1 
22 
3 
22 
2 
91 
2 
77 
2 
91 
Then, n(S)  = number of ways of drawing 3 balls out of 15  
= ^{15}C_{3}  


= 455. 
Let E = event of getting all the 3 red balls.
∴ n(E) = ^{5}C_{3} = ^{5}C_{2} =  (5 x 4)  = 10. 
(2 x 1) 
∴ P(E) =  n(E)  =  10  =  2  . 
n(S)  455  91 
1 
2 
2 
5 
8 
15 
9 
20 
9 
20 
Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
∴ P(E) =  n(E)  =  9  . 
n(S)  20 
1 
6 
5 
12 
1 
2 
7 
9 
5 
12 
Let E = Event that the sum is a prime number.
Then E  = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } 
∴ n(E) = 15.
∴ P(E) =  n(E)  =  15  =  5  . 
n(S)  36  12 
1 
13 
3 
13 
1 
4 
9 
52 
3 
13 
∴ P (getting a face card) =  12  =  3  . 
52  13 
3 
20 
29 
34 
47 
100 
13 
102 
13 
102 
Then, n(S) = ^{52}C_{2} =  (52 x 51)  = 1326. 
(2 x 1) 
Let E = event of getting 1 spade and 1 heart.
∴ n(E)  = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 
= (^{13}C_{1} x ^{13}C_{1})  
= (13 x 13)  
= 169. 
∴ P(E) =  n(E)  =  169  =  13  . 
n(S)  1326  102 
1 
13 
2 
13 
1 
26 
1 
52 
1 
26 
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
∴ P(E) =  n(E)  =  2  =  1  . 
n(S)  52  26 
1 
6 
1 
8 
1 
9 
1 
12 
1 
9 
Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.
∴ P(E) =  n(E)  =  4  =  1  . 
n(S)  36  9 
1 
2 
3 
4 
3 
8 
5 
16 
3 
4 
Then, E  = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} 
∴ n(E) = 27.
∴ P(E) =  n(E)  =  27  =  3  . 
n(S)  36  4 
21 
46 
25 
117 
1 
50 
3 
25 
21 
46 
Then, n(S)  = Number ways of selecting 3 students out of 25  
= ^{25}C_{3} `  


= 2300. 
n(E)  = (^{10}C_{1} x ^{15}C_{2})  


= 1050. 
∴ P(E) =  n(E)  =  1050  =  21  . 
n(S)  2300  46 
1 
15 
25 
57 
35 
256 
1 
221 
1 
221 
Then, n(S) = ^{52}C_{2} =  (52 x 51)  = 1326. 
(2 x 1) 
Let E = event of getting 2 kings out of 4.
∴ n(E) = ^{4}C_{2} =  (4 x 3)  = 6. 
(2 x 1) 
∴ P(E) =  n(E)  =  6  =  1  . 
n(S)  1326  221 
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