since the experiment has exactly the three possible mutually exclusive outcomes A, B and C, they must be exhaustive events.
\(\Rightarrow S=A\cup B\cup C\)
Therefore, by axioms of probability
\(P(A)\ge 0,P(B)\ge P(C)\ge 0\) and
\(P(A\cup B\cup C)=P(A)+P(B)+P(C)=P(S)=1\)
The assignment is permissible because
\(P(A)=\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(B)-1-\frac { 1 }{ \sqrt { 3 } } \ge 0,\quad P(C)-0\ge 0\)
\(P(S)=P(A)+P(B)+P(C)=\frac { 1 }{ \sqrt { 3 } } +1-\frac { 1 }{ \sqrt { 3 } } +0-1\)

9

tens	one's
4	5

The one's place can be filled up in 5 ways using 1,2,3,4,5 and tens place can be filled up in 4 ways.
∴ Number of two digit numbers using the digits 1,2,3,4,5 is 4 x 5 = 20.

Let p(n) = 1 + 3 + 5 + ... + (2n- 1). Therefore P(1) = 1 = 1² is true.
We assume that P(k) = 1 + 3 + 5 +: .. (2k-1) is true for n = k.That is P(k) = k²
We need to prove P(k + 1) = (k+ 1)²
P(k+1) = 1+3+5+...(2(k+1)-1)
\(=\underbrace{1+3+5+7+.....+(2k+1)}+2k+1\)
= P(k) + 2k+1
= k²+2k+1 = (k+1)²
This implies, P(k + 1) is true. Hence, by the principle of a mathematical induction, P(n) is true for all natural numbers.

Note that any such path comprises 6 horizontal unit lengths and 4 vertical unit lengths. This each path consists of 10 unit lengths where 6 are of one kind (horizontal) and 4 are of another kind (vertical).
Thus the total number of paths is \(\frac { 10! }{ 4!\times 6! } \) = 210.

In the letters of the word, ARTICLE, there are three vowels namely A, I, E.
There are 3 even places.
3 vowels can occupy the even places in 3P₃ = 3! ways.
Remaining 4 letters can occupy 4 places in 4! ways.
Hence, total number of ways of arrangement = 4! \(\times\) 3!
=  \(4\times 3\times 2\times 3\times 2\)
=144

(i) Since each question can be answered in 4 ways, the maximum number of different answers
= \(4\times 4\times 4\times 4\times 4={ 4 }^{ 5 }\)
(ii) When each question has more than one correct answer the maximum number of different answers \(=(5 \times 3)^{5}=15^{5}\)

(i)Repetition of digits is allowed

hundreds	tens	unit
4	4	4

The unit place can be filled in 4 ways.
Since repetition is allowed, the tens place and hundreds place can also be filled in 4 ways each.
∴ Total number one-digit numbers = 4 x 4 x 4 = 64
(ii) Repetition of digits is not allowed

hundreds	tens	unit
2	3	4

The unit place can be filled in 4 ways.
Since repetition of digits is not allowed, the tens place can be filled in 3 ways.
Hundreds place can be filled in 2 ways .
∴ Total number of 3-digit numbers without repetition = 4 \(\times\) 3 \(\times\) 2 = 24

Let A, B, and C denote the envelopes and 1, 2, and 3 denote the corresponding letters
The different combination of letters put into the envelopes are shown in the table Let c_i denote the outcomes of the events. Let X be the event of putting the letters into the exactly only one right envelopes Let Y be the event of putting none of the letters into the right envelope 
S = {C₁, C₂, C₃, C₄, C₅, C₆}, n(S) = 6
X = {C₂, C₃, C₆},  n(X) = 3
Y = {C₄, C₅}   n(Y) = 2
P(X) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) P(Y) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

Let A₁, A₂ and A₃ be the event of A, B, C becoming managers of the company respectively. Let X be the event that the office canteen will be improved.
Then, \(P\left(A_1\right)=\frac{5}{10}=0.5 \)
\(P\left(A_2\right)=\frac{3}{10}=0.3 \)
\(P\left(A_3\right)=\frac{2}{10}=0.2 \)
\(P\left(X / A_1\right)=0.4 \)
\(P\left(X / A_2\right)=0.5\)
\(P\left(X / A_3\right)=0.3\)
\(P\left(A_2 / X\right)=\frac{P\left(A_2\right) P\left(X / A_2\right)}{P\left(A_1\right) P\left(X / A_1\right)+P\left(A_2\right) P\left(X / A_2\right)}+P\left(A_3\right) P\left(X / A_3\right)\)
\(=\frac{0.3(0.5)}{0.5(0.4)+0.3(0.5)+0.2(0.3)} \)
\(=\frac{0.15}{0.2+0.15+0.06} \)
\(=\frac{0.15}{0.41}=\frac{15}{41}\)

