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

Values of random variable X	0	1	2	Total
Number of elements in inverse images	325	676	325	1326

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

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

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

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}

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)
= \(\frac { 2 }{ 8 } +\frac { 5 }{ 8 } \)
\(p(x\ge 1)=\frac { 7 }{ 8 } \)

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

x	1	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 30 } \)	\(\cfrac { 2 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 5 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 10 }{ 30 } \)

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

Observe that the given distribution is continuous
By definition \(\mu =E(X)=\int _{ -\infty }^{ \infty }{ xf(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -2x } \right) dx } +\int _{ 0 }^{ \infty }{ x\left( { \lambda e }^{ -\lambda x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ x\left( { e }^{ -\lambda x } \right) dx } \) 
=  \(0+\lambda \left( \frac { 1 }{ { \lambda }^{ 2 } } \right) \) (using Gamma integral for positive integer n,\(\int _{ 0 }^{ \infty }{ { x }^{ n } } { e }^{ -ax }dx=\cfrac { n }{ { a }^{ n+1 } } \))
= \(\frac { 1 }{ \lambda } \) 
Variance :
By definition,\(E\left( { X }^{ 2 } \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f(x) } dx\) (We can also use integration by parts or Bernoulli’s formula) 
= \(\int _{ -\infty }^{ 0 }{ 0\left( \lambda { e }^{ -\lambda x } \right) } dx+\int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( \lambda { e }^{ -2x } \right) } dx\) 
= \(0+\lambda \int _{ 0 }^{ \infty }{ { x }^{ 2 }\left( { e }^{ -2x } \right) dx } \)
 (using Gamma integral for positive integer)
Therefore Var(X ) = E(X² )- E(X )²
= \(\frac { 2 }{ { \lambda }^{ 2 } } -\left( \frac { 1 }{ \lambda } \right) ^{ 2 }=\frac { 1 }{ { \lambda }^{ 2 } } \) 
Hence the mean and variance are respectively \(\frac { 1 }{ \lambda } \)  and \(\frac { 1 }{ { \lambda }^{ 2 } } \)

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

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

 

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

Given \(p=\frac { 3 }{ 4 } \) 
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]

= \(\frac { 135 }{ 512 } \)

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

Values of the Random Variable X	0	1	2	Toatal
Number of elements in inverse images	25	10	1	36

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

x	0	1	2
f(x)	\(\frac { 25 }{ 36 } \)	\(\frac { 10 }{ 36 } \)	\(\frac { 1 }{ 36 } \)

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