n(S) = 52;\(P(J)=\cfrac { 4 }{ 52 } \)
(i) \(P(JJ)=\cfrac { 4 }{ 52 } \times \cfrac { 4 }{ 52 } =\cfrac { 1 }{ 13 } \times \cfrac { 1 }{ 13 } =\cfrac { 1 }{ 169 } \) 
(ii) \(P(JJ)=\cfrac { 4 }{ 52 } \times \cfrac { 3 }{ 51 } =\cfrac { 1 }{ 13 } \times \cfrac { 1 }{ 17 } =\cfrac { 1 }{ 221 } \)

S = {1, 2, 3, ..... , 50} ∴ n(S) = 50
Let A be the event of getting prime number.
∴ A = {2, 3, 5, 7,11,13,17,19,23,29,31,37,41,43, 47}
n(A) = 15, so P(A) = 15/50
Let B be the event of getting number multiple of 4
∴ B = {4, 8, 12, 16,20,24,28,32,36,40,44, 48}
n(B) = 12, so p(B) = 12/50
Here A and B are mutually exclusive. (i.e.,) A ∩ B = Φ
∴ P(A U B) = P(A) + P(B) = 15/50 + 12/50 = 27/50.

Let E₁ and E₂ be the events of choosing a boy and a girl respectively from the class.
Given that the number of boys to the number of girls = 1: 2
\(\therefore P({ E }_{ 1 })=\frac { 1 }{ 1+2 } =\frac { 1 }{ 3 } \) and P(E₂) = \(\frac { 2 }{ 1+12 } =\frac { 2 }{ 3 } \)
Let A be the event that a student chosen will get first class
Given P(A/E₁) = 0.28 and P(A/E₂) = 0.25
\(\therefore\)  By theorem of total probability, 
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂)
\(\Rightarrow \quad =\frac { 1 }{ 3 } \times 0.28+\frac { 2 }{ 3 } \times 0.25\)
\(=\frac { 28 }{ 300 } +\frac { 50 }{ 300 } =\frac { 78 }{ 300 } =0.26\) 

Probability of getting odd number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
Probability of getting even numbers = 1-\(\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
Now, probability of getting no odd number when the die is tossed twice = Probability of getting even number when the die is tossed thrice.
 \(\Rightarrow \quad \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \times \frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
\(\therefore\)  P(Odd number atleast once) = \(1-\frac { 1 }{ 8 } =\frac { 7 }{ 8 } \)

Let A and B the events of drawing a black card in first draw and second draw respectively.
In the first draw, there are 26 black cards out of 52 cards.
\(\therefore P(A)=\frac { 26 }{ 52 } =\frac { 1 }{ 2 } \)
In the second draw, there are 25 black cards out of 51 cards.
\(\therefore P(B/A)=\frac { 25 }{ 51 } \)
\(\therefore P(A\cap B)=P(A).P(B/A)=\frac { 1 }{ 2 } \times \frac { 25 }{ 51 } =\frac { 25 }{ 102 } \)