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 absolute error = 2%
⇒ \(\frac { dl }{ l } =2 \% =\frac { 2 }{ 100 } =0.02\)
Given T = 2ㅠ\(\sqrt { \frac { 1 }{ g } } \)
Taking logarithm on both sides,
log T = log 2ㅠ + \(\frac12\) log l - \(\frac12\) log g
Taking differential on both sides we get,
\(\frac{1}{T}dT=0+\frac{1}{2}.\frac{1}{l}.dl\)
\(\frac { \Delta T }{ T } =\frac { 1 }{ 2 } (.02)\)
\(\frac { \Delta T }{ T } =0\)
ஃ Percentage error = \(\frac { \Delta T }{ T } \times100=.01\times100=1 \%\)

Let X be the random variable denotes the amount win in the lottery
Values of the random variable -2   48   98 198  Total No. of elements in inverse images 589  6 4 1 600 Probability mass function
x -2 48 98 198 f(x) \(\\ \cfrac { 589 }{ 600 } \) \(\\ \cfrac { 6 }{ 600 } \) \(\\ \cfrac { 4}{ 600 } \) \(\\ \cfrac { 1}{ 600 } \) \(E(X)=-2\times \cfrac { 589 }{ 600 } +48\times \cfrac { 6 }{ 600 } +98\times \cfrac { 4 }{ 600 } +198\times \cfrac { 1 }{ 600 } \) 
= - 0.5
Loss = Rs. 0.50

\(S=\left\{\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right\}\)
(i) The sample space
S = {1, 2, 3, 4, 5, 6}\(\times\){1, 2, 3, 4, 5, 6}
consists of 36 ordered pairs (α, β) where α and β can take any integer value between 1 and 6 as shown. X is assigned to each point (α, β) the sum of the numbers on the dice .
That is X (α, β) = α + β
Therefore
X (1,1) = 1+1 = 2
X (1, 2) = X (2,1) = 3
X (1,3) = X (2,2) = X (3,1)= 4
X (1, 4) = X (2,3) = X (3, 2)  X (4,1) = 5
X (1,5) = X (2,4) = X (3,3) = X (4, 2) = X (5,1) = 6
X (1,6) = X (2,5) = X (3, 4) = X (4,3 = X (5, 2) X (6,1) = 7
X (2,6) = X (3,5) = X (4,4) = X (5,3) = X (6,2) = 8
X (3,6) = X (4,5) = X (5,4)  X (6,3) = 9
X (4,6) = X (5,5) X (6,4) = 10
X (5,6) = (6,5) = 11
X (6,6) = 12
(ii) Then the random variable X takes on the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
(iii) The inverse images of 10 is {(4, 6), (5, 5), (6, 4)}.
(iv) The number of inverse images are given below

Values of the random variable	2	3	4	5	6	7	8	9	10	11	12	Total
Number of elements in inverse image	1	2	3	4	5	6	5	4	3	2	1	36

The sample space S = {H,T} \(\times\) {H,T}
That is S = {TT, TH, HT, HH}
Let X be the random variable denoting the number of heads.
Therefore
X (TT ) = 0 , X (TH ) = 1,
X (HT) = 1, and X (HH) = 2 .
Then the random variable X takes on the values 0, 1 and 2

Values of the Random Variable	0	1	2	Total
Number of elements in inverse images	1	2	1	4

The probabilities are given by

\(f(0)=P(X=0)=\cfrac { 1 }{ 4 } \) 
\(f(1)=P(X=1)=\cfrac { 1 }{ 2 } \) 
and \(f(2)=P(X=2)=\cfrac { 1 }{ 4 } \) 
The function f (x) satisfies the conditions
(i) f (x) ≥ 0 , for x = 0, 1, 2
(ii) \(\underset { x }{ \Sigma } f(x)=\sum _{ x=0 }^{ x=2 }{ f(x) } =f(0)+f(1)+f(2)\) 
= \(\cfrac { 1 }{ 4 } +{ \cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } =1 }\) 
Therefore f (x) is a probability mass function.
The probability mass function is given by

x	0	1	2
f(x)	\(\cfrac { 1 }{ 4 } \)	\(\cfrac { 1 }{ 2 } \)	\(\cfrac { 1 }{ 4 } \)

(or)
\(f(x)\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & forx=0 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 2 } & forx=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 4 } & forx=2 \end{matrix} \end{cases}\)

