Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

n (S) = 36
X = {4,5,6,7,8}
From the sample space

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.

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

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

(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)
= \(\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2+1+1 }{ 10 } =\frac { 4 }{ 10 } \)
= \(\frac { 2 }{ 5 } \)

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

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

(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)
= \(\left( 1-\frac { \left( 3-2.5 \right) ^{ 2 } }{ 2 } \right) -\left( \frac { \left( 1.5-1 \right) ^{ 2 } }{ 2 } \right) \) 
= \(\cfrac { 1.75-0.25 }{ 2 } =0.75\) 
\(P\left( 1.5\le X\le \right) =\int _{ 1.5 }^{ 2.5 }{ f(x)dx } =\int _{ 1.5 }^{ 2 }{ (x-1) } dx+\int _{ 2 }^{ 2.5 }{ (-x+3) } dx=0.75\) 

Let X denote the winning amount. The possible events of selection are 
(i) both balls are black, or
(ii) one white and one black or
(iii) both are white 
Therefore X is a random variable that can be defined as
X (both are black balls) = Rs. 2(20) = Rs. 40
X (one black and one white ball) = Rs. 20 − Rs. 10 = Rs. 10
X (both are white balls) = (Rs. 20) = - Rs. 20
Therefore X takes on the values 40,10 and −20
Total number of balls n = 12
Total number of ways of selecting 2 balls = \(\left( \begin{matrix} 12 \\ 2 \end{matrix} \right) =\frac { 12\times 11 }{ 1\times 2 } =66\)
Number of ways of selecting 2 black balls =  \(\left( \begin{matrix} 4 \\ 2 \end{matrix} \right) =6\) 
Number of ways of selecting one black ball and one white ball = \(\left( \begin{matrix} 8 \\ 1 \end{matrix} \right) \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) =32\) 
Number of ways of selecting 2 white balls = \(\left( \begin{matrix} 8 \\ 2 \end{matrix} \right) =28\)

Probability mass function is

Mean :
\(E(X)\Sigma xf(x)=40.\left( \frac { 6 }{ 66 } \right) +10.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) .\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
That is expected winning amount is 0
 Variance :
\(\Sigma x^{ 2 }=\Sigma { x }^{ 2 }f(x)=40^{ 2 }.\left( \frac { 6 }{ 66 } \right) +10^{ 2 }.\left( \frac { 32 }{ 66 } \right) +\left( -20 \right) ^{ 2 }.\left( \frac { 28 }{ 66 } \right) =\frac { 4000 }{ 11 } \)
 (E(X )² = 0² = 0
This gives  \(V(X)=E({ X }^{ 2 })-\left( E(X))^{ 2 } \right) =\frac { 4000 }{ 11 } -0=\frac { 4000 }{ 11 } \)
Therefore E(X ) = 0 and \(V(x)=\frac { 4000 }{ 11 } \)

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

Also the above result can be proved without using truth tables. This proof will be provided after studying the logical equivalence

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
= \(\cfrac { 10 }{ 3 } -\left( \cfrac { 5 }{ 3 } \right) ^{ 2 }=\cfrac { 10 }{ 3 } -\cfrac { 25 }{ 9 } \) 
= \(\cfrac { 30-25 }{ 9 } =\cfrac { 5 }{ 9 } =0.54\) 

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

Let p be the probability of getting heads
q = 1-p
 \(p=\frac { 1 }{ 2 } \)
\(Mean=np=100\times \frac { 1 }{ 2 } =50\) 
\(Variance=npq=100\times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =25\) 
(ii)Let p be the probability of getting 4 when a die is thrown
n = 240 
\(p=\frac { 1 }{ 6 } \) [ஃ 4 appears only one]
\(\therefore Mean=np =240\times \frac { 1 }{ 6 } =40\)
\(Variance=npq= 40\times \frac { 5 }{ 6 } \)  
\(Variance=\frac { 100 }{ 3 } \)

(1) ¬p: It is not cold.
(2) p ∧ q: It is cold and raining.
(3) p ∨ q: It is cold or raining.
(4) q ∨ ¬p: It is raining or it is not cold
Observe that the statement formula ¬ p has only 1 variable p and its truth table has 2 = ( 2¹ ) rows. Each of the statement formulae p ∧ q and p ∨ q has two variables p and q. The truth table corresponding to each of them has 4 = (2² ) rows. In general, it follows that if a statement formula involves n variables, then its truth table will contain 2ⁿ rows.

(( p ∧ q) ∨ (¬r ∨¬s)) ∧ (¬ t ∧ v)) contains 6 variables p, q, r, s, t, and v. Hence the corresponding truth table will contain 2⁶ = 64 rows.

Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

Given A = {a, b, c, d} and * is defined as follows.

1) Commulative Property:
\(a * b=c\)  but  \(b \times a=d \Rightarrow a * b \neq b * a\)
\(a * c=b\) but \(c * a=c \Rightarrow a * c \neq c * a\) 
\(\therefore\) \(\text { * }\) is not commutative on A..
2) Associative Property: 
\((a+b) * c=c * c=a\) 
\(a *(b * c)=a * b=c\) 
\(\therefore(a * b) * c \neq a *(b * c)\) 
\(\text { * }\) is not associative on A.