4 cards can be drawn from a pack of cards in ⁵²C₄ ways 

∴ Exhaustive number of cases = n(S) = ⁵²C₄ 

(a) There are 4 suits, each containing 13 cards. 

Let A : Event of drawing one card from each suit 

⇒ Favourable number of cases = n(A) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^{13}C_1\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_4}\) = \(\frac{13\times13\times13\times13}{\frac{52\times51\times50\times49}{4\times3\times2\times1}}\)               ∵ \(\bigg[\,^nC_r = \frac{|\underline{n}}{|\underline{n-r}|\underline{r}}\bigg]\)

= \(\frac{2197}{20825}\)

(b) Let A : Drawing 4 spade cards of which one is king of spades. 

Then, Favourable number of cases = n(A) = ¹²C₃ x 1

(∵ There is only one king ofspades and the rest of the three spades we draw from remaining 12 spade cards) n(S) = ⁵²C₄

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{^{12}C_1\times1}{^{52}C_4}\) = \(\frac{\frac{12\times11\times10}{3\times2\times1}}{\frac{52\times51\times50\times49}{4\times3\times2\times1}}\) = \(\frac{12\times11\times10\times4}{52\times51\times50\times49}\) = \(\frac{44}{54145}\)

(c) Let A : Drawing at least one ace. 

Now since there are 4 aces in the pack of 52 cards, therefore, the number of ways of drawing 4 cards so that no card is an ace = ⁴⁸C₄ 

∴ Probability of drawing four cards so that none is an ace

P(\(\bar{A}\)) = \(\frac{^{48}C_4}{^{52}C_4}\) = \(\frac{48\times47\times46\times45}{52\times51\times50\times49}\) = \(\frac{38916}{54145}\)

[Here \(\bar{A}\) denotes the complement of event A, i.e, non-happening of event A]

∴ P(A) = 1 - P(\(\bar{A}\)) = 1 - \(\frac{38916}{54145}\) = \(\frac{15229}{54145}\)             

(∵ P(Event) + P(complement of event) = 1)