(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

​​​​​​Truth Table for ~(p ∧ ~q)

p	q	~q	p ∧ ~q	​​​​​~(p ∧ ~q)
T	T	F	F	T
T	F	T	T	F
F	T	T	F	T
F	F	T	F	T

Truth Table for ( p V q) ∧ ~q

p	q	p V q	~q	( p V q) ∧ ~q
T	T	T	F	T
T	F	T	T	T
F	T	T	F	T
F	F	F	T	T

Truth Table for (~p ⟶ r) ∧ ( p ↔️ q)

p	q	r	~ p	~ p ⟶ r	p ↔️ q	(~p ⟶ r) ∧ ( p ↔️ q)
T	T	T	F	T	T	T
T	T	F	F	T	T	T
T	F	T	F	T	F	F
T	F	F	F	T	F	F
F	T	T	T	T	F	F
F	T	F	T	F	F	F
F	F	T	T	T	T	T
F	F	F	T	F	F	F

p	q	q ➝ p	~p	~q	~q ➝ ~p
T	T	T	F	F	T
T	F	T	F	T	T
F	T	F	T	F	F
F	F	T	T	T	T

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

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 probability mass function is

x	0	1	2	3	4
f(x)	\(\frac{1}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)	\(\frac{2}{36}\)	\(\frac{3}{36}\)

(i) Since f(x) is a probability mass function.
\(\sum _{ i=1 }^{ 5 }{ f({ x }_{ i }) } =1\) 
⇒ k² + 2k² + 3k² + 2k + 3k = 1
⇒ 6k² + 5k = 1
⇒ 6k² + 5k - 1 = 0 
⇒ (k + 1) (6k - 1) = 0
⇒ k = -1 or  ⇒ \(k=\frac { 1 }{ 6 } \) 
⇒ \(k=\frac { 1 }{ 6 } \)
(ii) p(2 ≤ x < 5)
= p(x = 2) + p(x = 3) + p(x = 4)
= 2k² + 3k² + 2k = 5k² + 2k
= \(5\left( \frac { 1 }{ 36 } \right) +2\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 36 } +\frac { 1 }{ 3 } =\frac { 5+12 }{ 36 } \) 
= \(\frac { 17 }{ 36 } \) 
(iii) p(3 < x) = p(x > 3)
= p(x = 4) + p(x = 5)
= 2k + 3k = 5k
= \(5\left( \frac { 1 }{ 6 } \right) \) 
= \(\frac { 5 }{ 6 } \)

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

Let X b the random variable denotes number of heads when four coins are tossed once.
Then X take the values 0,1,2,3,4.
n(S) = 16

Values of random variable	0	1	2	3	4	Total
Number of elem in inverse image	1	4	6	4	1	16

The probability mass function is

x	0	1	2	3	4
f(x)	\(\cfrac { 1 }{ 16 } \)	\(\cfrac { 4 }{ 16 } \)	\(\cfrac { 6 }{ 16 } \)	\(\cfrac { 4 }{ 16 } \)	\(\cfrac { 1 }{ 16 } \)

Mean
= \(\\ E(x)=E(X)=\Sigma xf(x)\) 
= \(0\left( \frac {4 }{ 16 } \right) +1\left( \frac {24 }{ 16 } \right) +2\left( \frac { 36 }{ 16 } \right) +3\left( \frac { 16}{ 16 } \right) = \frac { 80 }{ 16 } \) 
= \(\frac { 1 }{ 4 } +\frac { 3 }{ 4 } +\frac { 3 }{ 4 } +\frac { 1 }{ 4 } =\frac { 8 }{ 4 } =2\) 
Variance
\(E({ X }^{ 2 })=\Sigma { x }^{ 2 }f(x)\) 
= \(0^{ 2 }\left( \frac { 1 }{ 16 } \right) +1^{ 2 }\left( \frac { 1 }{ 4 } \right) +2^{ 2 }\left( \frac { 3 }{ 8 } \right) +3^{ 2 }\left( \frac { 1 }{ 4 } \right) +4^{ 2 }\left( \frac { 1 }{ 16 } \right) \)
\(0\left( \frac {4 }{ 16 } \right) +1\left( \frac {24 }{ 16 } \right) +2\left( \frac { 36 }{ 16 } \right) +3\left( \frac { 16}{ 16 } \right) = \frac { 80 }{ 16 } \) 
\( =5\) 
(X) = E(X²) - [E(X)]²
= 5 - 2² = 5 - 4 = 1

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

Values of the Random Variable	2	3	4	5	6	Total
Number of elements in inverse images	1	4	10	12	9	36

\(P(X=2)=\frac { 1 }{ 36 } \), \(P(X=3)=\frac { 4 }{ 36 } \) 
\(P\left( X=4 \right) =\frac { 10 }{ 36 } \) , \(P(X=5)=\frac { 12 }{ 36 } \) and 
\(P(X=6)=\frac { 9 }{ 36 } \)
(i) Probability mass function is

x	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12}{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

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

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

It is known that Z₅ = {[0], [1], [2], [3], [4]}. The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0], [1], [2], [3], [4]}.