Let A₁ be the event that the items are produced by Machine-1, A₂ be the event that items are produced by Machine-II.
Let B be the event of drawn a defective item.
Then, \(P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)\)
We have, \(P\left(A_1\right)=0.60, P\left(B / A_1\right)=0.02\)
\(P\left(A_2\right)=0.40, P\left(B / A_2\right)=0.04\)
\(P(B)=P\left(A_1\right) \cdot P\left(B / A_1\right)+P\left(A_2\right) \cdot P\left(B / A_2\right)\)
\(=0.60 \times 0.02+0.40 \times 0.04 \)
\(=0.012+0.016 \)
\(=0.028 \)
\(=\frac{.28}{1000}=\frac{7}{250}\)

(i) There are as many strings as filling the 4 vacant places by the 4 letters, keeping in mind that repetition is not allowed. The first place can be filled in 4 different ways by any one of the letters B, I, R, D. Following which, the second place can be filled in by any one of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways, following which fourth place can be filled in 1 way. Thus the number of ways in which the 4 places can be filled, by the rule of product is 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 24. Hence, the required number of strings is 24.
(ii) There are 5 different letters with which 5 places are to be filled. The first place can be filled in 5 ways as any one of the five letters P, R, I, M, E can be placed there. Having filled the first place with any of the 5 letters, 4 letters are left to be placed in the second place, three letters are left for the third place and 2 letters are left to be put in the fourth place. The remaining 1 letter has to be placed in the fifth place. Hence, the total number of ways filling up five places is 5 \(\times\) 4 \(\times\) 3 \(\times\) 2 \(\times\) 1 = 120.

In the word THING, there are 5 letters
The lexicographic order of the word is G, H, I,N, T
Number of words starting with G = 4! = 24
Number of words starting with H =  4! = 24
Number of words starting with I 4! = 24
Number of words starting with NG 3! = 6
Number of words starting with NGH 2! = 2
Number of words starting with NGHI = 1!
Number of words starting with NGHIT = 1!
85^th string NGHIT 

Let p(n) be the statement 10^{n + 3} \(\times\) 4ⁿ⁺² + 5 is divisible by 9.
Step 1: Putting n = 1, we get,
10¹ + 3 \(\times\) 4^{1 +2} + 5 = 10 + 3(4³) + 5 = 15 + 3 (64)
= 207 is divisible by 9.
∴ p(1) is true
Step 2: Let us assume that p(k) is true.
∴ 10k + 3 x 4^{k+ 2} + 5 is divisible by 9
⇒ 10^k + 3 x 4^{k + 2} + 5 = 9 m
⇒ 10^k = 9 m - 5 - 3 x 4^k+2 for some scalar m.
Step 3: To prove thatp(k+ 1) is true
ie to p. T. 10^k+1+ 3 x 4^k+3 + 5 is divisible by 9
Consider 10^k+1+ 3 x 4^k+3 + 5
⇒ 10^k·10+3.4^k+2·4¹+5
= 10 [9m - 5 - 3 x 4^k+2] + 124^k+2 + 5 [using (1)]
= 90m - 50 - 30 x 4^k+2 + 12 4^k+2 + 5
= 90m - 45 - 18 (4^k+2)
= 9 [10m - 5 - 2 (4^k+2)] which is divisible by 9
∴ p(k+ 1) is true.
Hence, by the principle of mathematical induction, p(n) is true for all values of n.

Let P(n) be the statement \(\left(1-{1\over 2^2} \right)\left(1-{1\over 3^2} \right)\left(1-{1\over 4^2} \right)...\left(1-{1\over n^2} \right)={n+1\over 2n}\)
Step 1:
Putting n = 2, we get
\(\left(1-{1\over2^2}\right)\left(1-{1\over3^2}\right)={2+2\over 2(2)+2}\)
\(⇒\ \left(1-{1\over4}\right)\left(1-{1\over9}\right)={4\over 6}\)
\(⇒\ \left(3\over4\right)\left(8\over9\right)={2\over3}⇒{2\over3}={2\over3}\)
∵ p(1) is true
Step 2:
Let we assume that p(k) is true
\(∵\ \left(1-{1\over 2^2}\right)\left(1-{1\over 2^2}\right)...\left(1-{1\over (K+1^2)}\right)={K+2\over 2K+2}\)
Step 3:
To prove that p(K+1) is true

i.e to P.T \(\left(1-{1\over2^2}\right)\left(1-{1\over 3^2}\right)...\left(1-{1\over (K+1)^2}\right)\left(1-{1\over (K+2)^2}\right)={K+3\over 2(K+1)+2}={k+3\over 2K+4}\)
\(LHS=\left(1-{1\over 2^2}\right)\left(1-{1\over 3^2}\right)...\left(1-{}1\over (K+1)^2\right)\left(1-{1\over (K+2)^2}\right)\)
\(={K+2\over 2K+2}\left(1-{1\over (K+2)^2}\right)\)
\({K+2\over 2K+2}\left({(K+2)^2-1\over (KK+2)^2}\right)\) [Using (1)]
\(={K+2\over 2K+2}\left(K^2+4+4K-1\over (K+2)^2\right)\)
\(={K+2\over 2K+2}\left(K^2+4K+3\over (K+2)^2\right)={K+2\over 2(K+1)}{{(K+1)(K+3)\over (K+2)^2}}\)
\(={K+3\over 2(K+2)}={K+3\over 2K+4}=RHS\)
∵ p(K+1) is true
Hence by mathematical induction, p(n) is true for all values of n.

