 3\)
i) Since f(x) is a probability density function,
\(\int _{ 1 }^{ 3 }{ f(x)dx=1 } \)
\(\Rightarrow \int _{ 1 }^{ 3 }{ { 4k(x-1) }^{ 3 }dx=1 } \)

\(\Rightarrow k[{ (3-1) }^{ 4 }-{ (0) }^{ 4 }]=1\)
\(\Rightarrow k({ 2 }^{ 4 })=1\Rightarrow k(16)=1\)
\(\Rightarrow k=\frac { 1 }{ 16 } \)
ii) \(\therefore\)p.d.f
\(f(x)=\frac { 4\times 1 }{ 16 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)
\(f(x)=\frac { 1 }{ 4 } { (x-1) }^{ 3 }for\quad 1\le x\le 3\)

(a) We know that
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\)
\(\int _{ 0 }^{ 10 }{ Axdx } +\int _{ 10 }^{ 20 }{ A(20-x)dx=1 } \)
\(A\left\{ { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }+{ \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 } \right\} =1\)
A[(50-0)+(400-200)-(200-50)] = 1
\(A=\frac{1}{100}\)
(b) (i) The probability that the number of pounds of bread that will be sold tomorrow is more than 10 pounds is given by
\(P(10\le X\le 20)=\int _{ 10 }^{ 20 }{ \frac { 1 }{ 100 } (20-x) } dx\)
\(=\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 20 }\)
\(=\frac { 1 }{ 100 } [(400-200)-(200-50)]\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is less than 10 pounds, is given by
\(P(0\le X\le 20)=\int _{ 0 }^{ 10 }{ \frac { 1 }{ 100 } } xdx\)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 10 }\)
\(=\frac { 1 }{ 100 } (50-0)\)
= 0.5
(ii) The probability that the number of pounds of bread that will be sold tomorrow is between 5 and 15 pounds is
\(P(5\le X \le15)=\int _{ 5 }^{ 10 }{ \frac { 1 }{ 100 } xdx } +\int _{ 10 }^{ 15 }{ \frac { 1 }{ 100 } (20-x)dx } \)
\(=\frac { 1 }{ 100 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 5 }^{ 10 }+\frac { 1 }{ 100 } { \left[ 20x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 10 }^{ 15 }\)
= 0.75

(i) Since the condition of probability mass function is
\(\sum _{ i=1 }^{ \infty }{ p({ x }_{ i }) } =1\)
\(\therefore\) \(\sum _{ i=0 }^{ 7 }{ p({ x }_{ i }) } =1\)
0+a+2a+2a+3a+a²+2a²+7a²+a = 1
10a²+9a–1 = 0
(10a–1)(a+1) = 0
a = \(\frac{1}{10}\)and -1
Since p(x) cannot be negative, a = – 1 is not applicable. Hence, a = \(\frac{1}{10}\)
ii) P(X<3) = P(X = 0)+P(X=1)+P(X = 2)
= 0+a+2a
= 3a
\(\\ =\frac { 3 }{ 10 } \left( \because a=\frac { 1 }{ 10 } \right) \)
(iii) P(X>2) = 1-P(X\(\le\)2)
= 1-[P(X = 0)+P(X=1)+P(X=2)
= 1-\(\frac{3}{10}\)
= \(\frac{7}{10}\)
iv) P(2< x\(\le\)5) = P(X = 3)+P(X = 4)+P(X = 5)
= 2a+3a+a²
= 5a+a²
= \(\frac{5}{10}+\frac{1}{100}\)
\(=\frac{51}{100}\)

Given probability distribution function is 
X=x 1 3 5 P(X=x) 2k 3k 4k a) Since the given function is a p.d.f. each p_i≥0 and Σp_i=1
⇒2k + 3k+4k = 1 ⇒ 9k = 1 ⇒ k = \(\frac{1}{9}\)
b) P(X>2) = P(X = 3) + P(X = 5)
= 3k + 4k = 7k = 7(\(\frac{1}{9}\))
\(\left[ \because k=\frac { 1 }{ 9 } \right] \)
\(\therefore P(X\ge 2)=\frac { 7 }{ 9 } \)

When a coin is tossed, sample space S = {H, T}
Since he is receiving Rs. 4 for a head and pays Rs. 2 for a tail,
∴ X take values 4 and -2.
∴ Probability for getting a head is \(\frac{1}{2}\) and probability for getting a tail is \(\frac{1}{2}\).
The probability mass function is

X = x	4	-2
P(X = x)	\(\frac{1}{2}\)	\(\frac{1}{2}\)

∴ Expectation E(X) = Σxp(x)
= 4(\(\frac{1}{2}\)) - 2(\(\frac{1}{2}\))
= 2-1 = 1
∴ His expectation is Rs. 1
E(x²) = ∑x²p(x)
= 4²(\(\frac{1}{2}\)) + (-2)²(\(\frac{1}{2}\))
= 16(\(\frac{1}{2}\)) + 4(\(\frac{1}{2}\))
= 8 + 2 = 10
Variance (X) = E(X²)-[E(X)]²
= 10-1² = 9
∴ Variance of his gains Rs. 9

Expected value of the random variable is
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\)
\(=\int _{ 0 }^{ 25 }{ x \frac{1}{25}dx } \)
\(=\frac { 1 }{ 25 } \int _{ 0 }^{ 25 }{ xdx } \)
\(=\frac { 1 }{ 25 } { \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 25 }\)
= 12.5
Therefore, the expected waiting time of the commuter is 12.5 minutes.

Probability of getting a blue ball = (p) =\(\frac{1}{4}\) = 0.25
Total experiments (N) = 30
Expected value = Number of experiments × Probability
= N x p
= 30 x 0.25
= 7.50
Therefore, the expected value of getting blue ball is approximately 8.

Given probability distribution function is
\(= \begin{cases}kx & x =2, 4, 6 \\ k(x - 2), & x = 8 \\ 0, & \text { otherwise } \\ \end{cases}\)
where k is a constant

X = x	2	4	6	8
P(X=x)	2k	4k	6k	k(8-2) = 6k

Since the given function is a probability distribution function, each ρ_i>0Σ ρ_i=1
⇒ 2k+4k+6k+6k = 1
⇒ 18 k = 1
⇒ k = \(\frac{1}{18}\)

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ 0 }^{ -\infty }{ ax\quad dx } \Rightarrow a\int _{ 0 }^{ 1 }{ xdx=1 } \)
\(\Rightarrow a{ \left( \frac { { x }^{ 2 } }{ 2 } \right) }^{ 1 }=1\)
\(\Rightarrow \frac { a }{ 2 } (1-0)=1\)
\(\Rightarrow\)a = 2
\(P\left[ x\le \frac { 1 }{ 2 } \right] =\int _{ -\infty }^{ \frac { 1 }{ 2 } }{ f(x)dx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ axdx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 2xdx } \)
\(=\frac { 1 }{ 4 } \)

The sample space (S) = { (HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
X takes the values: 3, 2, 2, 1, 2, 1, 1, and 0

Range of X(Rx)	0	1	2	3
Px(x)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)
Fx(x)	\(\frac{1}{8}\)	\(\frac{4}{8}\)	\(\frac{7}{8}\)	1

Thus, we have