+5	0	1	2	3	4
0	0	1	2	3	4
1	1	2	3	4	0
2	2	3	4	0	1
3	3	4	0	1	2
4	4	0	1	2	3

(i) Since each box in the table is filled by exactly one element of Z₅, the output a +₅ b is unique and hence +₅ is a binary operation.
(ii) The entries are symmetrically placed with respect to the main diagonal. So +₅ has commutative property
(iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual
For instance, (2+₅3)+₅ 4 = 0+₅ 4 = 4(mod 5)
and 2+₅(3+₅4) = 2 +₅ 2 = 4(mod5)
Hence (2+₅3)+₅4 = 2+₅(3+₅4)
Proceeding like this one can verify this for all possible triples and ultimately it can be shown that +₅ is associative
(iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.
(v) The existence of inverse is guaranteed provided the identity 0 exists in each row and each column. From Table, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z₅, say 2 is outlined below.
First find the position of the identity element 0 in the III row headed by 2. Move horizontally along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2+₅ 5+ = 0 (mod5). In this way, the inverse of each and every element of Z₅ can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2, and, that of 4 is 1.

The table for the operation x₁₁ is as follows.

x11	1	3	4	5	9
1	1	3	4	5	9
3	3	9	1	4	5
4	4	1	5	9	3
5	5	4	9	3	1
9	9	5	3	1	4

Following the same kind of procedure as explained in the previous example, a brief outline of the process of verification of the properties of ×₁₁ on A is given below.
(i) Since each box has an unique element of A, ×₁₁ is a binary operation on A.
(ii) The entries are symmetrical about the main diagonal. Hence ×₁₁ has commutative property.
(iii) As usual, the associative property can be seen to be true.
(iv) The entries of both the row and column headed by the element 1 are identical. Hence 1 is the identity element.
(v) Since the identity 1 exists in each row and each column, the existence of inverse property is assured for ×₁₁. The inverse of 1 is 1, that of 3 is 4, that of 4 is 3, 5 is 9, and, that of 9 is 5.

It can be obtained by using examples 12.15 and 12.16 that
p↔q ≡ (¬ p ∨ q)  ∧ (¬q ∨ p) ... (1)
≡ (¬p∨q)  ∧ ( p∨ ¬q) (by Commutative Law) ... (2)
≡ (¬p ∧ ( p ∨ ¬q)) ∨ (q ∧ ( p ∨ ¬q)) (by Distributive Law)
≡ (¬p ∧ p) ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ (q ∧ ¬q) (by Distributive Law)
≡ F ∨ (¬p ∧ ¬q) ∨ (q ∧ p) ∨ F; (by Complement Law)
≡ (¬p ∧ ¬q) ∨ (q ∧ p) ; (by Identity Law)
≡ ( p ∧ q) ∨ (¬p ∧ ¬q) ; (by Commutative Law)
Finally (1) becomes p ↔️ q ≡ ( p ∧ q) v (ㄱp ∧ ㄱq)

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and * be the matrix multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.

~(p V q) V (~p ∧ q) and ~p

p	q	p V q	~(p ∧ q)	~p	~p ∧ q	~(p V q) V (~p ∧ q)
T	T	T	F	F	F	F
T	F	T	F	F	F	F
F	T	T	F	T	T	T
F	F	F	T	T	F	T

The entries in column (5) and column (7) are identical.
∴ ~(p V q) V (~p ∧ q) and ~p are logically equivalent.

p	q	r	~ q	~q V r	p➝(¬qVr)	~p	~pV(~qVr)
T	T	T	F	T	T	F	T
T	T	F	F	F	F	F	F
T	F	T	T	T	T	F	T
T	F	F	T	T	T	F	T
F	T	T	F	T	T	T	T
F	T	F	F	F	T	T	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

From the table, it is clear that the column of p➝(¬q V ~r) and ~pV(~q V r) are identical
∴  p➝(¬q V ~r) ≡ ~pV(~q V r) 
Hence proved.

