Let B denotes a boy and G denotes a girl. Then, all possible genders are expressed as
S={BBB,BBG,BGB,GBB,BGG,GBG,GGB,GGG}

(a) Let E₁ denotes the event that exactly one child is a girl.
Then, E₁={BBG,BGB,GBB}
\(\Rightarrow \)  \(P({ E }_{ 1 })=\frac { 3 }{ 8 } \)
(b) Let E₂ denotes the event that atleast two children are girls.
Then, E_{2 =} {BGG,GBG,GGB,GGG}
\(\Rightarrow \) \(P({ E }_{ 2 })=\frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)
(c) Let E₃ denotes the event that no child is a girl.
then, E₃ = {BBB}
\(\Rightarrow \)  \(P({ E }_{ 3 })=\frac { 1 }{ 8 } \)

We have, \(P(A\cup B)\) = 0.6 and \(P(A\cap B)\) = 0.2
We know, \(P(A\cup B)\) = P(A) + P(B) -\(P(A\cap B)\)
\(\therefore \)  0.6 = P(A) + P(B) - 0.2
P(A) + P(B) = 0.6 + 0.2 = 0.8 ...(i)
Now,
\(P(\overline { A } )+P(\overline { B } )\) 
=  P(A) +1-P(B)
= 2 - (P(A) + P(B))
= 2 - 0.8
= 1.2

Clearly, the sample space of the experiment is 
S= {B₁B₂, B₁W, B₂B_1, B₂W, WB₁, WB₂, BW₁, BW₂, W₁B, W₁W₂, W₂B, W₂W₁}
Let E be the event that balls of opposite colour are chosen, then
E={B₁W, WB_1, B₂W, WB_2, BW₁,​​​​​​​  BW₂, W₁B,​​​​​​​ W₂B​​​​​​​}
\(\therefore \)   \(P({ E })=\frac { 8 }{ 12 } =\frac { 2 }{ 3 } \quad \quad \)

\(We\ have,\ P(A\cup B)=\frac { 1 }{ 2 } \ and\ P(\overline { A } )=\frac { 2 }{ 3 } \)
\(\Rightarrow 1-P(A)=\frac { 2 }{ 3 } \)
\(\Rightarrow P(A)=1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
\(Also,\ we\ have\ P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 3 } +P(B\cap \overline { A } )\)
\(\Rightarrow P(B\cap \overline { A } )=\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } \)

Probability that the missing cards to be different colours
\(=\frac { ^{ 26 }C_{ 1 }\times ^{ 26 }C_{ 1 }\quad }{ ^{ 52 }C_{ 2 } } \)
[in a pack of cards, there are two colours,each of them contain 26 cards]
\(=\frac { 26\times 26\quad }{ \frac { 52\times 51 }{ 2\times 1 } } =\frac { 26 }{ 51 } \)

We have, P(A) = 0.5, P(B) = 0.4, and \(P(A\cup B)\) = 0.8
Clearly, \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\)
\(=0.5+0.4-0.8=0.1\)
\(Here,\ P(A\cap B)\le P(A)\)
\(and\ P(A\cap B)\le P(B)\)
So, given probabilities are consistently defined.

There are 9 letters in the word 'ALGORITHM'
These 9 letters can be arranged in a row  in 9! ways
\(\therefore\) Total number of possible outcomes = 9!
Now, if the letters GOR remain together, then we have 7 letters (considering GOR as 1 letter i.e. ALITHMGOR)
which can be arranged in 7!
Number of favourable outcomes = 7!
Hence, required probability =  \(\frac { 7! }{ 9! } =\frac { 1 }{ 9\times 8 } =\frac { 1 }{ 72 } \)

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]
= \(\frac { 1 }{ 11 } +\frac { 2 }{ 11 } +\frac { 1 }{ 11 } =\frac { 4 }{ 11 } \)            

Since a 3digit number cannot start with 0, therefore hundredth place can be filled up in 4 ways.
Now as repetition of  digits is allowed, therefore each of  ten's and unit's place can be filled in 5 ways.
Thus , the total number of possible3 -digit number
= 4 x 5 x 5 = 100
Note that, there are three digits numbers, which has the same digits viz. 222,444,666 and 888
\(\Rightarrow \)  Number of favourable outcomes = 4
Hence ,P (3digit number with same digits) = \(\frac { 4 }{ 100 } =\frac { 1 }{ 25 } \)