Total number of bulbs = 10
Number of defective bulbs = 2 .
ஃ Number of good bulbs = 10 - 2 = 8
Now selecting 5 from the 10 bulbs can be done in ¹⁰C₅ ways.
(i.e.,) \(n(S)=^{ 10 }{ C }_{ 5 }=\cfrac { 10\times 9\times 8\times 7\times 6 }{ 5\times 4\times 3\times 2\times 1 } =252\) 
Let A be the event of selecting 5 good bulbs (from 8 good bulbs)
\(\therefore n(A)={ 8C }_{ 5 }=8{ C }_{ 3 }=\cfrac { 8\times 7\times 6 }{ 3\times 2\times 1 } =56\)
\(\therefore P(A)=\cfrac { n(A) }{ n(S) } =\cfrac { 56 }{ 252 } =\cfrac { 2 }{ 9 } \)

(i) If A is a subset of B, then
A\(\cap \)B = A
\(\Rightarrow\) n(A\(\cap \)B) = n(A)
\(\Rightarrow\) P(A\(\cap \)B) = P(A)
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { P(A) }{ P(A) } =1\)
(ii) If \(A\cap B=\phi \) then n(A\(\cap \)B) = 0 \(\Rightarrow\) P(A\(\cap \)B) = 0
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { 0 }{ P(A) } =0\)

Consider the following events.
A: Person gets an electric contract
B: Person gets plumbing contract
Given P(A) = \(\frac { 2 }{ 5 } ,P(\bar { B } )=\frac { 4 }{ 7 } \)and P(AUB) = \(\frac { 2 }{ 3 } \)
We know, P(AUB) = P(A) + P(B) - P(\(A\cap B\))
\(\Rightarrow \frac { 2 }{ 3 } =\frac { 2 }{ 5 } +\left( 1-\frac { 4 }{ 7 } \right) -P(A\cap B)\)
\(\Rightarrow \frac { 2 }{ 3 } -\frac { 2 }{ 5 } =\frac { 3 }{ 7 } -P(A\cap B)\)
\(\Rightarrow P(A\cap B)=\frac { 2 }{ 5 } +\frac { 3 }{ 7 } -\frac { 2 }{ 3 } =\frac { 17 }{ 105 } \)

Out of 30 items, 2 can be selected in ³⁰C₂ ways,
\(\therefore\) n(S) = ³⁰C₂
Consider the events:
A: getting 2 apples
B: getting 2 good items
2 apples can be drawn from 20 apples in ²⁰C₂ ways.
\(\therefore P(A)=\frac { ^{ 20 }{ C }_{ 2 } }{ ^{ 30 }{ C }_{ 2 } } \)
Out of 22 good pieces, 2 can be selected in ²²C₂ ways
\(\Rightarrow \therefore P(B)=\frac { ^{ 20 }{ C }_{ 2 } }{ ^{ 30 }{ C_{ 2 } } } \)
Since there are 15 pieces which are good apples, out of which 2 can be selected in ¹⁵C₂ ways.
\(\therefore P(A\cap B)=\frac { ^{ 15 }{ { C }_{ 2 } } }{ ^{ 30 }{ { C }_{ 2 } } } \)
\(\therefore\) P(AUB) = P(A) + P(B) - P(A\(\cap \)B)
\(=\frac { ^{ 20 }{ C_{ 2 } } }{ ^{ 30 }{ { C }_{ 2 } } } +\frac { ^{ 22 }{ C_{ 2 } } }{ ^{ 30 }{ { C }_{ 2 } } } -\frac { ^{ 15 }{ { C }_{ 2 } } }{ ^{ 30 }{ { C }_{ 2 } } } \)
\(=\frac { \frac { 20\times 19 }{ 2\times 1 } }{ 435 } +\frac { \frac { 22\times 21 }{ 2\times 1 } }{ \frac { 2\times 1 }{ 435 } } -\frac { \frac { 15\times 14 }{ 2\times 1 } }{ 435 } \)
\(=\frac { 190 }{ 435 } +\frac { 231 }{ 435 } -\frac { 105 }{ 435 } =\frac { 190+231-105 }{ 435 } =\frac { 316 }{ 435 } \)

Here n(s) = 36
Let A be the event of getting 3 on first die and B be the event of getting the sum of 8 or greater.
\(\therefore\) A = {(3, 1) (3, 2) (3, 3)(3, 4) (3, 5) (3, 6)}
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2) (3, 6) (4, 5) (5, 4) (6, 3) (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)}
and \(A\cap B\) = {(3, 5), (3, 6)}
\(\therefore P(A)=\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)   \(P(B)=\frac { 15 }{ 36 } =\frac { 5 }{ 12 } \) and \(P(A\cap B)=\frac { 2 }{ 36 } =\frac { 1 }{ 18 } \)
\(\therefore P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 18 } }{ \frac { 1 }{ 6 } } =\frac { 1 }{ 18 } \times \frac { 6 }{ 1 } =\frac { 1 }{ 3 } \)