There are 11 letters in the word PROBABILITY.
Since, a letter is selected randomly, therefore the possible outcomes are P, R, O, B, A, I, L, T, Y.
But here, outcomes are not equally likely
as P(letter P) = \(\frac { 1 }{ 11 } ;\) P (letter R) = \(\frac { 1 }{ 11 } \)
P(letter O)= \(\frac { 1 }{ 11 } ;\) P (letter B) = \(\frac { 2 }{ 11 } \)
[\(\because \) B repeated twice in the word]
P(letter A) = \(\frac { 1 }{ 11 } ;\)   P(letter I) = \(\frac { 2 }{ 11 } \)
P(letter L) = \(\frac { 1 }{ 11 } ;\)  P(letter T) = \(\frac { 1 }{ 11 } \)
and  P(letter Y) = \(\frac { 1 }{ 11 } \)
Now, let E be event of getting a vowel. Then outcomes favourable to E are A, I and O.
\(\Rightarrow \)  E={A,I,O}
Now,\(P(E)=P(A)+P(I)+P(O)\)
[by aximatixc approach to probability]
= \(\frac { 1 }{ 11 } +\frac { 2 }{ 11 } +\frac { 1 }{ 11 } =\frac { 4 }{ 11 } \)            

\(P(A\cap B')=P(A)-P(A\cap B)=0.54-0.35=0.19\)

The given statement can be rewritten as 'The necessary and sufficient condition that the integer n is odd, is n² must be odd'.
Let p and q be statements given by
p : The integer n is odd,
and q : The integer n² is odd.
Then, the given statement is 'p if and only if q'.
In order to check its validity, we have to check the validity of the following statements
(i) 'If p, then q'.
(ii) 'If q, then p'.
(i) Validity of 'If p, then q'
The statement 'if p, then q' is given by
'If the integer n is odd, then n² is odd'.
Let n be odd. Then,
n = 2m + 1
n² = (2m + 1)² = n²= 4m (m + 1)+1
n² is odd integer. Thus, n is odd=n² is odd integer.
Hence, 'If p, then q' is true.
(ii) Validity of 'If q, then p'
The statement 'if q, then p' is given by
 'If n is an integer and n² is odd, then n is odd'.
so, let n be an even integer.
Then,   n=2k, for some integer k.
n²=4k²
n² is an even integer
n² is not an odd integer.
Thus, n is not odd.
n² is not odd.
'If q, then p' is true.
Hence,'p if and only if q' is true
So, the given statement is valid.

\(\lim_{ h\rightarrow 0 } \frac { (a+h)^{ 2 }sin(a+h)-a^{ 2 }sin\quad a }{ h } \)
\(=\lim_ { lim }{ h\rightarrow 0 } \frac { (a^{ 2 }+h^{ 2 }+2ah)[sin \ a\ cos \ h+cos\ a\ sin \ h)-a^{ 2 }sin\quad a }{ h } \)
\(\left[ \because \ sin(C+D)=sinCcosD+CosCsinD \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sin \ a(cos \ h-1) }{ h } +\frac { a^{ 2 }cos \ a \ sin \ h) }{ h } +(h+2a)(sin \ a \ cos \ h+cos\ a \ sin\ h) \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sina(-2sin^{ 2 }\frac { h }{ 2 } ) }{ \frac { h^{ 2 } }{ 2 } } .\frac { h }{ 2 } \right] +\overset { lim }{ h\rightarrow 0 } \frac { a^{ 2 }cosasinh }{ h } +\lim_{ h\rightarrow 0 } (h+2a)sin(a+h)\)
\(\left[ \because cosm \ h-1=-2sin^{ 2 } h/2\quad and\quad sinacos\quad h+cosasin\quad h=sin(a+h) \right] \)
\(=a^{ 2 }sin\quad a\times 0+a^{ 2 }\quad cosa(1)+2asina \quad \left[ \because \lim_{ x\rightarrow 0 } \frac { sin\quad x }{ x } =1 \right]\)
\( =a^{ 2 }cos \ a+2asin \ a\)

Here a=2,\(r=\frac { 4 }{ 2 } =2\) and n=10
\(\because \ { S }_{ 10 }=\frac { 2({ 2 }^{ 10 }-1) }{ 2-1 } =2({ 2 }^{ 10 }-1)=2(1024-1)=2046\)