Let p(n) be the statement
\(1^3+2^3+3^3+...+n^3=\left[{n(n+1)\over2}\right]\)
Step1: Put n = 1
\(⇒\ 1^3=\left[1(1+1)\over 2\right]^2=\left[1(2)\over2\right]^2\)
⇒ 1 = 1
∴ p(n) is true
Step 2: Let us assume that p(K) is true
∵ \(1^3 + 2^3 + 3^3 + ... +K^3​​​​​​​ =​​​​​​​\left[K(K+1)\over 2\right]^2\)
Step 3: To prove that p(K+1) is true
i.e to P.T \(1^3+2^3+...+K^3+(K+1)^3=\left[(K+1)(K+2)\over 2\right]^2\)
LHS = 1³ + 2³ + .... + K³ + (K + 1)³
\(=\left[K{(K+1)\over 2}\right]^2+(K+1)^3\)
\(=(K+1)^2\left[{K^2\over 4}+K+1\right]=(K+1)^2\left[K^2+4K+4\over 4\right]\)
\(={(K+1)^2(K+2)^2\over4}=\left[(k+1)(k+2)\over 2\right]=RHS\)
∵ p(K+1) is true
∵ By the principle of mathematical induction, p(n) is true for all values of n.

The repetition of digits is allowed

Hundreds	tens	unit
5	6	3

The unit place can be filled in 3 ways using the digits 1, 3, or 5 since we need 3 digit odd numeric Hundreds place can be filled in 5 ways excluding 0 and repetition of digits is allowed.
Tens place can be filled in 6 ways .
∴ By fundamental principle of multiplication, required number of 3 = digit odd numbers
= 5 \(\times\) 6 \(\times\) 3 = 30 \(\times\) 3 = 90.

\(P\left(A_{1}\right)=0.6, \quad P\left(A_{2}\right)=0.4\)
\(A_{1}=\text { Event of selecting a girl }\)
\(\mathrm{B}=\text { Event of student taller than } 1.8 \mathrm{~m}\)
\(A_{2}=\text { Event of selecting a boy }\)
\(P\left(B / A_{1}\right)=\frac{1}{100}, P\left(B / A_{2}\right)=\frac{4}{100}\)
To find \(=\frac{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(B / A_{1}\right)+P\left(A_{2}\right) \cdot P\left(B / A_{2}\right)} \)
\(=\frac{0.6 \times \frac{1}{100}}{0.6 \times \frac{1}{100}+0.4 \times \frac{4}{100}} \)
\(=\frac{6}{\frac{6}{1000}+\frac{16}{1000}} \)
\(=\frac{6}{22}=\frac{3}{11} \)

\(n(S)=n(A \times A) \quad=16\)
\(\text {To find } n(A)\)
\(\text {Let } a=4, b=1 \text { then } a^{2}-4 b \text { is }\)
A = 1,b = 1 then 16 - 4 > 0
b = 2. then 16 - 8 > 0
b=3 then 16-L2>0
b = 4 then 16 - 16 = 0
Let Q = 3,b = 1 thg,l, 9-4 > 0
b = 2 then 9 - 8 > 0
Let a = 2,b = l then 4 - 4 > 0
n(A).= 7,
\(P(A)=\frac{n(A)}{n(S)}=\frac{7}{16}\)

Number of 10 digit number that can be written by using the digits 2 and 3 is 2¹⁰

Number of rectangles in the chessboard is
\({ }^{9} C_{2} \times{ }^{9} C_{2} =\frac{9 \times 8}{1 \times 2} \times \frac{9 \times 8}{1 \times 2} \)
\(=36 \times 36=1296 \)

Number of ways of selecting 8 people from 12 in ¹²C₈ ways.
Let A and B both attend the party
Out of 10 remaining people 8 can attend  in ¹⁰C₆ ways.
.'. Number of ways in which two of them do not attend together = ¹²C₈ - ¹⁰C₆

\(\text { No. of lines }{ }^{10} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{2}+1=45-6+1=40\)

\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{4} \)
\(a^{2}-a^{a} C_{2} =a^{2}-a^{a} C_{a-a-4}\left(\because^{n} C_{r}={ }^{n} C_{n-r}\right) \)
\(a^{2}-a-4 =2 \)
\(a^{2}-a-6 =0 \)
\((a-3)(a+2) =0 \)
\(a=3 \text { or } a=-2 \text { which is impossible }\)