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

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

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 n = 12
P = 0.9
(i) P(X = 10) = 12C₁₀(0.9)¹⁰(1 - 0.9)²
= 12C₁₀(0.9)¹⁰ (0.1)²
(ii) P(X 2: 11) = R(X = 11) + P(X = 12)
= 12C₁₁(0.9)¹¹ (0.1) + 12C₁₂(0.9)¹²(0.1)⁶
= 12C (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)
(iii) P( at least 2 will not have a useful life of atleast 600 hours) = 1 - P( atleast 11 will have a useful life of atleast 600 hours)
= 1 - P(X ≥11)
= 1-2.1 (0.9)¹¹

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

(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

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

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
\(F(0)=\frac { 1 }{ 16 } \) 
\(F(1)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } =\frac { 5 }{ 16 } \) 
\(F(2)=\frac { 5 }{ 16 } +\frac { 3 }{ 8 } =\frac { 5 }{ 16 } +\frac { 6 }{ 16 } =\frac { 11 }{ 16 } \)
\(F(3)=\frac { 11 }{ 6 } +\frac { 1 }{ 4 } =\frac { 11 }{ 16 } +\frac { 4 }{ 16 } =\frac { 15 }{ 16 } \) 
\(F(4)=\frac { 15 }{ 16 } +\frac { 1 }{ 16 } =\frac { 16 }{ 16 } =1\)
\(F(x)=\left\{\begin{array}{lll} \frac{0}{16} & \text { for } & x<0 \\ \frac{1}{16} & \text { for } & x \leq 0 \\ \frac{5}{16} & \text { for } & x \leq 1 \\ \frac{11}{16} & \text { for } & x \leq 2 \\ \frac{15}{16} & \text { for } & x \leq 3 \\ 1 & \text { for } & x \leq 4 \end{array}\right.\)

 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

(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

\(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.