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 ∧ ~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

The possible events of selection are
(i) both balls may be black, or
(ii) one white and one black or
(iii) both are white.
Therefore X is a random variable that take the values,
X (both are black balls) = Rs. 2(30) = Rs. 60
X (one black and one white ball) = Rs. 30 − Rs. 20 = Rs. 10
X (both are white balls) = Rs. 2( − 20) = - Rs. 40
Therefore X takes on the values 60,10, and − 40.

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]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

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

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

A = {a+\(\sqrt5\) b:a,b ∈ z}
Let C = a+\(\sqrt5\) b
B = c+\(\sqrt5\)d∈A
where a, b, c, d ∈ Z
[∵ ac + 5bd∈Z and ad+bc ∈Z]
∴ B = (a+\(\sqrt5\)b).(c+\(\sqrt5\)d)
= ac+\(\sqrt5\)ad+cb\(\sqrt5\) + 5bd
= (ac+5bd)+\(\sqrt5\)(ad+bc)∈A
∴ C.B ∈A∀ a, b, c, d∈Z
∴ Usual multiplicaition is a binary operation on.

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”.

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]
\(\therefore Mean=np =240\times \frac { 1 }{ 6 } =40\)
\(Variance=npq= 40\times \frac { 5 }{ 6 } \)  
\(Variance=\frac { 100 }{ 3 } \)

\(f(x)= \begin{cases}2(x-1) & 1
\(Mean=E(X)=\int _{ 1 }^{ 2 }{ f(x)dx=\int _{ 1 }^{ 2 }{ 2((x-1)dx } } \) 
\( =2\left[\frac{8}{3}-\frac{4}{2}-\frac{1}{3}+\frac{1}{2}\right] \)
\( =2\left(\frac{7}{3}-\frac{3}{2}\right) \)
\( =2 \times \frac{5}{6} \)
\( =\frac{5}{3} \)
\(E({ x }^{ 2 })=\int _{ 1 }^{ 2 }{ { x }^{ 2 }f(x)dx } \) 
= \(\int _{ 1 }^{ 2 }{ { x }^{ 2 }.2\left( x-1 \right) } dx\) 
= \(2\int _{ 1 }^{ 2 }{ ({ x }^{ 3 }-{ x }^{ 2 })dx } \) 
= \(2\left[ \frac { { x }^{ 4 } }{ 4 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\) 
= \(2\left[ \left( 4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 4 } -\frac { 1 }{ 3 } \right) \right] \) 
= \(2\left[ \frac { 4 }{ 3 } +\frac { 1 }{ 12 } \right] =2\left[ \frac { 16+1 }{ 12 } \right] \) 
= \(\frac { 17 }{ 6 } \) 
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 M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and ∗ be the matrix multiplication.
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)and 
B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)∈M
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & xy \\ 2xy & 2xy \end{matrix} \right) \in M\\ \)
[∵ 2xy∈R-{0}]
∴ M is closed under M
Identity:
Since identity of 2\(\times\)2 matrices is I =\(\left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \)∉M 
∴ M has no identity under *.
Inverse:
Since it has no identity, it won't have inverse also.

(( p V q)∧ ~p)) ➝ q

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

The statement (( p V q)∧ ~p) ➝ q is a tautology.

(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.

Since f (x) is a probability density function, f (x) ≥ 0 and \(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
That is \(\int _{ -\infty }^{ 1 }{ 0dx } +\int _{ 1 }^{ 5 }{ kdx } +\int _{ 5 }^{ \infty }{ 0dx } =1\) 
\(0+k\left( x \right) _{ 1 }^{ 5 }+0=1\Rightarrow 4k=1\Rightarrow k=\frac { 1 }{ 4 } \) 
Therefore the probability density function is
\(f\left( x \right) =\begin{cases} \begin{matrix} \frac { 1 }{ 4 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 0 & Otherwise \end{matrix} \end{cases}\)
(i) Distribution function
The distribution function 
\(F(x)=P\left( X\le x \right) =\int _{ -\infty }^{ x }{ f(u)dx } \) 
When x < 1, \(F(x)=\int _{ -\infty }^{ x }{ f(u)du } =\int _{ -\infty }^{ x }{ oldu } =0\) 
When 1 ≤ x ≤ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u)du=\int _{ -\infty }^{ x }{ 0du } +\int _{ 1 }^{ x }{ odu } +\int _{ 1 }^{ x }{ \frac { 1 }{ 4 } du } =\frac { 1 }{ 4 } (x-1) } \) 
When x ≥ 5 \(F(x)=\int _{ -\infty }^{ x }{ f(u) } du=\int _{ -\infty }^{ x }{ odu } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 1 }^{ 5 }{ \frac { 1 }{ 4 } du } +\int _{ 5 }^{ 5 }{ odu } =1\) 
Thus \(F(x)=\begin{cases} \begin{matrix} 0 & x<1 \end{matrix} \\ \begin{matrix} \frac { x-1 }{ 1 } & 1\le x\le 5 \end{matrix} \\ \begin{matrix} 1 & x>5 \end{matrix} \end{cases}\) 
(ii) P(X < 3) = P(X ≤ 3) = F(3) = \(\frac { 3-1 }{ 2 } =\frac { 1 }{ 2 } \) (Since F(x) is continuous)
(iii) P(2 < X < 4) = P(2 ≤ X ≤ 4) F(4) - F(2) = \(\frac { 3 }{ 4 } -\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \) 
(iv) P(3 ≤ X ) = P(X ≥ 3) = 1− P(X < 3) = 1 - \(1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

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

Since X is a discrete random variable, from the given data, X takes on the values
−2, −1, 0, and 1.
For discrete random variable X, by definition, we have f (x) = P(X = x)
Therefore left hand limit of f(x) at x = -2 is F(− 2⁻ ) 
f (−2) = P(X =-2 ) = F(-2 ) - F(- 2^- )= 0.25-0 = 0.25
Similarly for other jump points, we have
f (−1) = P(X = -1) = F(-1) - F(-2) = 0.60 - 0.25 = 0.35.
f (0) = P(X ) 0) = F(0) - F(-1) = 0.90 - 0.60 = 0.30 ,
f (1) = P(X =1) = F(1) - F(0) 1- 0.90 = 0.10 .
Therefore the probability mass function is

x	-2	-1	0	1
f(x)	0.25	0.35	0.30	0.10

The distribution function F(x) has jumps at x = -2, -1, 0, and 1. The jumps are respectively 0.25, 0.35, 0.30, and 0.1 is shown in the figure given below.
These jumps determine the probability mass function

(i) \(P(X<0)=\sum _{ -\infty }^{ -1 }{ P(X=x)=P(X=-1)=0.25+0.35 } =0.60\)
(ii) \(P(X\ge -1)=\sum _{ -1 }^{ 1 }{ P(X=x)=P(x=-1) } +P(X=0)+P(X=1)=0.35+030+0.10=0.75\)