Truth table for \(\neg(p \wedge q)\)

P	Q	\(p \wedge q\)	\(\neg(p \wedge q)\)
T	T	T	F
T	F	F	T
F	T	F	T
F	F	F	T

Truth table for \(\neg p \vee \neg q\)

P	Q	\(\neg p\)	\(\neg q\)	\(\neg p \vee \neg q\)
T	T	F	F	F
T	F	F	T	T
F	T	T	F	T
F	F	T	T	T

To prove  \(\neg(p \rightarrow q) \equiv p \wedge \neg q\)
Truth table for \(\neg(p \rightarrow q)\)

p	q	\(p \rightarrow q \)	\(\neg(p \rightarrow q)\)
T	T	T	F
T	F	F	T
F	T	T	F
F	F	T	F

Truth table for \(p \wedge \neg q\)

p	q	\(\neg q\)	\(p \wedge \neg q\)
T	T	F	F
T	F	T	T
F	T	F	F
F	F	T	T

The entries in the column \(\neg(p \rightarrow q) \text { and } p \wedge \neg q\) are identical and they are equivalent.

x	0	1	2	3	4	5
f(x)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 10 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 10 } \)

\(\therefore Mean=E(x)=\Sigma xf(x)=0\left( \frac { 1 }{ 5 } \right) +1\left( \frac { 1 }{ 5 } \right) +2\left( \frac { 1 }{ 10 } \right) +3\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 1 }{ 5 } \right) +5\left( \frac { 1 }{ 10 } \right) \) 
\(\frac { 1 }{ 5 } +\frac { 1 }{ 5 } +\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { 1 }{ 2 } \) 
= \(\frac { 2+2+6+8+5+ }{ 10 } =\frac { 23 }{ 10 } =2.3\) 
= \(f({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\)
= \({ 0 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 1 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +2^{ 2 }\left( \frac { 1 }{ 10 } \right) +{ 3 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 4 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 5 }^{ 2 }\left( \frac { 1 }{ 10 } \right) \) 
= \(0+\frac { 1 }{ 5 } +\frac { 4 }{ 10 } +\frac { 9 }{ 5 } +\frac { 16 }{ 5 } +\frac { 25 }{ 10 } \) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\)
Variance = E(X²) - [E(X)]²
= 8.1- (2.3)2
= 8.1- 5.29
= 2.81

 

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

\(S=\left|\begin{array}{l} (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{array}\right|\)
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) 
\(f(x)=\begin{cases} \begin{matrix} \frac { 25 }{ 36 } & for \ x=0 \end{matrix} \\ \begin{matrix} \frac { 10 }{ 36 } & for \ x=1 \end{matrix} \\ \begin{matrix} \frac { 1 }{ 36 } & for \ x=2 \end{matrix} \end{cases}\)

Then A∨ B =\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\vee \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\vee 1 & 1\vee 1 \\ 1\vee 0 & 1\vee 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)
\(A\wedge B=\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\wedge \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 0\wedge 1 & 1\wedge 1 \\ 1\wedge 0 & 1\wedge 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}\)

Then the four types of conditional statements corresponding to p and q are respectively listed below.
(i) p →q : (conditional statement) “If the number of primes is infinite then Ooty is in Kerala”.
(ii) q → p : (converse statement) “If Ooty is in Kerala then the number of primes is infinite”
(iii) ¬p → ¬q (inverse statement) “If the number of primes is not infinite then Ooty is not in Kerala”.
(iv) ¬q → ¬p (contrapositive statement) “If Ooty is not in Kerala then the number of primes is not infinite”.

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

Given a*b = a + b + ab -7, ∀ a,b ∈R
If a ∈R, b∈R then ab ∈ R
(a*b) = a +b+ ab - 7 ∈R
For example, let 1, 2 ∈ R
(1*2) = 1+2+(1)(2)-7
= 2 ∈ R
* a binary operation on R

[Here a = 3, b = \(\frac{-7}{15}\)]
\(=3-\frac { 7 }{ 15 } -\frac { 21 }{ 15 } -7\)
\(\therefore 3*\left( \frac { -7 }{ 15 } \right) =\frac { -88 }{ 15 } \)

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 \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 1\vee 1 & 0\vee 0 \end{matrix}\begin{matrix} 1\vee 0 & 0\vee 1 \\ 0\vee 1 & 1\vee 0 \\ 0\vee 0 & 1\vee 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 1 \\ 1 & 1 \\ 0 & 1 \end{matrix} \right) \) [∵ a∨b=max(a,b)]

Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\((A\wedge B)\vee C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)