There are three ways of drawing four balls.
(i) A : 2 red, 1 green and 1 black ball
(ii) B : 1 red, 2 green and 1 black ball
(iii) C : 1 red, 1 green and 2 black balls.
Since A, B, C are mutually exclusive events
\(\therefore \ P(A\cup B\cup C)=P(A)+P(A)+P(B)+P(C)\)
\(=\frac { ^{ 9 }C_{ 2 }\times ^{ 6 }C_{ 1 }\times ^{ 5 }C_{ 1 } }{ ^{ 20 }C_{ 4 } } +\frac { ^{ 9 }C_{ 1 }\times ^{ 6 }C_{ 2 }\times ^{ 5 }C_{ 1 } }{ ^{ 20 }C_{ 4 } } +\frac { ^{ 9 }C_{ 1 }\times ^{ 6 }C_{ 1 }\times ^{ 5 }C_{ 2 } }{ ^{ 20 }C_{ 4 } } \)
\(=\frac { 9\times 4\times 6\times 5\times 9\times 3\times 5\times 5\times 9\times 6\times 5\times 2 }{ 5\times 19\times 6\times 17 } \)
\(=\frac { 1080+675+540 }{ 9690 } =\frac { 225 }{ 9690 } =\frac { 153 }{ 646 } \)

There are 7 letters in the word
\(\therefore \) Total events = 7!
Now there are 3 vowels in the word 'SOCIETY' when we put three vowels altogether and consider as one letter. Then favourable events = 5! x 3!
\(\text{thus required probability }=\frac { 5!\times 3! }{ 7! } =\frac { 1 }{ 7 } \)

(i) out of 10 red balls, one ball is drawn in \(^{ 10 }C_{ 1 }\)=10 ways
out of 6 green balls, one ball is drawn is \(^{ 6 }C_{ 1 }\) = 6 ways
\(\text{ Favourable events}=10\times 6=60\)
\(\text{Total events}=^{ 20 }C_{ 2 }=190\)
\(\text{thus required probability }=\frac { 60 }{ 190 } =\frac { 6 }{ 19 } \)
(ii) out of 10 red balls, one ball are drawn in \(^{ 10 }C_{ 2 }\)=45 ways
out of 6 green balls, two balls is drawn in \(^{ 6 }C_{ 2 }\) = 15 ways
out of 4 black balls, two balls is drawn in \(^{ 4 }C_{ 2 }\)=6 ways
\(\therefore \text{ Favourable events}=45+15+6=66\ ways\)
\(\text{Total events}=^{ 20 }C_{ 2 }=190\)
\(\text{thus required probability }=\frac { 66 }{ 190 } =\frac { 33 }{ 195 } \)

i) there are 2 cards which are black and king
\(\therefore \text{Favourable events }=^{ 52 }C_{ 1 }=52\)
\(\text{thus required probability }=\frac { 2 }{ 52 } =\frac { 1 }{ 26 } \)
(ii) There are 26 cards which are red including 2 kings and 2 black king. So there are 28 cards which are either red or king
\(\therefore \text{Favourable events }=^{ 52 }C_{ 1 }=52\)
\(\text{thus required probability }=\frac { 28 }{ 52 } =\frac { 7 }{ 13 } \)

A = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}
B = {(I, 3), (1, 4), (2, 3), (2, 4), (3, 4), (4, 3)}
C = {(2, 3), (3, 4), (4, 2), (4, 3)}
D = {(I, 2), (2, 4)}
A and Dare mutually exclusive events

(i) {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2),
(6, 3), (6, 4), (6, 5), (6, 6)}
(ii) {(I, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 4), (4, 5), (4, 6), (5, 1), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3),
(6, 4), (6, 5), (6, 6)}
(iii) {(I, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 5), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 3),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1)}
(iv) \(\phi \)
(v) (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (5, 1), (5, 2)}
(vi) {(I, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, I)}

C = {(2,6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, ID, (6, 6)}

B = {(2, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

A = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2),(5, 3), (5, 4), (5, 5), (5, 6)}

Total number of possible outcomes = \(^{ 52 }C_{ 4 }\)
We know that, there are 4 suits each having 13 cards.
\(\therefore\) Number of favourable outcomes
=\(^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }\times ^{ 13 }C_{ 1 }=\left( 13 \right) ^{ 4 }\)
Hence, P(getting one card from each suit) = \(\frac { \left( 13 \right) ^{ 4 } }{ ^{ 52 }C_{ 4 } } \)

Total number of possible outcomes = \(^{ 52 }C_{ 4 }\)
We know that, there are 26 red cards an d 26 black cards 
\(\therefore\) Number of favourable outcomes = \(^{ 26 }C_{ 2 }\times ^{ 26 }C_{ 2 }\)
Hence, p (getting 2 red and 2 black cards) \(=\frac { \left( ^{ 26 }C_{ 2 } \right) ^{ 2 } }{ ^{ 52 }C_{ 4 } } \)

When a die is thrown, then sample space
s={1,2,3,4,5,6}
i)A: a number less than 7={1,2,3,4,5,6}
ii)B: a number greater than 7={}=\(\phi \)
iii)C: a multiple of 3={3,6}
iv)D: a number less than 4={1,2,3}
v)E: an even number greater than 4={6}
vi)F: a number not less than3={3,4,5,6}
Now, \(A\cup B\) = The elements which are in A or B or both
={1,2,3,4,5,6}\(\cup \phi \)
={1,2,3,4,5,6}
\(A\cap B\) = The element which are common in both
A and B
={1,2,3,4,5,6}\(\cap \phi =\phi \)
\(B\cup C\) =The element which are common in both B or C
={}\(\cup \){3,6}={3,6}
\(E\cap F\) =The element which are common in both E and F
={6}\(\cup \){3,4,5,6}={6}
\(D\cap E\) =The element which are common in both D and E
={1,2,3}\(\cup \){6}=\(\phi \)
A - C   =The element which is in A but not in C
={1,2,3,4,5,6}-{3,6}
={1,2,4,5}
D - E   =The element which is in D but in E
={1,2,3} -{6}={1,2,3}
F^'=The element which is in not in F
=(S - F)
={1,2,3,4,5,6}-{3,4,5,6}={1,2}
and \(E\cap { F }^{ ' }\)=\(E\cap (S-F)\)
=\(E\cap \)({1,2,3,4,5,6}-{3,4,5,6})
={6}\(\cap \){1,2}=\(\phi \)

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

\(P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)\)
\(=1-0.88=0.12\)

Let R denotes a red ball and B denotes a black ball. Then, possible outcomes of the experiment are RR, RB, BR and BB