Let A_l and A₂ be the events that the items produced by Machine-I and II respectively.
Let B be the event of selecting a defective item
\(P\left( { A }_{ 1 } \right) =\cfrac { 45 }{ 100 } =\cfrac { 9 }{ 20 } \) ;
\(P\left( A_{ 2 } \right) =\cfrac { 55 }{ 100 } =\cfrac { 11 }{ 20 } \) 
\(P(B/{ A }_{ 1 })=\cfrac { 10 }{ 100 } =\cfrac { 1 }{ 10 } \) 
\(P(B/{ A }_{ 2 })=\cfrac { 5 }{ 100 } =\cfrac { 1 }{ 20 } \) 
(i) P(B)= P(B/A_I) P(A_I) + P(B/A₂) P(A₂)
= \(\cfrac { 1 }{ 10 } \times \cfrac { 9 }{ 20 } +\cfrac { 1 }{ 20 } \times \cfrac { 11 }{ 20 } =\cfrac { 9 }{ 200 } +\cfrac { 11 }{ 400 } =\cfrac { 18+11 }{ 400 } =\cfrac { 29 }{ 400 } \) 
= \(P\left( { A }_{ 2 }/B \right) =\cfrac { P\left( B/{ A }_{ 2 } \right) P\left( { A }_{ 2 } \right) }{ P\left( B/{ A }_{ 1 } \right) P\left( { A }_{ 1 } \right) +P\left( B/A_{ 2 } \right) P\left( { A }_{ 2 } \right) } =\cfrac { P(B/A_{ 2 })P({ A }_{ 2 }) }{ P(B) } \) 
= \(\cfrac { \frac { 1 }{ 20 } \times \cfrac { 11 }{ 20 } }{ \frac { 29 }{ 400 } } \) 
= \(\cfrac { 11/400 }{ 29/400 } =\cfrac { 11 }{ 29 } \)

Let A, Band C be the three possible events of selections. The number of combinations are shown below:

n(S) = Selecting 4 from 10 students = \(^{ 10 }{ C }_{ 4 }=\cfrac { 10\times 9\times 8\times 7 }{ 4\times 3\times 2\times 1 } =210\) 
\(n\left( A\cup B\cup \right) C=n(A)+n(B)+n(C)\)
= ⁴C₂C⁴C₂+⁴C1X⁶C1+⁴C₀X⁶C₄
\(^{ 4 }{ C }_{ 2 }=\cfrac { 4\times 3 }{ 2\times 1 } =6;^{ 4 }{ C }_{ 1 }=4;^{ 4 }{ C }_{ 0 }=1\) 
\(^{ 6 }{ C }_{ 2 }=\cfrac { 6\times 5 }{ 2\times 1 } =15;^{ 6 }{ C }_{ 3 }=\cfrac { 6\times 5\times 4 }{ 3\times 2\times 1 } =20.^{ 6 }{ C }_{ 4 }=^{ 6 }{ C }_{ 2 }=15\) 
\(n\left( A\cup B\cup C \right) =(6)(15)+(4)(20)+(1)(15)=90+80+15=185\) 
\(\therefore P(A\cup B\cup C)=\cfrac { 185 }{ 210 } =\cfrac { 37 }{ 42 } \)

A = {(1, 1) (2, 2) (3, 3) 94, 4) (5, 5) (6, 6)}
B = {(4, 6) (6, 4) (5, 5) (6, 5) (5, 6) (6, 6)}
A\(\cap \)B = {(6, 6) (5, 5)}
\(\therefore\) P(A) = P(1, 1) +P(2, 2)+P(3, 3)+P(4, 4) +P(5, 5) +P96, 6)
= P(1). P(1)+P(2).P(2) + P(3).P(3) +P(4).P(4) +P(5).P(5)+P(6).P(6)
\(=\frac { 2 }{ 10 } \times \frac { 2 }{ 10 } +\frac { 2 }{ 10 } \times \frac { 2 }{ 10 } \times \frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 3 }{ 10 } \times \frac { 3 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } \)
\(=\frac { 4 }{ 100 } +\frac { 4 }{ 100 } +\frac { 1 }{ 100 } +\frac { 9 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
P(B) = P(4, 6) + p(6, 4) + P(5, 5) + P(6, 5) + P(5, 6) + P(6, 6)
\(=\frac { 3 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 3 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } \)
\(=\frac { 3 }{ 100 } +\frac { 3 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 10 }{ 100 } =\frac { 1 }{ 10 } \)
\(P(A\cap B)\) = P95, 5) + P(6, 6) = P(5).P(5)+P(6).P(6)
\(=\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } =\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 2 }{ 100 } =\frac { 1 }{ 50 } \)
\(\therefore\) P(A\(\cap \)B) = P(A) \(\times\) P(B)
\(\frac { 1 }{ 50 } =\frac { 1 }{ 5 } \times \frac { 1 }{ 10 } =\frac { 1 }{ 50 } \)
Hence A and B are independent events. 

Consider the following events.
E_1: I purse is chosen
E2: II purse is chosen
A: Coin pulled out is silver
\(\Rightarrow \quad \therefore P(E_{ 1 })=P({ E }_{ 2 })=\frac { 1 }{ 2 }\)
There are 3 silver and 4 copper coins in I purse
\(\Rightarrow \quad P(A/{ E }_{ 1 })=\frac { 3 }{ 7 } \)
There are 4 silver and 3 copper coins in II Purse
\(\therefore\) P(A/E₁) =  \(\frac { 4 }{ 7 } \)
By the theorem of total probability
P(A) = P(E₁).P(A/E₁)+P(E₂).P(A/E₂) = \(\frac { 1 }{ 2 } \times \frac { 3 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 4 }{ 7 } =\frac { 3 }{ 14 } +\frac { 4 }{ 17 } =\frac { 7 }{ 14 } =\frac { 1 }{ 2 } \)

Out of integers from 1 to 11, there are 5 even integers and 6 odd integers.
Let A; Both the numbers chosen are odd
B: Sum of he numbers chosen is even 
[Number of ways of selecting odd nos = ⁶C₂ No. of ways of getting sum as an even number = ⁵C₂ + ⁶C₂]
\(\therefore P(A/B)=\frac { p(A\cap B) }{ P(B) } =\frac { ^{ 6 }{ { C }_{ 2 } } }{ ^{ 5 }{ { C }_{ 2 }+ }^{ 6 }{ { C }_{ 2 } } } =\frac { \frac { 6\times 5 }{ 2\times 1 } }{ \frac { 5\times 4 }{ 2\times 1 } +\frac { 6\times 5 }{ 2\times 1 } } =\frac { 15 }{ 10+15 } =\frac { 15 }{ 25 } =\frac { 3 }{ 5 } \)
\(\therefore P(A/B)=\frac { 3 }{ 5 } \)