The thousands place of the 4-digit number is to be filled with any of the digits from 1 to 9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands place can be filled is 9.The hundreds, tens, and units place can be filled by any of the digits from 0 to 9. However, the digits cannot be repeated in the 4-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and unitsplace is to be filled by the remaining 9 digits.Therefore, there will be as many such 3-digit numbers as there are permutations of 9 different digits taken 3 at a time. Number of such 3-digit numbers  
\(={ }^{9} \mathrm{P}_{3}=\frac{9 !}{(9-3) !}=\frac{9 !}{6 !} \)
\(=\frac{9 \times 8 \times 7 \times 6 !}{6 !}=9 \times 8 \times 7=504\)
Thus, by multiplication principle, the required number of 4-digit numbers is 9 × 504 = 4536

\(Let\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } } =K\)
\(\Rightarrow a=K\sin { A,b=K\sin { B\quad and } } c=K\sin { C } \)
\(Now,\frac { \sin { \left( A-B \right) } }{ \sin { \left( A+B \right) } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } =\frac { { K }^{ 2 }\sin { ^{ 2 }A-{ K }^{ 2 }\sin { ^{ 2 }B } } }{ { K }^{ 2 }\sin { ^{ 2 }A+{ K }^{ 2 }\sin { ^{ 2 }B } } } \)
\( \Rightarrow \quad \quad \frac { \sin { \left( A-B \right) } }{ \sin { \left( A+B \right) } } =\frac { \sin { ^{ 2 }A-\sin { ^{ 2 }B } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { \left( \pi -C \right) } } =\frac { \sin { \left( A+B \right) } \sin { \left( A-B \right) } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \quad \left[ \therefore A+B=\pi -C \right] \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { C } } =\frac { \sin { C\sin { \left( A-B \right) } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \)
\(\Rightarrow \frac { \sin { \left( A-B \right) } }{ \sin { C } } =\frac { \sin { C\sin { \left( A-B \right) } } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } =0\)
\(\Rightarrow \sin { A-B\left[ \frac { 1 }{ \sin { C } } -\frac { \sin { C } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } \right] } =0\)
\(\Rightarrow \sin { \left( A-B \right) =0 } \quad or\quad \frac { 1 }{ \sin { C } } -\frac { \sin { C } }{ \sin { ^{ 2 }A+\sin { ^{ 2 }B } } } =0\)
\(\Rightarrow A=B\quad or\quad \sin { ^{ 2 }A+\sin { ^{ 2 }B } -\sin { ^{ 2 }C } } =0\)
\( \Rightarrow A=B\quad or\quad \frac { { a }^{ 2 } }{ { k }^{ 2 } } +\frac { { b }^{ 2 } }{ { k }^{ 2 } } -\frac { { c }^{ 2 } }{ { k }^{ 2 } } =0\)
\(\Rightarrow A=B\quad or\quad { a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }\)
\(\therefore \) Either the triangle is isosceles or right angled.
Hence proved.

Given A={1,2,3,4}, B = {1,2,3} And C={2,4}
Now, P(A)={  \(\phi\), {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}}  ......(i)
P(B) ={\(\phi\){1},{2},{3},{1,2},{1,3},{1,3},{2,3},{1,2,3}}    ....(ii)
and P(c) ={\(\phi\)),{2},{4},{2,4}}  ...(iii)
Given conditions is X \(\subseteq\) B and X\(\neq\) B and X \(\nsubseteq\) C..
\(\Rightarrow\) X \(\in\) P(B), X \(\neq\) B and X \(\in\) P(c)
\(\therefore\) X ={1},{3},{1,2},{1,3},{2,3}
[using eqs (ii) and (iii)]

It is false statement 
e.g Let P= {1,2} and Q ={2,3,4,5}
Clearly 1 \(\in\) P and P\(\nsubseteq\) Q but 1\(\nsubseteq\) Q.
Thus X \(\in\) P and P \(\nsubseteq\) Q Need not imply that x\(\in\)Q

All rectangles, Rhombus and square are parallelograms because its opposite sides are equal and parallel.
Therefore \({ F }_{ 2 }\subset { F }_{ 1 },{ F }_{ 3 }\subset { F }_{ 1 }\) and \({ F }_{ 4 }\subset { F }_{ 1 }\)
\({ F }_{ 1 }={ F }_{ 2 }\cup { F }_{ 3 }\subset { F }_{ 4 }\)