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) = \(rac{n(A)}{n(S)}\) = \(rac{^{13}C_1 imes^{13}C_1 imes^{13}C_1 imes^{13}C_1}{^{52}C_4}\) = \(rac{13 imes13 imes13 imes13}{rac{52 imes51 imes50 imes49}{4 imes3 imes2 imes1}}\) ∵ \(\bigg[\,^nC_r = rac{|\underline{n}}{|\underline{n-r}|\underline{r}}\bigg]\)= \(rac{2197}{20825}\)(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) = \(rac{n(A)}{n(S)}\) = \(rac{^{12}C_1 imes1}{^{52}C_4}\) = \(rac{rac{12 imes11 imes10}{3 imes2 imes1}}{rac{52 imes51 imes50 imes49}{4 imes3 imes2 imes1}}\) = \(rac{12 imes11 imes10 imes4}{52 imes51 imes50 imes49}\) = \(rac{44}{54145}\)(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 aceP(\(\bar{A}\)) = \(rac{^{48}C_4}{^{52}C_4}\) = \(rac{48 imes47 imes46 imes45}{52 imes51 imes50 imes49}\) = \(rac{38916}{54145}\)[Here \(\bar{A}\) denotes the complement of event A, i.e, non-happening of event A]∴ P(A) = 1 - P(\(\bar{A}\)) = 1 - \(rac{38916}{54145}\) = \(rac{15229}{54145}\) (∵ P(Event) + P(complement of event) = 1)