Given: pack of 52 cards 

Formula: P(E) = \(\frac{favourable\,outcomes}{total\,possible\,outcomes}\)

since a card is drawn from a pack of 52 cards, therefore number of elementary events in the sample space is 

n(S)= ⁵²C₁ = 52 

(i) let E be the event of drawing a black king 

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

P(E) = \(\frac{n(E)}{n(S)} \)

 P(E) = \(\frac{2}{52}\) = \(\frac{1}{26}\) 

(ii) let E be the event of drawing a black card or a king

n(E) = ²⁶C₁+⁴C₁- ²C₁= 28 

we are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice

P(E) = \(\frac{n(E)}{n(S)} \)

 P(E) = \(\frac{28}{52}\) = \(\frac{7}{13}\) 

(iii) let E be the event of drawing a black card and a king 

n(E)=²C₁ =2 (there are two black kings one of spade and other of club)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{5}{52}\) = \(\frac{1}{26}\)  

(iv) let E be the event of drawing a jack, queen or king 

n(E)=⁴C₁+ ⁴C₁+ ⁴C₁ = 12

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{12}{52}\) = \(\frac{3}{13}\)  

(v) let E be the event of drawing neither a heart nor a king now consider E’ as the event that either a heart or king appears 

n(E’) = ⁶C₁+ ⁴C₁ - 1 = 16 (there is a heart king so it is deducted)

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{16}{52}\) = \(\frac{4}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{4}{13}\) = \(\frac{9}{13}\) 

(vi) let E be the event of drawing a spade or king 

n(E)= ¹³C₁ + ⁴C₁ - 1 = 16

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{16}{52}\) = \(\frac{4}{13}\)  

(vii) let E be the event of drawing neither an ace nor a king now consider E’ as the event that either an ace or king appears

n(E’) = ⁴C₁+ ⁴C₁ = 8

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{8}{52}\) = \(\frac{2}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\) 

(viii) let E be the event of drawing a diamond card 

n(E)= ¹³C₁ = 13

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{13}{52}\) = \(\frac{1}{13}\)   

(ix) let E be the event of drawing not a diamond card now consider E’ as the event that diamond card appears 

n(E’) = ¹³C₁ =13

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{13}{52}\) = \(\frac{1}{4}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) 

(x) let E be the event of drawing a black card 

n(E)= ²⁶C₁= 26 (spades and clubs)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\)   

(xi) let E be the event of drawing not an ace now consider E’ as the event that ace card appears

n(E’) = ⁴C₁= 4

P(E') = \(\frac{n(E')}{n(S)} \)

 ​​​​​​​P(E') = \(\frac{4}{52}\) = \(\frac{1}{13}\)   

P(E) = 1- P(E’)

P(E) = 1 - \(\frac{1}{13}\) = \(\frac{12}{13}\)  

(xii) let E be the event of not drawing a black card 

n(E) = ²⁶C₁= 26 (red cards of hearts and diamonds)

P(E) = \(\frac{n(E)}{n(S)} \)

 ​​​​​​​P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\)