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

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

\( \mathrm{n}=10, \mathrm{p}=\frac{1}{5}, \mathrm{k}=4 \)
\( \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{p}^{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}}, \mathrm{x}=0,1,2, \ldots, \mathrm{n} \)
\( \mathrm{p}=\frac{1}{5} \)
\( \mathrm{q}=1-\mathrm{p}=1-\frac{1}{5}=\frac{4}{5} \)
\( \mathrm{P}(\mathrm{X}=4)={ }^{10} \mathrm{C}_{4}\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)
\( =210 \times\left(\frac{1}{5}\right)^{4}\left(\frac{4}{5}\right)^{6} \)
\( =210 \times \frac{4^{6}}{5^{10}}\)

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) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

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}

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

\(\)\(Mean=E(X)=\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ 2((x-1)dx } }\) 
\(=2\left[\frac{8}{3}-\frac{4}{2}-\frac{1}{3}+\frac{1}{2}\right]\)
\(=2\left(\frac{7}{3}-\frac{3}{2}\right)\)
\(=2 \times \frac{5}{6}\)
\(=\frac{5}{3}\)
\(E({ x }^{ 2 })=\int _{ 1 }^{ 2 }{ { x }^{ 2 }f(x)dx }\) 
= \(\int _{ 1 }^{ 2 }{ { x }^{ 2 }.2\left( x-1 \right) } dx\) 
= \(2\int _{ 1 }^{ 2 }{ ({ x }^{ 3 }-{ x }^{ 2 })dx }\) 
= \(2\left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\) 
= \(2\left[ \left( 4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right) \right]\) 
= \(2\left[ \frac { 4 }{ 3 } +\frac { 1 }{ 12 } \right] =2\left[ \frac { 16+1 }{ 12 } \right]\) 
= \(\frac { 17 }{ 6 }\) 
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
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\)